Question

The sides of a $△ABC$ lie on the lines $3x+4y=0,4x+3y=0$ and $x=3$. Let $(h,k)$ be the centre of the circle inscribed in $\Delta ABC$. The value of $(h+k)$ equals:

• A. $0$
• B. $\frac{1}{4}$
• C. $\frac{-1}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $0$

Let AB, BC and CA be $3x+4y=0$, $4x+3y=0$ and $x=3$ respectively.

Since AB and BC do not have any constant terms, $B=(0,0)=(x_2,y_2)$

Solving AB and CA gives $A=(3,-\frac{9}{4})=(x_1,y_1)$

Solving BC and CA gives $C=(3,-4)=(x_3,y_3)$

$a=BC=\sqrt{(3-0{)}^2+(0+4{)}^2}=\sqrt{3^2+4^2}=\sqrt{4+16}=5$

$b=CA=\sqrt{(3-3{)}^2+{(4-\frac{9}{4})}^2}=\sqrt{{\left(\frac{16-9}{4}\right)}^2}=\frac{7}{4}$

$c=AB=\sqrt{(3-0{)}^2+{(0+\frac{9}{4})}^2}=\sqrt{3^2+\frac{81}{16}}=\sqrt{\frac{164+81}{16}}=\frac{15}{4}$

In centre is given by $(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})$

$=(\frac{5\times 3+0+\frac{15\times 3}{4}}{5+\frac{7+15}{4}},\frac{5(-\frac{9}{4})+0+\frac{15}{4}(-4)}{5+\frac{7+15}{4}})$

$=(\frac{15+\frac{45}{4}}{5+\frac{22}{4}},\frac{-\frac{45}{4}-15}{5+\frac{22}{4}})=(h,k)$

$h+k=\frac{15+\frac{45}{4}-\frac{45}{4}-15}{5+\frac{22}{4}}=0$.