Question

Find the incentre of the triangle given by the points (-36,7),(20,7)and(0,-8)

• A. (-1,0)
• B. (1,0)
• C. (0,-1)
• D. (-1,-1)

Answer 1

Answer:

A. (-1,0)

$a=BC=\sqrt{(20-0)^2+(7+8)^2}$

$=\sqrt{400+225}=\sqrt{625}=25$

$b=AC=\sqrt{(-36-0)^2+(7+8)^2}$

$=\sqrt{1296+225}=\sqrt{1521}=39$

$c=AB=\sqrt{(-36-20)^2+(7-7)^2}$

$=56$

$x_1=-36,y_1=7,x_2=20,y_2=7x_3=0,y_3=-8,a=25,b=39,c=56$

$x=\frac{ax_1+bx_2+cx_3}{a+b+c}$

$x=\frac{25\times-36+39\times20+56\times0}{25+39+56}=\frac{-900+780+0}{120}=\frac{-120}{120}=-1$

$x=\frac{ay_1+by_2+cy_3}{a+b+c}$

$y=\frac{25\times7+39\times7+56\times-8}{25+39+56}=\frac{125+273-448}{120}=\frac{0}{120}=0$

$(x,y)=(-1,0)$