Straight Lines for JEE: Forms, Distances, Bisectors and Centres

A straight line is the simplest curve in the plane, and coordinate geometry begins with it. Once a point is a pair of numbers and a line is an equation of the first degree, every question about lengths, angles, areas and positions turns into algebra that can be checked line by line. The same results return in every later chapter: the circle, the parabola, the ellipse and the hyperbola all use the distance of a point from a line, the slope of a line and the family of lines through a point.

This note covers the distance and section formulas with harmonic conjugates, the area of a triangle and of a polygon, the locus of a point, the slope of a line, every standard form of the equation of a line, the position of a point relative to a line, the angle between two lines, the distance of a point from a line and between parallel lines, the foot of the perpendicular and the image of a point, the bisectors of the angles between two lines, the centres of a triangle, families of lines and concurrency, lines that form a figure, and the change of origin and rotation of axes. Each section ends with worked examples from JEE Main, JEE Advanced and IIT JEE papers, graded from easy to hard.

Distance and the section formula

Take two perpendicular number lines meeting at the origin $O$. A point $P$ is then the ordered pair $(x,y)$ of its signed distances from the $y$-axis and the $x$-axis.

Distance between two points

For $A(x_1,y_1)$ and $B(x_2,y_2)$, the horizontal and vertical gaps are $x_2-x_1$ and $y_2-y_1$, and Pythagoras gives

$$AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

The distance of $P(x,y)$ from the origin is $\sqrt{x^2+y^2}$.

If the points are given in polar form, $A(r_1,\theta_1)$ and $B(r_2,\theta_2)$, the cosine rule in triangle $OAB$ gives

$$AB^2=r_1^2+r_2^2-2r_1r_2\cos(\theta_1-\theta_2)$$

Section formula

Let $P$ divide the join of $A(x_1,y_1)$ and $B(x_2,y_2)$ in the ratio $m:n$, so that $AP:PB=m:n$.

For internal division ($P$ between $A$ and $B$),

$$P=\left(\frac{mx_2+nx_1}{m+n},\ \frac{my_2+ny_1}{m+n}\right)$$

For external division ($P$ on $AB$ produced), replace $n$ by $-n$:

$$P=\left(\frac{mx_2-nx_1}{m-n},\ \frac{my_2-ny_1}{m-n}\right)$$ Internal division with P between A and B, and external division with P on AB produced beyond B

Reason. Drop perpendiculars from $A$, $P$, $B$ to the $x$-axis. The horizontal gaps are in the same ratio as $AP:PB$ (similar triangles), so $x-x_1:x_2-x=m:n$, which gives $x=\frac{mx_2+nx_1}{m+n}$.

Ratio in which a line divides a segment

The line $L\equiv ax+by+c=0$ meets the join of $A(x_1,y_1)$ and $B(x_2,y_2)$ at the point dividing $AB$ in the ratio

$$\frac{m}{n}=-\frac{ax_1+by_1+c}{ax_2+by_2+c}$$

A positive ratio means internal division ($A$ and $B$ on opposite sides of the line); a negative ratio means external division.

Harmonic conjugates

If $R$ divides $PQ$ internally and $S$ divides $PQ$ externally in the same ratio, then $R$ and $S$ are harmonic conjugates of each other with respect to $P$ and $Q$:

$$\frac{PR}{RQ}=\frac{PS}{SQ}\ \text{ (as lengths)}$$

Examples

Example 1 (JEE Main 2014). Let $PS$ be the median of the triangle with vertices $P(2,2)$, $Q(6,-1)$ and $R(7,3)$. The equation of the line passing through $(1,-1)$ and parallel to $PS$ is

▸ SHOW SOLUTION

Step 1. The mid-point of $QR$

$$S=\left(\frac{6+7}{2},\frac{-1+3}{2}\right)=\left(\frac{13}{2},1\right)$$

Step 2. Slope of $PS$

$$m=\frac{1-2}{\frac{13}{2}-2}=-\frac{2}{9}$$

Step 3. The parallel line through $(1,-1)$

$$y+1=-\frac29(x-1)$$$$2x+9y+7=0$$

Answer: option (4)

Example 2. Find the point where the line $2x+y-5=0$ meets the join of $A(-1,2)$ and $B(4,7)$, and the harmonic conjugate of that point with respect to $A$ and $B$.

▸ SHOW SOLUTION

Step 1. The ratio

$L(A)=-2+2-5=-5$ and $L(B)=8+7-5=10$, so

$$\frac{m}{n}=-\frac{-5}{10}=\frac12$$

The ratio is positive, so the line cuts $AB$ internally in the ratio $1:2$.

Step 2. The point of division

$$R=\left(\frac{1\cdot4+2(-1)}{3},\ \frac{1\cdot7+2\cdot2}{3}\right)$$$$R=\left(\frac23,\frac{11}{3}\right)$$

Check: $2\cdot\frac23+\frac{11}{3}-5=0$.

Step 3. The harmonic conjugate

The conjugate $S$ divides $AB$ externally in the same ratio $1:2$:

$$S=\left(\frac{4-2(-1)}{1-2},\ \frac{7-2\cdot2}{1-2}\right)=(-6,-3)$$

Answer: $R\left(\frac23,\frac{11}{3}\right)$ and $S(-6,-3)$

Example 3 (JEE Advanced 2025, Paper 1). Let $P$, $Q$ and $R$ be three distinct points in the $XY$-plane, and let $S$ be a point inside the triangle $PQR$ such that $\overrightarrow{SP}+5\overrightarrow{SQ}+6\overrightarrow{SR}=\vec 0$. Let $E$ and $F$ be the mid-points of the sides $PR$ and $QR$ respectively. Then the value of $\dfrac{EF}{ES}$ is ______.

▸ SHOW SOLUTION

Step 1. Coordinates of $S$

Write $P(x_1,y_1)$, $Q(x_2,y_2)$, $R(x_3,y_3)$ and $S(x,y)$. The $x$-components of the given relation give $(x_1-x)+5(x_2-x)+6(x_3-x)=0$, so

$$x=\frac{x_1+5x_2+6x_3}{12}$$

and the same for $y$. So $S$ is the point with weights $1,5,6$ at $P,Q,R$.

Step 2. The vector $EF$

$E=\frac{P+R}{2}$ and $F=\frac{Q+R}{2}$, so $F-E=\frac{Q-P}{2}$ and

$$EF=\frac12\,PQ$$

Step 3. The vector $ES$

$$S-E=\frac{P+5Q+6R}{12}-\frac{6P+6R}{12}$$$$S-E=\frac{5(Q-P)}{12}$$

So $ES=\frac{5}{12}\,PQ$.

Step 4. The ratio

$$\frac{EF}{ES}=\frac{1/2}{5/12}=\frac65$$

Answer: $1.2$

Area of a triangle and of a polygon

Area of a triangle

For vertices $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$,

$$\Delta=\frac12\left|\,S\,\right|$$

where $S=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$.

In determinant form,

$$\Delta=\frac12\left|\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\right|$$

Reason. Drop perpendiculars from the three vertices to the $x$-axis. The triangle is one trapezium minus the other two, and the area of a trapezium is half the sum of the parallel sides times the width.

When one vertex is the origin, $\Delta=\frac12\lvert x_1y_2-x_2y_1\rvert$ for the other two vertices.

Collinearity

Three points are collinear exactly when the area is zero:

$$\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=0$$

Equivalently, the slope of $AB$ equals the slope of $BC$, or $AB+BC=AC$ for the three distances.

Area of a polygon

For a polygon with vertices $(x_1,y_1),(x_2,y_2),\dots,(x_n,y_n)$ taken in order round the boundary,

$$\text{Area}=\frac12\left|\sum_{k=1}^{n}(x_ky_{k+1}-x_{k+1}y_k)\right|$$

with $(x_{n+1},y_{n+1})=(x_1,y_1)$. Each term is twice the signed area of the triangle formed by the origin and one side.

A quadrilateral with vertices taken in order round the boundary, beside the same four points joined in a crossing order

Examples

Example 4 (IIT JEE 2008, Paper 2). Consider three points $P=\big(-\sin(\beta-\alpha),\,-\cos\beta\big)$, $Q=\big(\cos(\beta-\alpha),\,\sin\beta\big)$ and $R=\big(\cos(\beta-\alpha+\theta),\,\sin(\beta-\theta)\big)$, where $0<\alpha,\beta,\theta<\dfrac{\pi}{4}$. Then

▸ SHOW SOLUTION

Step 1. Expand the coordinates of $R$

$$x_R=\cos(\beta-\alpha)\cos\theta-\sin(\beta-\alpha)\sin\theta$$$$y_R=\sin\beta\cos\theta-\cos\beta\sin\theta$$

Step 2. Read them as a combination of $P$ and $Q$

$x_R=x_Q\cos\theta+x_P\sin\theta$ and $y_R=y_Q\cos\theta+y_P\sin\theta$, so

$$R=\cos\theta\cdot Q+\sin\theta\cdot P$$

Step 3. Test for collinearity

A point $\lambda Q+\mu P$ lies on the line $PQ$ only when $\lambda+\mu=1$. Here

$$\cos\theta+\sin\theta=\sqrt2\sin\left(\theta+\frac{\pi}{4}\right)>1$$

for $0<\theta<\frac{\pi}{4}$. So $R$ is not on line $PQ$, and the three points are not collinear.

Answer: option (D)

Example 5. Find the area of the quadrilateral with vertices $A(1,1)$, $B(5,2)$, $C(4,6)$ and $D(-1,4)$, taken in this order.

▸ SHOW SOLUTION

Step 1. The terms $x_ky_{k+1}-x_{k+1}y_k$

Side	Term
$AB$	$1\cdot2-5\cdot1=-3$
$BC$	$5\cdot6-4\cdot2=22$
$CD$	$4\cdot4-(-1)\cdot6=22$
$DA$	$(-1)\cdot1-1\cdot4=-5$

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Step 2. Add and halve

$$\text{Area}=\frac12\lvert-3+22+22-5\rvert=18$$

The vertices go round the boundary anticlockwise, so the order is right.

