If the line 3x + 4y - 24 = 0 intersects the x -axis is at the point...

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

$⇒ x = 8$

Hence, the point $A$ is \left(8, 0\right) .$</p><p>Now, the$y$intercept of the line$3x + 4y - 24 = 0$can be obtained by putting$x = 0$i.e.$4y - 24 = 0$</p><p>$⇒ y = 6$</p><p>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

$I ≡ \left(\dfrac{a{x}_{1} + b{x}_{2} + c{x}_{3}}{a + b + c}, \dfrac{a{y}_{1} + b{y}_{2} + c{y}_{3}}{a + b + c}\right)$

$⇒ I ≡ \left(\dfrac{6\left(8\right) + 8\left(0\right) + 10\left(0\right)}{24}, \dfrac{6\left(0\right) + 8\left(6\right) + 10\left(0\right)}{24}\right)$

$⇒ I = \left(2, 2\right) .$