Question

The centre of the incircle of the triangle formed by the line 3x+4y=243x+4y=24 with the axes is

• A. $(3,3)$
• B. $(2,2)$
• C. $(-2,2)$
• D. $(2,-2)$

Answer 1

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

Since the equation of line is given as

$3x+4y=24\Rightarrow \frac{3x}{24}+\frac{4y}{24}=1\Rightarrow \frac{x}{8}+\frac{y}{6}=1⟶\left(1\right)$

since the line 3x+4y=24 meets on x-axis at A and on y-axis at B such that OA=8 and OB=6

$\therefore AB=\sqrt{{OA}^2+{OB}^2}=\sqrt{64+36}=10areaof\Delta AOB=\frac{1}{2}.b.h=\frac{1}{2}.8.6=24squnit$

Let the coordinate of centre of incircle be (r,r)

$also,s=\frac{a+b+c}{2}=\frac{8+6+10}{2}=12Now,r=\frac{ar\Delta }{s}=\frac{24}{12}=2$

therefore coordinate of incircle is (2,2)