Answer: $18$ square units

Example 6 (JEE Main 2017). Let $k$ be an integer such that the triangle with vertices $(k,-3k)$, $(5,k)$ and $(-k,2)$ has area $28$ square units. Then the orthocentre of this triangle is at the point

▸ SHOW SOLUTION

Step 1. Area in terms of $k$

$$2\Delta=\lvert k(k-2)+5(2+3k)+4k^2\rvert$$$$2\Delta=\lvert5k^2+13k+10\rvert=56$$

Step 2. Find the integer $k$

$5k^2+13k+10=-56$ has no real root. From $5k^2+13k-46=0$,

$$(k-2)(5k+23)=0\Rightarrow k=2$$

The vertices are $A(2,-6)$, $B(5,2)$, $C(-2,2)$.

Step 3. Two altitudes

$BC$ is horizontal ($y=2$), so the altitude from $A$ is $x=2$. The slope of $CA$ is $\frac{-6-2}{2+2}=-2$, so the altitude from $B$ has slope $\frac12$:

$$y-2=\frac12(x-5)$$

Step 4. Intersect

At $x=2$: $y=2-\frac32=\frac12$.

Answer: $\left(2,\frac12\right)$, option (4)

Example 7 (JEE Main 2015). The number of points, having both coordinates as integers, that lie in the interior of the triangle with vertices $(0,0)$, $(0,41)$ and $(41,0)$ is

▸ SHOW SOLUTION

Step 1. Describe the interior

The triangle is $x\ge0$, $y\ge0$, $x+y\le41$. A point is strictly inside when

$$x\ge1,\quad y\ge1,\quad x+y\le40$$

Step 2. Count by columns

For $x=k$ ($1\le k\le39$), $y$ runs from $1$ to $40-k$, which is $40-k$ values. So the count is

$$39+38+\dots+1=\frac{39\cdot40}{2}=780$$

Answer: option (4)

Locus of a point

A locus is the path traced by a point that moves under a given condition. Its equation is the relation satisfied by the coordinates of every position of the point, and by no other point.

Method

1. Call the moving point $(h,k)$.
2. Write the given condition as equations in $h$, $k$ and any parameter (an angle, a slope, a second point).
3. Eliminate the parameter to get one relation between $h$ and $k$.
4. Replace $h$ by $x$ and $k$ by $y$.

Examples

Example 8 (JEE Main 2026, 8 April Shift 2). If a straight line drawn through the point of intersection of $4x+3y-1=0$ and $3x+4y-1=0$ meets the coordinate axes at $P$ and $Q$, then the locus of the mid-point of $PQ$ is

▸ SHOW SOLUTION

Step 1. The fixed point

Subtracting the equations gives $x=y$, and then $7x=1$:

$$\left(\frac17,\frac17\right)$$

Step 2. The variable line

Let the mid-point be $(h,k)$, so $P(2h,0)$ and $Q(0,2k)$, and the line is

$$\frac{x}{2h}+\frac{y}{2k}=1$$

Step 3. Make it pass through the fixed point

$$\frac{1}{14h}+\frac{1}{14k}=1\Rightarrow h+k=14hk$$

Step 4. Replace $(h,k)$ by $(x,y)$

$$x+y-14xy=0$$

Answer: option (2)

Example 9 (JEE Main 2018). A straight line through the fixed point $(2,3)$ intersects the coordinate axes at distinct points $P$ and $Q$. If $O$ is the origin and the rectangle $OPRQ$ is completed, find the locus of $R$.

▸ SHOW SOLUTION

Step 1. Coordinates

Let $P(a,0)$ and $Q(0,b)$. The fourth vertex of rectangle $OPRQ$ is $R(a,b)$.

Step 2. The line passes through $(2,3)$

$$\frac{2}{a}+\frac{3}{b}=1$$

Step 3. Replace $(a,b)$ by $(x,y)$

$$\frac2x+\frac3y=1\Rightarrow 3x+2y=xy$$

Answer: $3x+2y=xy$

Example 10 (JEE Main 2020, 7 Jan Shift 2). From any point $P$ on the line $x=2y$, a perpendicular is drawn to the line $y=x$; let $Q$ be its foot. The locus of the mid-point of $PQ$ is

▸ SHOW SOLUTION

Step 1. Coordinates of $P$ and $Q$

Let $P(2t,t)$. The foot on $y=x$ is the point of $y=x$ with the same value of $x+y$:

$$Q=\left(\frac{3t}{2},\frac{3t}{2}\right)$$

Step 2. Mid-point of $PQ$

$$M=\left(\frac{7t}{4},\frac{5t}{4}\right)$$

Step 3. Eliminate $t$

$\frac{x}{7}=\frac{y}{5}$, that is $5x=7y$.

Answer: option (2)

Example 11 (IIT JEE 2009, Paper 2). The locus of the orthocentre of the triangle formed by the lines $(1+p)x-py+p(1+p)=0$, $(1+q)x-qy+q(1+q)=0$ and $y=0$, where $p\ne q$, is

▸ SHOW SOLUTION

Step 1. The vertices

With $y=0$ the first two lines give $B(-p,0)$ and $C(-q,0)$. Subtracting the first two equations, $x(p-q)-y(p-q)+(p-q)(1+p+q)=0$, so their intersection $A$ lies on

$$x-y+1+p+q=0$$

Putting this in the first line gives $A=\big(pq,\ (1+p)(1+q)\big)$.

Step 2. Two altitudes

$BC$ is the $x$-axis, so the altitude from $A$ is $x=pq$. The side $AC$ is the second line, with slope $\frac{1+q}{q}$, so the altitude from $B(-p,0)$ is

$$y=-\frac{q}{1+q}(x+p)$$

Step 3. The orthocentre

At $x=pq$: $y=-\frac{q\cdot p(q+1)}{1+q}=-pq$. So $H=(pq,-pq)$ and

$$x+y=0$$

Answer: option (D)

Example 12 (IITianForum). A point $P$ lies inside the triangle formed by the lines $y=0$, $4x-3y=0$ and $3x+4y-15=0$, and the sum of its distances from the three lines is $3$. The locus of $P$ is a part of the line

▸ SHOW SOLUTION

Step 1. Signs inside the triangle

The vertices are $(0,0)$, $(5,0)$ and $\left(\frac95,\frac{12}{5}\right)$. At the inside point $(2,1)$: $y>0$, $4x-3y>0$ and $3x+4y-15<0$.

Step 2. Remove the modulus signs

$$y+\frac{4x-3y}{5}+\frac{15-3x-4y}{5}=3$$

Step 3. Simplify

$$5y+x-7y+15=15\Rightarrow x-2y=0$$

Answer: option (A)

Slope of a line

The inclination of a line is the angle $\theta$, $0\le\theta<\pi$, that it makes with the positive direction of the $x$-axis, measured anticlockwise. Its slope is

$$m=\tan\theta,\qquad\theta\ne\frac{\pi}{2}$$ A line making the inclination theta with the positive x-axis, with the rise and run between two of its points

Parallel and perpendicular lines

For non-vertical lines with slopes $m_1$ and $m_2$:

$$\text{parallel}\iff m_1=m_2$$ $$\text{perpendicular}\iff m_1m_2=-1$$

Reason. Parallel lines have equal inclinations. If $\theta_2=\theta_1+\frac{\pi}{2}$, then $\tan\theta_2=-\cot\theta_1$, so $m_1m_2=-1$.

A line parallel to $ax+by+c=0$ is $ax+by+k=0$, and a line perpendicular to it is $bx-ay+k=0$.

Examples

Example 13 (JEE Main 2014). Let $a$, $b$, $c$ and $d$ be non-zero numbers. If the point of intersection of the lines $4ax+2ay+c=0$ and $5bx+2by+d=0$ lies in the fourth quadrant and is equidistant from the two axes, then

▸ SHOW SOLUTION

Step 1. Write the point

In the fourth quadrant and equidistant from the axes, the point is $(t,-t)$ with $t>0$.

Step 2. Put it in both lines

$$4at-2at+c=0\Rightarrow t=-\frac{c}{2a}$$$$5bt-2bt+d=0\Rightarrow t=-\frac{d}{3b}$$

Step 3. Equate

$$\frac{c}{2a}=\frac{d}{3b}\Rightarrow3bc-2ad=0$$

Answer: option (1)

Example 14 (JEE Main 2021, 26 Feb Shift 1). The intersection of the three lines $x-y=0$, $x+2y=3$ and $2x+y=6$ is a

▸ SHOW SOLUTION

Step 1. The vertices

$$x-y=0,\ x+2y=3\Rightarrow(1,1)$$$$x-y=0,\ 2x+y=6\Rightarrow(2,2)$$$$x+2y=3,\ 2x+y=6\Rightarrow(3,0)$$

Step 2. The sides

The squared sides are $2$, $5$ and $5$. Two sides are equal, and $2+5\ne5$, so there is no right angle.

Answer: option (3)

Forms of the equation of a line

Every straight line has an equation of the first degree in $x$ and $y$, and every such equation is a straight line. A line is fixed by two conditions: two points, or a point and a direction, or two intercepts.

The intercepts $a$, $b$ and $c$ carry signs: a line cutting the negative $y$-axis has $c<0$.

Two-point form as a determinant

$$\begin{vmatrix}x&y&1\\x_1&y_1&1\\x_2&y_2&1\end{vmatrix}=0$$

This is the collinearity test with $(x,y)$ as the third point.

Intercept form and normal form

The line $\frac xa+\frac yb=1$ cuts off the triangle $OAB$ with $A(a,0)$, $B(0,b)$, of area $\frac12\lvert ab\rvert$.

In the normal form, $p>0$ is the length of the perpendicular $ON$ from the origin to the line, and $\alpha$ is the angle $ON$ makes with the positive $x$-axis, $0\le\alpha<2\pi$.

