JEE Main 2024 — Straight Lines Question with Solution

Given: Line $AC: 2x-y=2$ and $BD: x=y$.

Line $AC \text{and} BD$ intersects at point $D$.

$\Rightarrow 2x-x=2$

$\Rightarrow x=y=2$

$\Rightarrow D\equiv \left(2,2\right)$

We know that, by angle bisector theorem, the angle bisector of a triangle divides the base in the same ratio as the other two sides.

$\Rightarrow \frac{CD}{AD}=\frac{BC}{AB}$

It is given that, $2AB=BC$.

$\Rightarrow \frac{CD}{AD}=\frac{2AB}{AB}$

$\Rightarrow CD:AD=2:1$

Let the coordinates of $C$ be $\left(h,k\right)$.

By using section formula,

$D\equiv \left(\frac{h\times 1+2\times 4}{3},\frac{1\times k+2\times 6}{3}\right)=\left(2,2\right)$

$\Rightarrow h+8=6, k+12=6$

$\Rightarrow h=-2, k=-6$

$\Rightarrow C\equiv \left(-2,-6\right)$

Now, $2AB=BC$

$\Rightarrow 2\sqrt{{\left(4-\alpha \right)}^2+{\left(6-\beta \right)}^2}=\sqrt{{\left(\alpha +2\right)}^2+{\left(\beta +6\right)}^2}$

Also, $B$ lines on $x=y$.

$\Rightarrow 4\left[{\left(4-\alpha \right)}^2+{\left(6-\alpha \right)}^2\right]={\left(\alpha +2\right)}^2+{\left(\alpha +6\right)}^2$

$\Rightarrow 4\left(16+{\alpha }^2-8\alpha +36+{\alpha }^2-12\alpha \right)={\alpha }^2+4+4\alpha +{\alpha }^2+36+12\alpha$

$\Rightarrow 8{\alpha }^2-80\alpha +208=2{\alpha }^2+16\alpha +40$

$\Rightarrow 6{\alpha }^2-96\alpha +168=0$

$\Rightarrow {\alpha }^2-16\alpha +28=0$

$\Rightarrow \left(\alpha -14\right)\left(\alpha -2\right)=0$

$\Rightarrow \alpha =2,14$

But, coordinates of $B \text{and} D$ cannot be same.

$\Rightarrow \alpha =\beta =14$

$\Rightarrow \alpha +2\beta =42$