JEE Main 2019 — Mathematics Coordinate Geometry

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

$$\Rightarrow x=8$$⇒x=8

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

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

$$\Rightarrow y=6$$⇒y=6

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

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

And, we know that the co-ordinates of incentre $I$I of a triangle with vertices $({x}_{1},{y}_{1}),({x}_{2},{y}_{2})$(x1​,y1​),(x2​,y2​) and $({x}_{3},{y}_{3})$(x3​,y3​) and having length of the sides opposite to these vertices as $a,b$a,b and $c,$c, respectively, is

$$I\equiv (\frac{a{x}_{1}+b{x}_{2}+c{x}_{3}}{a+b+c},\frac{a{y}_{1}+b{y}_{2}+c{y}_{3}}{a+b+c})$$I≡(a+b+cax1​+bx2​+cx3​​,a+b+cay1​+by2​+cy3​​)

$$\Rightarrow I\equiv (\frac{6(8)+8(0)+10(0)}{24},\frac{6(0)+8(6)+10(0)}{24})$$⇒I≡(246(8)+8(0)+10(0)​,246(0)+8(6)+10(0)​)

$$\Rightarrow I=(2, 2).$$⇒I=(2,2).