Intercept form with intercepts a and b, beside the normal form with the perpendicular ON of length p at angle alpha

Reason. $N=(p\cos\alpha,p\sin\alpha)$, and the line is perpendicular to $ON$, so its slope is $-\cot\alpha$. The point-slope form through $N$ simplifies to $x\cos\alpha+y\sin\alpha=p$.

Parametric form

The point at signed distance $r$ from $A(x_1,y_1)$ along the line of inclination $\theta$ is

$$(x_1+r\cos\theta,\ y_1+r\sin\theta)$$

$r>0$ on one side of $A$ and $r<0$ on the other. This form turns a question about a length along a line into a question about one number $r$.

A line through A at inclination theta, with the point P at distance r along it and the horizontal and vertical steps r cos theta and r sin theta

From the general form $ax+by+c=0$

For the normal form write $-ax-by=c$ when $c>0$ and $ax+by=-c$ when $c<0$, then divide by $\sqrt{a^2+b^2}$. The right side is then $p=\frac{\lvert c\rvert}{\sqrt{a^2+b^2}}$.

Examples

Example 15 (JEE Main 2025, 2 April Shift 2). Let the area of the triangle formed by a straight line $L:x+by+c=0$ with the coordinate axes be $48$ square units. If the perpendicular drawn from the origin to the line $L$ makes an angle of $45^\circ$ with the positive $x$-axis, then the value of $b^2+c^2$ is

▸ SHOW SOLUTION

Step 1. Use the normal form

With $\alpha=45^\circ$ the line is $\frac{x}{\sqrt2}+\frac{y}{\sqrt2}=p$, that is $x+y=p\sqrt2$. Comparing with $x+by+c=0$ gives $b=1$.

Step 2. Use the area

The intercepts are both $-c$, so

$$\frac12c^2=48\Rightarrow c^2=96$$

Step 3. Answer

$$b^2+c^2=1+96=97$$

Answer: option (3)

Example 16 (JEE Main 2021, 31 Aug Shift 1). If $p$ and $q$ are the lengths of the perpendiculars from the origin on the lines $x\operatorname{cosec}\alpha-y\sec\alpha=k\cot2\alpha$ and $x\sin\alpha+y\cos\alpha=k\sin2\alpha$ respectively, then $k^2$ is equal to

▸ SHOW SOLUTION

Step 1. Bring the first line to normal form

Multiply by $\sin\alpha\cos\alpha$:

$$x\cos\alpha-y\sin\alpha=k\cot2\alpha\sin\alpha\cos\alpha$$

The right side is $\frac{k\cos2\alpha}{2}$, and $\cos^2\alpha+\sin^2\alpha=1$, so

$$p=\frac{\lvert k\cos2\alpha\rvert}{2}$$

Step 2. The second line

It is already in normal form: $q=\lvert k\sin2\alpha\rvert$.

Step 3. Eliminate $\alpha$

$$4p^2+q^2=k^2\cos^22\alpha+k^2\sin^22\alpha=k^2$$

Answer: option (1)

Example 17 (JEE Main 2026, 24 Jan Shift 2). Let the angles made with the positive $x$-axis by two straight lines drawn from the point $P(2,3)$ and meeting the line $x+y=6$ at a distance $\sqrt{\dfrac23}$ from $P$ be $\theta_1$ and $\theta_2$. Then the value of $\theta_1+\theta_2$ is

▸ SHOW SOLUTION

Step 1. The point at distance $r$

With $r=\sqrt{\frac23}$, the point is $\left(2+r\cos\theta,\ 3+r\sin\theta\right)$. It lies on $x+y=6$:

$$5+r(\cos\theta+\sin\theta)=6$$$$\cos\theta+\sin\theta=\sqrt{\frac32}$$

Step 2. Square

$$1+\sin2\theta=\frac32\Rightarrow\sin2\theta=\frac12$$

Step 3. The two angles

$2\theta=\frac{\pi}{6}$ or $\frac{5\pi}{6}$, so $\theta_1=\frac{\pi}{12}$ and $\theta_2=\frac{5\pi}{12}$. Both make $\cos\theta+\sin\theta$ positive, so neither is extra.

$$\theta_1+\theta_2=\frac{\pi}{2}$$

Answer: option (3)

Position of a point relative to a line

Write $L(x,y)=ax+by+c$. The line $L=0$ divides the plane into two half-planes, and $L$ keeps one sign on each of them.

Reason. On a vertical line $x=x_0$ (when $b\ne0$), $L$ is a first-degree function of $y$ that vanishes only at the point of the line. So $L$ has one sign above the line and the opposite sign below it.

The line L equal to zero with the half-plane where L is positive and the half-plane where L is negative

Inside a triangle

A point $P$ is inside the triangle formed by three lines when, for each line, $P$ lies on the same side as the opposite vertex. In practice: find the three vertices, then for each line compare the sign of $L$ at $P$ and at the vertex not on that line.

Examples

Example 18 (JEE Main 2020, 2 Sep Shift 2). The set of all possible values of $\theta$ in the interval $(0,\pi)$ for which the points $(1,2)$ and $(\sin\theta,\cos\theta)$ lie on the same side of the line $x+y=1$ is

▸ SHOW SOLUTION

Step 1. Sign at $(1,2)$

With $L=x+y-1$, $L(1,2)=2>0$.

Step 2. The same sign at the other point

$$\sin\theta+\cos\theta-1>0$$$$\sqrt2\sin\left(\theta+\frac{\pi}{4}\right)>1$$

Step 3. Solve on $(0,\pi)$

$\theta+\frac{\pi}{4}$ runs over $\left(\frac{\pi}{4},\frac{5\pi}{4}\right)$, and the sine exceeds $\frac{1}{\sqrt2}$ for $\theta+\frac{\pi}{4}<\frac{3\pi}{4}$, so $0<\theta<\frac{\pi}{2}$.

Answer: option (1)

Example 19 (JEE Main 2025, 24 Jan Shift 2). Let the points $\left(\dfrac{11}{2},\alpha\right)$ lie on or inside the triangle with sides $x+y=11$, $x+2y=16$ and $2x+3y=29$. Then the product of the smallest and the largest values of $\alpha$ is equal to

▸ SHOW SOLUTION

Step 1. The vertices

$$x+y=11,\ x+2y=16\Rightarrow(6,5)$$$$x+y=11,\ 2x+3y=29\Rightarrow(4,7)$$$$x+2y=16,\ 2x+3y=29\Rightarrow(10,3)$$

Step 2. Which sides cross $x=\frac{11}{2}$

The value $\frac{11}{2}$ lies between $4$ and $6$ and between $4$ and $10$, so the vertical line meets the sides $x+y=11$ and $2x+3y=29$:

$$y=11-\frac{11}{2}=\frac{11}{2}$$$$y=\frac{29-11}{3}=6$$

Step 3. The range of $\alpha$

$\frac{11}{2}\le\alpha\le6$, and the product of the extremes is $\frac{11}{2}\cdot6=33$.

Answer: option (3)

Angle between two lines

If two lines have slopes $m_1=\tan\theta_1$ and $m_2=\tan\theta_2$, the angle $\theta$ between them is $\lvert\theta_1-\theta_2\rvert$ or its supplement, and

$$\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$

Reason. $\tan(\theta_1-\theta_2)=\frac{\tan\theta_1-\tan\theta_2}{1+\tan\theta_1\tan\theta_2}$. Taking the modulus gives the acute angle; the obtuse angle is $\pi-\theta$.

Two intersecting lines with inclinations theta one and theta two and the angle theta between them

For $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$,

$$\tan\theta=\left|\frac{a_1b_2-a_2b_1}{a_1a_2+b_1b_2}\right|$$

Lines making a given angle with a given line

Through $(x_1,y_1)$ there are two lines making an angle $\alpha$ with the line of slope $m$. Their slopes are

$$m'=\frac{m+\tan\alpha}{1-m\tan\alpha}$$ $$m''=\frac{m-\tan\alpha}{1+m\tan\alpha}$$

Equivalently, add and subtract $\alpha$ to the inclination of the given line.

Examples

Example 20 (IIT JEE 2011, Paper 1). A straight line $L$ through the point $(3,-2)$ is inclined at an angle $60^\circ$ to the line $\sqrt3x+y=1$. If $L$ also intersects the $x$-axis, then the equation of $L$ is

▸ SHOW SOLUTION

Step 1. Inclination of the given line

The slope is $-\sqrt3$, so its inclination is $120^\circ$.

Step 2. The two lines at $60^\circ$

Their inclinations are $120^\circ\pm60^\circ$, that is $60^\circ$ or $180^\circ$. The second is parallel to the $x$-axis and never meets it, so $L$ has slope $\tan60^\circ=\sqrt3$.

Step 3. Equation

$$y+2=\sqrt3(x-3)$$$$y-\sqrt3x+2+3\sqrt3=0$$

Answer: option (B)

Example 21 (JEE Main 2021, 18 March Shift 1). The equation of one of the straight lines which passes through the point $(1,3)$ and makes an angle $\tan^{-1}\sqrt2$ with the straight line $y+1=3\sqrt2x$ is

▸ SHOW SOLUTION

Step 1. The two slopes

With $m=3\sqrt2$ and $\tan\alpha=\sqrt2$, $m\tan\alpha=6$:

$$m'=\frac{3\sqrt2+\sqrt2}{1-6}=-\frac{4\sqrt2}{5}$$$$m''=\frac{3\sqrt2-\sqrt2}{1+6}=\frac{2\sqrt2}{7}$$

Step 2. The line of slope $-\frac{4\sqrt2}{5}$

$$5(y-3)=-4\sqrt2(x-1)$$$$4\sqrt2x+5y-(15+4\sqrt2)=0$$

Answer: option (1)

Example 22 (JEE Main 2025, 3 April Shift 1). A line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines $L_1:2x+y+6=0$ and $L_2:4x+2y-p=0$, $p>0$, at the points $A$ and $B$ respectively. If $AB=\dfrac{9}{\sqrt2}$ and the foot of the perpendicular from the point $A$ on the line $L_2$ is $M$, then $\dfrac{AM}{BM}$ is equal to

▸ SHOW SOLUTION

Step 1. The points $A$ and $B$

The line is $y=x$. With $L_1$: $3x+6=0$, so $A(-2,-2)$. With $L_2$: $B\left(\frac p6,\frac p6\right)$.

