Question

If the line $3x+4y-24=0$ meets the coordinates axes at A and B, then incentre of the $\Delta OAB$ is (where O is origin)

• A. $(1,2)$
• B. $(2,2)$
• C. $(12,12)$
• D. $(2,12)$

Answer 1

The correct option is B $(2,2)$

Given the line $3x+4y-24=0$

Put $x=0\Rightarrow y=6$

and when $y=0x=8$

$\therefore$ the points are (0,6) and (8,0)

Let o be the origin An in centre is the point where the triangle three angles bisection intersect.The coordinates of the in center o are

$OX=\frac{a{A}_x+b{B}_x+c{C}_x}{p}$

$OY=\frac{a{A}_y+b{B}_y+c{C}_y}{p}$

where $A_x$ and$A_y$ are the $xandy$ coordinates of the point A a,b,c the side lengths opposite vertex A,B and C

p is the perimeter of the triangle

$\therefore OX=\frac{0\times 10+8\times 6+0\times 8}{24}=2$

$OY=\frac{0\times 10+0\times 6+6\times 8}{24}=2$

$\therefore$ incentre $=(2,2)$