Line Through A(4,3), Slope>1, Meets x−y−2=0 at B — Find B’s Line

JEE Main 2022 — 25 July, Morning Shift. Previous Year Question.

Problem

A line, with slope greater than one, passes through the point $A(4,3)$ and intersects the line $x-y-2=0$ at the point $B$. If the length of the line segment $AB$ is $\dfrac{\sqrt{29}}{3}$, then $B$ also lies on the line:

Key insight. Rather than guessing the unknown line’s equation, it’s more direct to describe any point on a line through $A$ using the parametric (“distance”) form — a single parameter $r$ representing signed distance from $A$. Plugging that parametric point into the given line $x-y-2=0$ turns the whole problem into solving for the slope, and then locating $B$ exactly.

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Approach

Write any point on the line through $A(4,3)$ at signed distance $r$ as $(4+r\cos\theta,\ 3+r\sin\theta)$, where $\tan\theta$ is the unknown slope. Substitute this into $x-y-2=0$ to get $r$ in terms of $\theta$, then use the given length $AB$ to solve for the slope. Once the slope is known, find $B$’s actual coordinates and check which of the four given lines it satisfies.

Solution

Step 1 — Set up the parametric point on the line

Let the slope be $\tan\theta = t$, with $t>1$. A general point on the line through $A(4,3)$ at signed distance $r$ is:

$$(x,y) = \left(4+\frac{r}{\sqrt{1+t^2}},\ 3+\frac{rt}{\sqrt{1+t^2}}\right)$$

using $\cos\theta = \dfrac{1}{\sqrt{1+t^2}}$, $\sin\theta = \dfrac{t}{\sqrt{1+t^2}}$.

Step 2 — Substitute into the given line

Plugging into $x-y-2=0$:

$$\left(4+\frac{r}{\sqrt{1+t^2}}\right) – \left(3+\frac{rt}{\sqrt{1+t^2}}\right) – 2 = 0$$

$$-1 + \frac{r(1-t)}{\sqrt{1+t^2}} = 0 \implies r = \frac{\sqrt{1+t^2}}{1-t}$$

Step 3 — Use the given length to solve for t

Since $AB = |r| = \dfrac{\sqrt{29}}{3}$, squaring both sides:

$$\frac{1+t^2}{(1-t)^2} = \frac{29}{9}$$

Cross-multiplying:

$$9(1+t^2) = 29(1-t)^2 \implies 9+9t^2 = 29 – 58t + 29t^2$$

$$20t^2 – 58t + 20 = 0 \implies 10t^2 – 29t + 10 = 0$$

Step 4 — Solve the quadratic and pick the valid root

$$t = \frac{29 \pm \sqrt{841-400}}{20} = \frac{29\pm21}{20} = \frac{5}{2} \text{ or } \frac{2}{5}$$

Since the slope must be greater than $1$: $t = \dfrac{5}{2}$.

Step 5 — Find the coordinates of B

With $t=\dfrac{5}{2}$: $\sqrt{1+t^2} = \sqrt{1+\dfrac{25}{4}} = \dfrac{\sqrt{29}}{2}$, and:

$$r = \frac{\sqrt{1+t^2}}{1-t} = \frac{\frac{\sqrt{29}}{2}}{1-\frac{5}{2}} = \frac{\frac{\sqrt{29}}{2}}{-\frac{3}{2}} = -\frac{\sqrt{29}}{3}$$

So:

$$x_B = 4 + \frac{r}{\sqrt{1+t^2}} = 4 + \frac{-\frac{\sqrt{29}}{3}}{\frac{\sqrt{29}}{2}} = 4 – \frac{2}{3} = \frac{10}{3}$$

$$y_B = 3 + \frac{rt}{\sqrt{1+t^2}} = 3 + \left(-\frac{2}{3}\right)\cdot\frac{5}{2} = 3 – \frac{5}{3} = \frac{4}{3}$$

So $B = \left(\dfrac{10}{3},\dfrac{4}{3}\right)$.

Step 6 — Check which given line B satisfies

$$x_B + 2y_B = \frac{10}{3}+2\times\frac{4}{3} = \frac{10}{3}+\frac{8}{3} = \frac{18}{3} = 6$$

This matches $x+2y=6$ exactly.

Answer

Common mistakes

• Picking $t=\dfrac{2}{5}$ instead of $t=\dfrac{5}{2}$. The quadratic naturally produces both roots; only the one satisfying “slope greater than one” is valid here.
• Getting the sign of $r$ wrong. Since $1-t$ is negative when $t>1$, $r$ comes out negative — meaning $B$ lies in the opposite direction from what a naive positive-$r$ assumption would suggest. Carrying the sign through correctly is what gives the right coordinates for $B$.

Practise next

• A line through $P(2,1)$ with slope greater than $2$ meets $2x-y-1=0$ at $Q$, with $PQ=5$. Find the coordinates of $Q$, using the same parametric-form method.