Step 2. Find $p$

$$AB=\sqrt2\left|\frac p6+2\right|=\frac{9}{\sqrt2}\Rightarrow\frac p6=\frac52$$

(the other root makes $p<0$). So $p=15$.

Step 3. The right triangle $AMB$

$AM\perp L_2$, so $\frac{AM}{BM}=\tan\angle ABM$, where $\angle ABM$ is the angle between $y=x$ (slope $1$) and $L_2$ (slope $-2$):

$$\tan\angle ABM=\left|\frac{1-(-2)}{1+1\cdot(-2)}\right|=3$$

Answer: option (4)

Distance of a point from a line

The length of the perpendicular from $P(x_1,y_1)$ to the line $ax+by+c=0$ is

$$d=\frac{\lvert ax_1+by_1+c\rvert}{\sqrt{a^2+b^2}}$$ The perpendicular PN from a point P to a line, beside two parallel lines with the distance d between them

Reason. Let the perpendicular meet the line at $N$, so $N=P+t(a,b)$ for some $t$, because $(a,b)$ is perpendicular to the line. Putting $N$ in the line gives $t=-\frac{ax_1+by_1+c}{a^2+b^2}$, and $PN=\lvert t\rvert\sqrt{a^2+b^2}$.

Examples

Example 23 (JEE Main 2026, 23 Jan Shift 1). A rectangle is formed by the lines $x=0$, $y=0$, $x=3$ and $y=4$. Let the line $L$ be perpendicular to $3x+y+6=0$ and divide the area of the rectangle into two equal parts. Then the distance of the point $\left(\dfrac12,-5\right)$ from the line $L$ is equal to

▸ SHOW SOLUTION

Step 1. The slope of $L$

The given line has slope $-3$, so $L$ has slope $\frac13$.

Step 2. $L$ passes through the centre

A line halves the area of a rectangle when it passes through its centre $\left(\frac32,2\right)$:

$$y-2=\frac13\left(x-\frac32\right)\Rightarrow2x-6y+9=0$$

Step 3. The distance

$$d=\frac{\lvert1+30+9\rvert}{\sqrt{4+36}}=\frac{40}{2\sqrt{10}}=2\sqrt{10}$$

Answer: option (4)

Example 24 (JEE Main 2026, 21 Jan Shift 1). Let a point $A$ lie between the parallel lines $L_1$ and $L_2$ such that its distances from $L_1$ and $L_2$ are $6$ and $3$ units respectively. Then the area (in sq. units) of the equilateral triangle $ABC$, where $B$ and $C$ lie on $L_1$ and $L_2$, is

▸ SHOW SOLUTION

Step 1. Place the lines

Take $L_2$ as $y=0$ and $L_1$ as $y=9$, so that $A(0,3)$ is $3$ above $L_2$ and $6$ below $L_1$. Let the side be $a$.

Two parallel lines with the point A between them at distances 6 and 3, and the equilateral triangle ABC with B and C on the lines

Step 2. Two heights

Let $AC$ go down to $L_2$ at an angle $\theta$ below the horizontal. Turning $AC$ through $60^\circ$ towards $L_1$ gives $AB$, which rises at $60^\circ-\theta$:

$$a\sin\theta=3$$$$a\sin(60^\circ-\theta)=6$$

Step 3. Expand the second

$$\frac{\sqrt3}{2}a\cos\theta-\frac12a\sin\theta=6$$$$\sqrt3\,a\cos\theta=15$$

Step 4. Eliminate $\theta$

$$a^2=(a\sin\theta)^2+(a\cos\theta)^2$$$$a^2=9+75=84$$

Step 5. Area

$$\frac{\sqrt3}{4}a^2=\frac{\sqrt3}{4}\cdot84=21\sqrt3$$

Answer: option (3)

Example 25 (JEE Main 2020, 5 Sep Shift 1). If the line $2x-y+3=0$ is at a distance $\dfrac{1}{\sqrt5}$ and $\dfrac{2}{\sqrt5}$ from the lines $4x-2y+\alpha=0$ and $6x-3y+\beta=0$ respectively, then the sum of all possible values of $\alpha$ and $\beta$ is ______.

▸ SHOW SOLUTION

Step 1. Match coefficients

Write the given line as $4x-2y+6=0$ and as $6x-3y+9=0$.

Step 2. First distance

$$\frac{\lvert\alpha-6\rvert}{\sqrt{20}}=\frac{1}{\sqrt5}\Rightarrow\lvert\alpha-6\rvert=2$$

So $\alpha=4$ or $8$.

Step 3. Second distance

$$\frac{\lvert\beta-9\rvert}{\sqrt{45}}=\frac{2}{\sqrt5}\Rightarrow\lvert\beta-9\rvert=6$$

So $\beta=3$ or $15$.

Step 4. Add

$$4+8+3+15=30$$

Answer: $30$

Example 26 (JEE Advanced 2013, Paper 1). For $a>b>c>0$, the distance between $(1,1)$ and the point of intersection of the lines $ax+by+c=0$ and $bx+ay+c=0$ is less than $2\sqrt2$. Then

▸ SHOW SOLUTION

Step 1. The point of intersection

Subtracting the equations gives $(a-b)(x-y)=0$, so $x=y$, and then

$$x=y=-\frac{c}{a+b}$$

Step 2. The distance

$$\sqrt2\left(1+\frac{c}{a+b}\right)<2\sqrt2\Rightarrow\frac{c}{a+b}<1$$

So $a+b-c>0$.

Step 3. The other option

$a>b$ and $c>0$ give $a-b+c>0$ directly.

Answer: (A) and (C)

Example 27 (JEE Advanced 2014, Paper 1). For a point $P$ in the plane, let $d_1(P)$ and $d_2(P)$ be the distances of the point $P$ from the lines $x-y=0$ and $x+y=0$ respectively. The area of the region $R$ consisting of all points $P$ lying in the first quadrant of the plane and satisfying $2\le d_1(P)+d_2(P)\le4$ is ______.

▸ SHOW SOLUTION

Step 1. Remove the modulus signs

In the first quadrant $d_2=\frac{x+y}{\sqrt2}$. For $x\ge y$, $d_1=\frac{x-y}{\sqrt2}$, so

$$d_1+d_2=\sqrt2x$$

Step 2. The region below $y=x$

$2\le\sqrt2x\le4$ gives $\sqrt2\le x\le2\sqrt2$, with $0\le y\le x$. This is the region between two vertical lines, cut by the axis and the line $y=x$.

The region in the first quadrant between x equal to root 2 and 2 root 2 below y equal to x, and its mirror image above y equal to x

Step 3. Its area

It is a trapezium with parallel sides $\sqrt2$ and $2\sqrt2$ and width $\sqrt2$:

$$\frac12(\sqrt2+2\sqrt2)\sqrt2=3$$

Step 4. Use the symmetry

Above $y=x$ the roles of $x$ and $y$ swap, giving another $3$.

Answer: $6$

Foot of the perpendicular, image and reflection

For the point $P(x_1,y_1)$ and the line $ax+by+c=0$, the foot of the perpendicular $N(h,k)$ is given by

$$\frac{h-x_1}{a}=\frac{k-y_1}{b}=-\frac{ax_1+by_1+c}{a^2+b^2}$$

The image $Q(h,k)$ of $P$ in the line, the point with the line as the perpendicular bisector of $PQ$, is given by

$$\frac{h-x_1}{a}=\frac{k-y_1}{b}=-\frac{2(ax_1+by_1+c)}{a^2+b^2}$$

Reason. $PN$ is along the normal $(a,b)$, so $N=P+t(a,b)$, and $N$ on the line fixes $t$ as in the distance formula. The image is twice as far along the same normal.

The point P, the foot N of its perpendicular on a line, and the image Q with N the mid-point of PQ

Image of a line and reflection of a ray

To reflect a whole line in a mirror, reflect two of its points. If the line meets the mirror at $M$, then $M$ is its own image, so one more point is enough.

A ray of light reflected by a line mirror obeys angle of incidence = angle of reflection. Equivalently, the reflected ray, extended backwards, passes through the image of the source. This turns a reflection question into an image question.

A ray from S striking a mirror at M, the reflected ray, and the image S prime of S lying on the reflected ray extended backwards

Examples

Example 28 (JEE Main 2021, 25 Feb Shift 1). The image of the point $(3,5)$ in the line $x-y+1=0$ lies on

▸ SHOW SOLUTION

Step 1. Use the formula

$L(3,5)=3-5+1=-1$ and $a^2+b^2=2$, so

$$\frac{h-3}{1}=\frac{k-5}{-1}=-\frac{2(-1)}{2}=1$$

The image is $(4,4)$.

Step 2. Test the options

$(4-2)^2+(4-4)^2=4$, so it lies on the first circle.

Answer: option (1)

Example 29 (JEE Main 2025, 22 Jan Shift 1). Let the triangle $PQR$ be the image of the triangle with vertices $(1,3)$, $(3,1)$ and $(2,4)$ in the line $x+2y=2$. If the centroid of $\triangle PQR$ is the point $(\alpha,\beta)$, then $15(\alpha-\beta)$ is equal to

▸ SHOW SOLUTION

Step 1. Reflect the centroid only

Reflection keeps mid-points and ratios, so the centroid of the image is the image of the centroid $G\left(2,\frac83\right)$.

Step 2. Image of $G$

$L(G)=2+\frac{16}{3}-2=\frac{16}{3}$ and $a^2+b^2=5$:

$$\frac{\alpha-2}{1}=\frac{\beta-\frac83}{2}=-\frac{32}{15}$$

So $\alpha=-\frac{2}{15}$ and $\beta=\frac83-\frac{64}{15}=-\frac{24}{15}$.

