JEE Main 2019 — Straight Lines Question with Solution

The $x$ intercept of the line $3x+4y-24=0$ can be obtained by putting $y=0$ i.e. $3x-24=0$

$\Rightarrow x=8$

Hence, the point $A$ is $\left(8, 0\right).$

Now, the $y$ intercept of the line $3x+4y-24=0$ can be obtained by putting $x=0$ i.e. $4y-24=0$

$\Rightarrow y=6$

Hence, the point $B$ is $\left(0, 6\right).$

In $∆OAB,$ we have $OA=b=8, OB=a=6$ and $AB=c=\sqrt{{\left(8-0\right)}^2+{\left(0-6\right)}^2}=\sqrt{64+36}=10$ units.

And, we know that the co-ordinates of incentre $I$ of a triangle with vertices $\left(x_1, y_1\right), \left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ and having length of the sides opposite to these vertices as $a, b$ and $c,$ respectively,  is

$I\equiv \left(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c}, \frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}\right)$

$\Rightarrow I\equiv \left(\frac{6\left(8\right)+8\left(0\right)+10\left(0\right)}{24}, \frac{6\left(0\right)+8\left(6\right)+10\left(0\right)}{24}\right)$

$\Rightarrow I=\left(2, 2\right).$