Step 3. Answer

$$15(\alpha-\beta)=-2+24=22$$

Answer: option (4)

Example 30 (JEE Main 2026, 21 Jan Shift 2). $P$ is a point on the circle $x^2+y^2=4$, $Q$ is a point on the line $5x+y+2=0$, and $x-y+1=0$ is the perpendicular bisector of $PQ$. Then $13$ times the sum of the abscissae of all such points $P$ is ______.

▸ SHOW SOLUTION

Step 1. $P$ is the image of $Q$

The perpendicular bisector of $PQ$ is the mirror, so $P$ is the image of $Q$ in $x-y+1=0$. The image of $(x,y)$ in this line is $(y-1,\ x+1)$.

Step 2. Image of the whole line

If $Q(a,b)$ is on $5x+y+2=0$, then $P=(b-1,a+1)$. Writing $P=(X,Y)$ gives $a=Y-1$, $b=X+1$, and

$$5(Y-1)+(X+1)+2=0$$$$X+5Y-2=0$$

Step 3. Meet the circle

Put $x=2-5y$ in $x^2+y^2=4$:

$$26y^2-20y=0\Rightarrow y=0\ \text{or}\ y=\frac{10}{13}$$

The abscissae are $2$ and $2-\frac{50}{13}=-\frac{24}{13}$.

Step 4. Answer

$$13\left(2-\frac{24}{13}\right)=2$$

Answer: $2$

Example 31 (JEE Main 2026, 5 April Shift 2). From the point $(-1,-1)$, two rays are sent making angles of $45^\circ$ with the line $x+y=0$. These rays get reflected from the mirror $x+2y=1$. If the equations of the reflected rays are $ax+by=9$ and $cx+dy=7$, $a,b,c,d\in\mathbb Z$, then the value of $ad+bc$ is ______.

▸ SHOW SOLUTION

Step 1. The two incident rays

The line $x+y=0$ has inclination $135^\circ$, so the rays have inclinations $90^\circ$ and $180^\circ$: they are $x=-1$ and $y=-1$.

Step 2. Points where they meet the mirror

$x=-1$ meets $x+2y=1$ at $(-1,1)$, and $y=-1$ meets it at $(3,-1)$.

Step 3. Image of the source

The reflected rays pass through the image $S'$ of $S(-1,-1)$ in $x+2y-1=0$. With $L(S)=-4$ and $a^2+b^2=5$,

$$S'=\left(-1+\frac85,\ -1+\frac{16}{5}\right)=\left(\frac35,\frac{11}{5}\right)$$

Step 4. The reflected rays

Through $(-1,1)$ and $S'$ the slope is $\frac{6/5}{8/5}=\frac34$, giving $3x-4y+7=0$, that is $-3x+4y=7$.

Through $(3,-1)$ and $S'$ the slope is $\frac{16/5}{-12/5}=-\frac43$, giving $4x+3y=9$.

Step 5. Answer

$a=4$, $b=3$, $c=-3$, $d=4$, so

$$ad+bc=16-9=7$$

Answer: $7$

Example 32 (IITianForum). $P(4,1)$ is a fixed point. $Q$ and $R$ are points on the lines $y=x$ and $y=0$ respectively. The least possible value of the perimeter $PQ+QR+RP$ is

▸ SHOW SOLUTION

Step 1. Reflect $P$ in both lines

The image of $P$ in $y=x$ is $P_1(1,4)$, and in $y=0$ it is $P_2(4,-1)$. Then $PQ=P_1Q$ and $RP=RP_2$.

The point P between the lines y equal to x and y equal to 0, its two images, and the straight path from one image to the other meeting the lines at Q and R

Step 2. A broken path between the images

$$PQ+QR+RP=P_1Q+QR+RP_2$$$$P_1Q+QR+RP_2\ge P_1P_2$$

Equality holds when $Q$ and $R$ lie on the segment $P_1P_2$.

Step 3. The least value

$$P_1P_2=\sqrt{3^2+5^2}=\sqrt{34}$$

The segment meets $y=x$ at $Q\left(\frac{17}{8},\frac{17}{8}\right)$ and $y=0$ at $R\left(\frac{17}{5},0\right)$, both on the correct rays.

Answer: option (A)

Bisectors of the angles between two lines

Two intersecting lines $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ form two pairs of vertically opposite angles, and so they have two angle bisectors, perpendicular to each other:

$$\frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$$

Reason. A point is on a bisector exactly when it is equidistant from the two lines. The two signs give the two bisectors.

Two intersecting lines with the bisector of the angle containing the origin and the other bisector, the origin marked

Which bisector is which

First make $c_1$ and $c_2$ positive (multiply an equation by $-1$ if needed). Then:

A quicker check: take any point on one bisector and find the angle the bisector makes with one of the lines. If it is less than $45^\circ$, that bisector bisects the acute angle.

The internal bisector of an angle of a triangle

The internal bisector of angle $A$ in triangle $ABC$ meets $BC$ at $D$ with

$$\frac{BD}{DC}=\frac{AB}{AC}$$

so $D$ comes from the section formula. A vertex and its reflection in an internal bisector lie on the opposite sides of that angle: the reflection of $A$ in the bisector of angle $B$ lies on line $BC$.

Examples

Example 33 (IIT JEE 2007, Paper 2). Lines $L_1:y-x=0$ and $L_2:2x+y=0$ intersect the line $L_3:y+2=0$ at $P$ and $Q$ respectively. The bisector of the acute angle between $L_1$ and $L_2$ intersects $L_3$ at $R$.

Statement 1: The ratio $PR:RQ$ equals $2\sqrt2:\sqrt5$.

Statement 2: In any triangle, the bisector of an angle divides the triangle into two similar triangles.

▸ SHOW SOLUTION

Step 1. The points

$P(-2,-2)$ and $Q(1,-2)$, and the lines meet at $O(0,0)$.

Step 2. Is $\angle POQ$ acute?

$\overrightarrow{OP}\cdot\overrightarrow{OQ}=-2+4=2>0$, so $\angle POQ$ is acute and its bisector is the acute-angle bisector.

Step 3. The ratio

$$\frac{PR}{RQ}=\frac{OP}{OQ}=\frac{2\sqrt2}{\sqrt5}$$

So Statement 1 is true.

Step 4. Statement 2

A bisector gives two triangles with one equal angle and one common side, but they are not similar in general. Statement 2 is false.

Answer: option (C)

Example 34 (JEE Main 2026, 4 April Shift 1). Let the line $L_1:x+3=0$ intersect the lines $L_2:x-y=0$ and $L_3:3x+y=0$ at the points $A$ and $B$ respectively. Let the bisector of the obtuse angle between the lines $L_2$ and $L_3$ intersect the line $L_1$ at the point $C$. Then $\dfrac{BC^2}{AC^2}$ is equal to

▸ SHOW SOLUTION

Step 1. The points

$A(-3,-3)$ and $B(-3,9)$; $L_2$ and $L_3$ meet at $O(0,0)$.

Step 2. Is $\angle AOB$ obtuse?

$\overrightarrow{OA}\cdot\overrightarrow{OB}=9-27=-18<0$, so $\angle AOB$ is obtuse, and its bisector is the obtuse-angle bisector.

Step 3. Angle bisector theorem in $\triangle OAB$

$$\frac{BC^2}{AC^2}=\frac{OB^2}{OA^2}=\frac{90}{18}=5$$

Answer: option (1)

Example 35 (JEE Main 2026, 24 Jan Shift 1). Let $A(1,0)$, $B(2,-1)$ and $C\left(\dfrac73,\dfrac43\right)$ be three points. If the equation of the bisector of the angle $ABC$ is $\alpha x+\beta y=5$, then the value of $\alpha^2+\beta^2$ is

▸ SHOW SOLUTION

Step 1. The two sides at $B$

$$BA=\sqrt2,\qquad BC=\sqrt{\frac19+\frac{49}{9}}=\frac{5\sqrt2}{3}$$

Step 2. The point $D$ on $AC$

$AD:DC=BA:BC=3:5$, so

$$D=\frac{5A+3C}{8}=\left(\frac32,\frac12\right)$$

Step 3. The line $BD$

Slope $\frac{\frac12+1}{\frac32-2}=-3$, so $y+1=-3(x-2)$, that is $3x+y=5$.

Answer: $\alpha^2+\beta^2=10$, option (4)

Example 36 (IITianForum). In a triangle $ABC$ the vertex $A$ is $(4,2)$, and the internal bisectors of the angles $B$ and $C$ are $x+y-2=0$ and $x-1=0$ respectively. The equation of the side $BC$ is

▸ SHOW SOLUTION

Step 1. Reflect $A$ in the bisector of $B$

The image of $A$ in the bisector of angle $B$ lies on $BC$. With $L(A)=4$ and $a^2+b^2=2$,

$$A_1=(4-4,\ 2-4)=(0,-2)$$

Step 2. Reflect $A$ in the bisector of $C$

In the line $x=1$ the image is

$$A_2=(-2,2)$$

Step 3. The line $A_1A_2$

Slope $\frac{2+2}{-2-0}=-2$, so $y+2=-2x$:

$$2x+y+2=0$$

Check: it meets the bisectors at $B(-4,6)$ and $C(1,-4)$, and the bisectors meet at the incentre $(1,1)$.

Answer: option (A)

Centres of a triangle

Let the vertices be $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$ with opposite sides $a=BC$, $b=CA$, $c=AB$.

The excentres $I_2$ and $I_3$ come from changing the sign of $b$ or of $c$ instead of $a$.

Reason for the incentre. The bisector from $A$ meets $BC$ at $D$ with $BD:DC=c:b$, and the bisector from $B$ then divides $AD$ in the ratio $AI:ID=(b+c):a$. The section formula gives $I$.

When the angles are known, the circumcentre and orthocentre also have weighted forms:

$$O=\frac{\sum x_1\sin2A}{\sum\sin2A},\qquad H=\frac{\sum x_1\tan A}{\sum\tan A}$$

(with the same expressions for the $y$-coordinates, and $H$ valid for a triangle with no right angle).

The Euler line

In every triangle the orthocentre $H$, the centroid $G$ and the circumcentre $O$ are collinear, and

$$HG:GO=2:1$$ A scalene triangle with its orthocentre H, centroid G and circumcentre O on one straight line, G two thirds of the way from H to O

Examples

Example 37 (IIT JEE 2007, Paper 2). Let $O(0,0)$, $P(3,4)$, $Q(6,0)$ be the vertices of the triangle $OPQ$. The point $R$ inside the triangle $OPQ$ is such that the triangles $OPR$, $PQR$, $OQR$ are of equal area. The coordinates of $R$ are

▸ SHOW SOLUTION

Step 1. Equal areas give the centroid

$\triangle OQR$ has one third of the area of $\triangle OPQ$ on the same base $OQ$, so $R$ is at one third of the height of $P$ above $OQ$. The same holds for the other two sides, and the only such point is the centroid.

Step 2. Compute it

$$R=\left(\frac{0+3+6}{3},\frac{0+4+0}{3}\right)=\left(3,\frac43\right)$$

Answer: option (C)

Example 38 (JEE Main 2026, 2 April Shift 1). Let the mid-points of the sides of a triangle $ABC$ be $\left(\dfrac52,7\right)$, $\left(\dfrac52,3\right)$ and $(4,5)$. If its incentre is $(h,k)$, then $3h+k$ is equal to

▸ SHOW SOLUTION

Step 1. Recover the vertices

Each vertex is the sum of the two adjacent mid-points minus the opposite one:

$$(4,1),\quad(4,9),\quad(1,5)$$

Step 2. The sides

The side joining $(4,1)$ and $(4,9)$ is $8$; the other two are both $5$. The weight $8$ goes on the vertex $(1,5)$ opposite the side of length $8$.

Step 3. The incentre

$$h=\frac{8\cdot1+5\cdot4+5\cdot4}{18}=\frac{8}{3}$$$$k=\frac{8\cdot5+5\cdot1+5\cdot9}{18}=5$$

Step 4. Answer

$$3h+k=8+5=13$$

Answer: option (3)

Example 39 (JEE Main 2026, 4 April Shift 1). Let the vertex $A$ of a triangle $ABC$ be $(1,2)$, and the mid-point of the side $AB$ be $(5,-1)$. If the centroid of the triangle is $(3,4)$ and its circumcentre is $(\alpha,\beta)$, then $21(\alpha+\beta)$ is equal to

▸ SHOW SOLUTION

Step 1. The vertices $B$ and $C$

$B=2(5,-1)-(1,2)=(9,-4)$, and $C=3G-A-B$:

$$C=(9-1-9,\ 12-2+4)=(-1,14)$$

Step 2. Two perpendicular bisectors

$PA^2=PB^2$ and $PA^2=PC^2$ for $P(\alpha,\beta)$ give

$$4\alpha-3\beta=23$$$$-\alpha+6\beta=48$$

Step 3. Solve

$$\beta=\frac{215}{21},\qquad\alpha=\frac{282}{21}$$$$21(\alpha+\beta)=497$$

Answer: option (3)

Example 40 (JEE Main 2026, 5 April Shift 1). In an equilateral triangle $PQR$, let the vertex $P$ be at $(3,5)$ and the side $QR$ be along the line $x+y=4$. If the orthocentre of the triangle is $(\alpha,\beta)$, then $9(\alpha+\beta)$ is equal to

▸ SHOW SOLUTION

Step 1. Foot of the altitude from $P$

$L(P)=3+5-4=4$ and $a^2+b^2=2$, so the foot is

$$M=(3-2,\ 5-2)=(1,3)$$

Step 2. Orthocentre equals centroid

In an equilateral triangle $H=G$, which divides $PM$ in the ratio $2:1$:

$$H=\frac{P+2M}{3}=\left(\frac53,\frac{11}{3}\right)$$

Step 3. Answer

$$9(\alpha+\beta)=9\cdot\frac{16}{3}=48$$

Answer: option (4)

Example 41 (JEE Main 2026, 28 Jan Shift 1). Let $ABC$ be an equilateral triangle with orthocentre at the origin and side $BC$ on the line $x+2\sqrt2\,y=4$. If the coordinates of the vertex $A$ are $(\alpha,\beta)$, then the greatest integer less than or equal to $\left|\alpha+\sqrt2\,\beta\right|$ is

▸ SHOW SOLUTION

Step 1. $A$ is on the altitude through $O$

The altitude is perpendicular to $BC$ and passes through $O$, so it is $y=2\sqrt2x$, and $\beta=2\sqrt2\alpha$.

Step 2. Use $AO:OD=2:1$

$OD=\frac{4}{\sqrt{1+8}}=\frac43$, so $AO=\frac83$, and $A$ is on the opposite side of $O$ from $BC$. With $OA$ along the unit vector $-\frac13(1,2\sqrt2)$,

$$A=\frac83\left(-\frac13,-\frac{2\sqrt2}{3}\right)$$$$A=\left(-\frac89,-\frac{16\sqrt2}{9}\right)$$

Step 3. Answer

$$\alpha+\sqrt2\beta=-\frac89-\frac{32}{9}=-\frac{40}{9}$$

and $\left\lfloor\frac{40}{9}\right\rfloor=4$.

Answer: option (4)

Example 42 (JEE Main 2026, 4 April Shift 2). Let $A$, $B$ be points on the two half-lines $x-\sqrt3\,\lvert y\rvert=\alpha$, $\alpha>0$, at a distance $\alpha$ from their point of intersection $P$. The line segment $AB$ meets the angle bisector of the given half-lines at the point $Q$. If $PQ=\dfrac92$ and $R$ is the radius of the circumcircle of $\triangle PAB$, then $\dfrac{\alpha^2}{R}$ is equal to ______.

▸ SHOW SOLUTION

Step 1. The two half-lines

They are $x-\sqrt3y=\alpha$ and $x+\sqrt3y=\alpha$ for $x\ge\alpha$, meeting at $P(\alpha,0)$ with inclinations $\pm30^\circ$. The angle between them is $60^\circ$, and the bisector is the $x$-axis.

Step 2. The triangle is equilateral

$PA=PB=\alpha$ with the angle $60^\circ$ between them, so $\triangle PAB$ is equilateral with side $\alpha$, and $Q$ is the mid-point of $AB$.

Step 3. Find $\alpha$ and $R$

$$PQ=\frac{\sqrt3}{2}\alpha=\frac92\Rightarrow\alpha=3\sqrt3$$$$R=\frac{\alpha}{\sqrt3}=3$$

Step 4. Answer

$$\frac{\alpha^2}{R}=\frac{27}{3}=9$$

Answer: $9$

Example 43 (JEE Main 2025, 7 April Shift 2). If the orthocentre of the triangle formed by the lines $y=x+1$, $y=4x-8$ and $y=mx+c$ is at $(3,-1)$, then $m-c$ is

▸ SHOW SOLUTION

Step 1. The vertex on the first two lines

$x+1=4x-8$ gives $P(3,4)$.

Step 2. The third side is horizontal

The altitude from $P$ passes through $P(3,4)$ and $H(3,-1)$, so it is the vertical line $x=3$. The third side is perpendicular to it: $m=0$, and the side is $y=c$.

Step 3. Another altitude

The third side meets $y=x+1$ at $Q(c-1,c)$. The altitude from $Q$ is perpendicular to $y=4x-8$, so its slope is $-\frac14$, and it passes through $H$:

$$\frac{-1-c}{3-(c-1)}=-\frac14\Rightarrow c=0$$

Answer: $m-c=0$, option (1)

Example 44 (JEE Main 2019, 9 Jan Shift 2). Let the equations of two sides of a triangle be $3x-2y+6=0$ and $4x+5y-20=0$. If the orthocentre of this triangle is at $(1,1)$, then the equation of its third side is

▸ SHOW SOLUTION

Step 1. The common vertex

$$A=\left(\frac{10}{23},\frac{84}{23}\right)$$

The slope of $AH$ is $\frac{84/23-1}{10/23-1}=-\frac{61}{13}$, so the third side has slope $\frac{13}{61}$.

Step 2. A second vertex

Let $B$ be on $3x-2y+6=0$. The altitude from $B$ is perpendicular to $4x+5y-20=0$, so it has slope $\frac54$ through $H$: $5x-4y-1=0$. With $3x-2y+6=0$ this gives

$$B=\left(-13,-\frac{33}{2}\right)$$

Step 3. The third side

Through $B$ with slope $\frac{13}{61}$:

$$61\left(y+\frac{33}{2}\right)=13(x+13)$$$$26x-122y-1675=0$$

Answer: option (1)

Example 45 (JEE Advanced 2021, Paper 1). Consider a triangle $\Delta$ whose two sides lie on the $x$-axis and the line $x+y+1=0$. If the orthocentre of $\Delta$ is $(1,1)$, then the equation of the circle passing through the vertices of the triangle $\Delta$ is

▸ SHOW SOLUTION

Step 1. The vertices

The two sides meet at $A(-1,0)$. Let $B(b,0)$ be on the $x$-axis and $C(t,-1-t)$ on $x+y+1=0$.

Step 2. Altitude from $C$

It is perpendicular to the $x$-axis and passes through $H(1,1)$, so $t=1$ and $C(1,-2)$.

Step 3. Altitude from $B$

It is perpendicular to $x+y+1=0$ and passes through $H$, so it is $y=x$, and $B=(0,0)$.

Step 4. The circle

Through the origin, $x^2+y^2+Dx+Ey=0$. Then $(-1,0)$ gives $D=1$ and $(1,-2)$ gives $E=3$.

Answer: option (B)

Example 46 (JEE Advanced 2026, Paper 1). Let $P$ be the point on the parabola $y=x^2$ such that the slope of the tangent to the parabola at $P$ is $4$. Let $Q$ be the point in the first quadrant lying on the circle $x^2+y^2=2$ such that the slope of the tangent to the circle at $Q$ is $-1$. Let $R$ be the point in the first quadrant lying on the ellipse $x^2+4y^2=8$ such that the slope of the tangent to the ellipse at $R$ is $-\dfrac12$. Then the radius of the circle passing through the points $P$, $Q$ and $R$ is

▸ SHOW SOLUTION

Step 1. The three points

On $y=x^2$, $2x=4$ gives $P(2,4)$. On the circle the tangent is perpendicular to the radius $OQ$, so $OQ$ has slope $1$ and $Q(1,1)$. On the ellipse $2x+8yy'=0$ gives $y'=-\frac{x}{4y}=-\frac12$, so $x=2y$ and $R(2,1)$.

Step 2. A right angle

$PR$ is vertical ($x=2$) and $QR$ is horizontal ($y=1$), so the angle at $R$ is $90^\circ$.

Step 3. Circumradius

The circumcentre is the mid-point of the hypotenuse $PQ=\sqrt{1+9}=\sqrt{10}$:

$$\text{radius}=\frac{\sqrt{10}}{2}=\sqrt{\frac52}$$

Answer: option (3)

Example 47 (IITianForum). The line $L_1:4x+3y-24=0$ is rotated clockwise through the angle $\tan^{-1}\dfrac{24}{7}$ about the point where it meets the $x$-axis, and the new line is $L_2$. If $(h,k)$ is the incentre of the triangle formed by $L_1$, $L_2$ and the $y$-axis, then $3h+k$ is equal to ______.

▸ SHOW SOLUTION

Step 1. The new slope

$L_1$ meets the $x$-axis at $(6,0)$. Its slope is $-\frac43$. Turning clockwise through $\beta$ with $\tan\beta=\frac{24}{7}$ gives the slope

$$\frac{-\frac43-\frac{24}{7}}{1-\frac43\cdot\frac{24}{7}}=\frac{-100}{-75}=\frac43$$

Step 2. The new line and the triangle

$L_2$: $y=\frac43(x-6)$, that is $4x-3y-24=0$. With the $y$-axis the vertices are $(6,0)$, $(0,8)$, $(0,-8)$.

Step 3. The incentre

The side opposite $(6,0)$ is $16$; the other two sides are $10$ each.

$$h=\frac{16\cdot6+10\cdot0+10\cdot0}{36}=\frac83$$

By symmetry about the $x$-axis, $k=0$.

Answer: $3h+k=8$

Family of lines and concurrency

If $L_1\equiv a_1x+b_1y+c_1=0$ and $L_2\equiv a_2x+b_2y+c_2=0$ intersect at $P$, then for every real $\lambda$

$$L_1+\lambda L_2=0$$

is a line through $P$. Reason. $P$ makes both $L_1$ and $L_2$ zero, so it makes $L_1+\lambda L_2$ zero, and the equation is still of the first degree.

Several lines of the family L1 plus lambda L2 passing through the common point P of L1 and L2

Concurrency of three lines

The lines $a_rx+b_ry+c_r=0$ ($r=1,2,3$) are concurrent when two of them meet and the third passes through that point. This happens when

$$\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}=0$$

and two of the lines are not parallel. Equivalently, there are constants $\lambda_1,\lambda_2,\lambda_3$, not all zero, with $\lambda_1L_1+\lambda_2L_2+\lambda_3L_3\equiv0$.

Examples

Example 48 (JEE Main 2019, 9 Jan Shift 1). Consider the set of all lines $px+qy+r=0$ such that $3p+2q+4r=0$. Which one of the following statements is true?

▸ SHOW SOLUTION

Step 1. Rewrite the condition

Divide by $4$:

$$p\cdot\frac34+q\cdot\frac12+r=0$$

Step 2. Read off the point

This says that $\left(\frac34,\frac12\right)$ satisfies $px+qy+r=0$ for every such line.

Answer: option (3)

Example 49 (JEE Main 2026, 22 Jan Shift 2). Among the statements (S1): If $A(5,-1)$ and $B(-2,3)$ are two vertices of a triangle whose orthocentre is $(0,0)$, then its third vertex is $(-4,-7)$; and (S2): If positive numbers $2a$, $b$, $c$ are three consecutive terms of an A.P., then the lines $ax+by+c=0$ are concurrent at $(2,-2)$,

▸ SHOW SOLUTION

Step 1. (S1): two altitudes through $O$

Let $C(h,k)$. $OA\perp BC$ and $OB\perp AC$:

$$5(h+2)-(k-3)=0$$$$-2(h-5)+3(k+1)=0$$

Step 2. Solve

The equations are $5h-k=-13$ and $2h-3k=13$, giving $h=-4$, $k=-7$. (S1) is correct.

Step 3. (S2): the AP condition

$2b=2a+c$ means $2a-2b+c=0$, which is the line $ax+by+c=0$ at the point $(2,-2)$. So every such line passes through $(2,-2)$, and (S2) is correct.

Answer: option (4)

Example 50 (IITianForum). For all values of $\alpha$, the lines $(3\cos\alpha+\sin\alpha)x+(2\cos\alpha-3\sin\alpha)y=7\cos\alpha-5\sin\alpha$ pass through one fixed point. The image of that point in the line $x+y=5$ is

▸ SHOW SOLUTION

Step 1. Separate $\cos\alpha$ and $\sin\alpha$

With $L_1=3x+2y-7$ and $L_2=x-3y+5$ the lines are

$$L_1\cos\alpha+L_2\sin\alpha=0$$

Step 2. The fixed point

It is the intersection of $3x+2y=7$ and $x-3y=-5$: put $x=3y-5$ in the first to get $y=2$, so the point is $(1,2)$.

Step 3. The image in $x+y=5$

The image of $(x_1,y_1)$ in $x+y=k$ is $(k-y_1,k-x_1)$:

$$(5-2,\ 5-1)=(3,4)$$

Answer: option (A)

Example 51 (IITianForum). If $3a^2+5ab-2ac-2b^2+3bc-c^2=0$, then the family of straight lines $ax+by+c=0$ may be concurrent at

▸ SHOW SOLUTION

Step 1. Factor the condition

The left side splits into two linear factors:

$$3a^2+5ab-2ac-2b^2+3bc-c^2$$$$=(a+2b-c)(3a-b+c)$$

Expanding the right side confirms all six terms.

Step 2. Read each factor as a point

$a+2b-c=0$ is $ax+by+c=0$ at $(-1,-2)$, and $3a-b+c=0$ is the same line at $(3,-1)$.

So each line of the family passes through $(-1,-2)$ or through $(3,-1)$.

Answer: (A) and (B)

Example 52 (JEE Main 2025, 24 Jan Shift 1). Let the lines $3x-4y-\alpha=0$, $8x-11y-33=0$ and $2x-3y+\lambda=0$ be concurrent. If the image of the point $(1,2)$ in the line $2x-3y+\lambda=0$ is $\left(\dfrac{57}{13},\dfrac{-40}{13}\right)$, then $\lvert\alpha\lambda\rvert$ is equal to

▸ SHOW SOLUTION

Step 1. The mirror passes through the mid-point

The mid-point of $(1,2)$ and its image is $\left(\frac{35}{13},-\frac{7}{13}\right)$, which lies on $2x-3y+\lambda=0$:

$$\frac{70+21}{13}+\lambda=0\Rightarrow\lambda=-7$$

Step 2. Concurrency

$$\begin{vmatrix}3&-4&-\alpha\\8&-11&-33\\2&-3&-7\end{vmatrix}=0$$$$3(77-99)+4(-56+66)+2\alpha=0$$$$-26+2\alpha=0\Rightarrow\alpha=13$$

The first two lines are not parallel, so the three lines do meet.

Answer: $\lvert\alpha\lambda\rvert=91$, option (2)

Example 53 (JEE Main 2025, 3 April Shift 2). Consider the lines $x(3\lambda+1)+y(7\lambda+2)=17\lambda+5$, $\lambda$ being a parameter, all passing through a point $P$. One of these lines (say $L$) is farthest from the origin. If the distance of $L$ from the point $(3,6)$ is $d$, then the value of $d^2$ is

▸ SHOW SOLUTION

Step 1. The fixed point

$$(x+2y-5)+\lambda(3x+7y-17)=0$$

Solving $x+2y=5$ and $3x+7y=17$ gives $P(1,2)$.

Step 2. The farthest line

For any line through $P$, the distance from $O$ is at most $OP$, with equality when the line is perpendicular to $OP$. So $L\perp OP$, and $L$ is $x+2y=5$.

Step 3. Distance from $(3,6)$

$$d=\frac{\lvert3+12-5\rvert}{\sqrt5}=\frac{10}{\sqrt5}\Rightarrow d^2=20$$

Answer: option (1)

Lines that form a figure

Many questions give some sides or diagonals of a parallelogram, rhombus or square and ask for the rest. The facts used are few.

A line through the centre of a parallelogram (or rectangle) always divides its area into two equal parts.

Examples

Example 54 (JEE Main 2026, 23 Jan Shift 2). Let $A(1,2)$ and $C(-3,-6)$ be two diagonally opposite vertices of a rhombus, whose sides $AD$ and $BC$ are parallel to the line $7x-y=14$. If $B(\alpha,\beta)$ and $D(\gamma,\delta)$ are the other two vertices, then $\lvert\alpha+\beta+\gamma+\delta\rvert$ is equal to

▸ SHOW SOLUTION

Step 1. The diagonals bisect each other

The mid-point of $AC$ is $(-1,-2)$, so $B+D=2(-1,-2)$:

$$\alpha+\gamma=-2,\qquad\beta+\delta=-4$$

Step 2. Answer

$$\lvert\alpha+\beta+\gamma+\delta\rvert=6$$

Step 3. The vertices themselves

The rhombus ABCD with A at 1, 2 and C at minus 3, minus 6, its sides AD and BC parallel and its diagonals crossing at minus 1, minus 2

$BD$ is perpendicular to $AC$ (slope $2$) through $(-1,-2)$. Meeting the line through $A$ of slope $7$ gives $D\left(\frac13,-\frac83\right)$, and then $B\left(-\frac73,-\frac43\right)$. The sum is $-6$, as found.

Answer: option (3)

Example 55 (JEE Main 2016). Two sides of a rhombus are along the lines $x-y+1=0$ and $7x-y-5=0$. If its diagonals intersect at $(-1,-2)$, then which one of the following is a vertex of this rhombus?

▸ SHOW SOLUTION

Step 1. The vertex on both sides

$x-y=-1$ and $7x-y=5$ give $A(1,2)$, so the opposite vertex is $C=2(-1,-2)-A=(-3,-6)$.

Step 2. Sides through $C$

They are parallel to the given sides: $x-y=3$ and $7x-y=-15$.

Step 3. The other two vertices

$7x-y=5$ with $x-y=3$ gives $\left(\frac13,-\frac83\right)$, and $x-y=-1$ with $7x-y=-15$ gives $\left(-\frac73,-\frac43\right)$.

Answer: option (2)

Example 56 (JEE Main 2025, 8 April Shift 2). Let $a$ be the length of a side of a square $OABC$ with $O$ being the origin. Its side $OA$ makes an acute angle $\alpha$ with the positive $x$-axis and the equations of its diagonals are $(\sqrt3+1)x+(\sqrt3-1)y=0$ and $(\sqrt3-1)x-(\sqrt3+1)y+8\sqrt3=0$. Then $a^2$ is equal to

▸ SHOW SOLUTION

Step 1. Direction of the diagonal $OB$

The first diagonal passes through $O$ with slope $-\frac{\sqrt3+1}{\sqrt3-1}=-(2+\sqrt3)=\tan105^\circ$.

Step 2. Direction of $OA$

$OA$ makes $45^\circ$ with $OB$ and is acute, so $\alpha=60^\circ$ and $A=\left(\frac a2,\frac{\sqrt3a}{2}\right)$.

Step 3. $A$ is on the other diagonal

$$\frac{(\sqrt3-1)a}{2}-\frac{(\sqrt3+1)\sqrt3a}{2}+8\sqrt3=0$$$$-2a+8\sqrt3=0\Rightarrow a=4\sqrt3$$

Answer: $a^2=48$, option (1)

Example 57 (JEE Main 2025, 2 April Shift 2). Let $A(4,-2)$, $B(1,1)$ and $C(9,-3)$ be the vertices of a triangle $ABC$. Then the maximum area of the parallelogram $AFDE$, formed with vertices $D$, $E$ and $F$ on the sides $BC$, $CA$ and $AB$ of the triangle $ABC$ respectively, is ______.

▸ SHOW SOLUTION

Step 1. Area of the triangle

$$2\Delta=\lvert4\cdot4+1\cdot(-1)+9\cdot(-3)\rvert$$$$2\Delta=\lvert16-1-27\rvert=12\Rightarrow\Delta=6$$

Step 2. The two corner triangles

Let $BD:BC=t$. Since $DF\parallel CA$ and $DE\parallel AB$, the triangles $BFD$ and $DEC$ are similar to $BAC$ with ratios $t$ and $1-t$:

$$[AFDE]=\Delta\left(1-t^2-(1-t)^2\right)$$$$[AFDE]=2\Delta\,t(1-t)$$

Step 3. Maximise

$t(1-t)\le\frac14$, with equality at $t=\frac12$, so the maximum is $\frac{\Delta}{2}=3$.

Answer: $3$

Example 58 (IITianForum). $P$ is the point of intersection of the lines $3x-y=0$ and $y=0$, and $C$ is the point $(4,3)$. A variable line through $C$ meets the ray $y=3x$, $x>0$, at $A$ and the positive $x$-axis at $B$. The least possible area of the triangle $PAB$ is ______.

▸ SHOW SOLUTION

Step 1. Coordinates in terms of $B$

Let $B(b,0)$. The line $BC$ is $y=\frac{3(x-b)}{4-b}$. It meets $y=3x$ where

$$x=\frac{b}{b-3},\qquad y=\frac{3b}{b-3}$$

and $x>0$ needs $b>3$.

Step 2. The area

$$[PAB]=\frac12\cdot b\cdot\frac{3b}{b-3}=\frac{3b^2}{2(b-3)}$$

Step 3. Minimise with $t=b-3>0$

$$\frac{3(t+3)^2}{2t}=\frac32\left(t+6+\frac9t\right)$$$$\frac32\left(t+6+\frac9t\right)\ge\frac32(6+6)=18$$

by $t+\frac9t\ge6$, with equality at $t=3$. Then $b=6$, $A(2,6)$, and $C$ is the mid-point of $AB$.

Answer: $18$

Change of origin and rotation of axes

Shift of origin

If the origin is moved to $(h,k)$ with the axes kept parallel, a point with old coordinates $(x,y)$ has new coordinates $(X,Y)$ with

$$x=X+h,\qquad y=Y+k$$

The new equation of a curve comes from writing $X+h$ for $x$ and $Y+k$ for $y$. Distances, slopes and angles do not change.

Rotation of axes

If the axes are turned anticlockwise through $\theta$ about the origin, the old and new coordinates are related by

$$x=x'\cos\theta-y'\sin\theta$$ $$y=x'\sin\theta+y'\cos\theta$$

and, solving back,

$$x'=x\cos\theta+y\sin\theta$$ $$y'=-x\sin\theta+y\cos\theta$$ The old axes x and y and the new axes x prime and y prime turned through theta, with one point P and its two pairs of coordinates

For a shift to $(h,k)$ followed by a rotation through $\theta$, replace $x$, $y$ in the last two lines by $x-h$, $y-k$.

Rotating a point instead of the axes

Turning a point $P(x,y)$ anticlockwise through $\theta$ about the origin gives

$$P'=(x\cos\theta-y\sin\theta,\ x\sin\theta+y\cos\theta)$$

Examples

Example 59 (JEE Main 2021, 27 July Shift 2). The point $P(a,b)$ undergoes the following three transformations successively: reflection about the line $y=x$; translation through $2$ units along the positive direction of the $x$-axis; rotation through angle $\dfrac{\pi}{4}$ about the origin in the anticlockwise direction. If the coordinates of the final position of the point $P$ are $\left(-\dfrac{1}{\sqrt2},\dfrac{7}{\sqrt2}\right)$, then the value of $2a+b$ is equal to

▸ SHOW SOLUTION

Step 1. Reflection, then translation

Reflection in $y=x$ gives $(b,a)$; the translation gives $(b+2,a)$.

Step 2. Rotation through $\frac{\pi}{4}$

$$\frac{b+2-a}{\sqrt2}=-\frac{1}{\sqrt2}$$$$\frac{b+2+a}{\sqrt2}=\frac{7}{\sqrt2}$$

Step 3. Solve

$b-a=-3$ and $a+b=5$, so $a=4$, $b=1$.

Answer: $2a+b=9$, option (2)

Example 60 (JEE Main 2025, 8 April Shift 2). A line passing through the point $P(a,0)$ makes an acute angle $\alpha$ with the positive $x$-axis. Let this line be rotated about the point $P$ through an angle $\dfrac{\alpha}{2}$ in the clockwise direction. If in the new position, the slope of the line is $2-\sqrt3$ and its distance from the origin is $\dfrac{1}{\sqrt2}$, then the value of $3a^2\tan^2\alpha-2\sqrt3$ is

▸ SHOW SOLUTION

Step 1. The angle

After the clockwise turn the inclination is $\alpha-\frac{\alpha}{2}=\frac{\alpha}{2}$, and $\tan\frac{\alpha}{2}=2-\sqrt3=\tan15^\circ$. So $\alpha=30^\circ$.

Step 2. Distance from the origin

The new line passes through $P(a,0)$ at inclination $15^\circ$, so its distance from $O$ is $\lvert a\rvert\sin15^\circ$:

$$a^2\sin^215^\circ=\frac12$$

Step 3. Find $a^2$

$\sin^215^\circ=\frac{1-\cos30^\circ}{2}=\frac{2-\sqrt3}{4}$, so

$$a^2=\frac{2}{2-\sqrt3}=2(2+\sqrt3)$$

Step 4. Answer

$$3a^2\cdot\frac13-2\sqrt3=4+2\sqrt3-2\sqrt3=4$$

Answer: option (1)

Which method to use

What JEE asks

JEE Main asks about one straight-line question per shift, and many circle and conic questions use the same results. In 2026 the most frequent types were the centres of a triangle (Examples 38 to 42 and 49), angle bisectors and the angle bisector theorem (Examples 34 and 35), images and reflected rays (Examples 30 and 31), distances to parallel lines (Examples 23 and 24), the parametric form (Example 17), a locus through a family of lines (Example 8) and a rhombus from its diagonals (Example 54). The 2026 equilateral-triangle questions all used one fact: the orthocentre, centroid and circumcentre coincide and divide the altitude in the ratio $2:1$.

JEE Advanced asks fewer but deeper questions: the orthocentre locus as a straight line (Example 11), regions defined by distances from two lines (Example 27), conditions from a distance inequality (Example 26), the orthocentre used to recover a triangle and its circumcircle (Example 45), and a weighted point written by the section formula (Example 3, 2025). Statement questions on the angle bisector theorem (Example 33) and collinearity hidden in trigonometric coordinates (Example 4) appeared in earlier years.

Quick recall

Practise these methods on real exam questions in the doMath Straight Lines chapter. The same distance and image formulas run through the Family of Circles note, the vector form of these results is in the Complex Numbers in Coordinate Geometry note, and the trigonometric side of the centres of a triangle is in the Solution of Triangles note.