# The incentre of the triangle formed by the lines y=|x| and y=1 is A) (0, 2 - √2) B) (2 - √2 , 0) C) (2+ √2 , 0) D) (1,  1/√3)

embizze | High School Teacher | (Level 2) Educator Emeritus

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Given a triangle with vertices `(x_a,y_a),(x_b,y_b),(x_c,y_c)` and the sides opposite those vertices as a,b,c; the coordinates for the incenter are given by:

`((ax_a+bx_b+cx_c)/p,(ay_a+by_b+cy_c)/p)` where p is the perimeter of the triangle.

The triangle given is formed by y=|x| and y=1. The verices are A=(-1,1), B=(1,1) and C=(0,0) with sides `a=b=sqrt(2),c=2` ; the perimeter is `2+2sqrt(2)` .

The coordinates of the incenter are:

`((sqrt(2)(-1)+sqrt(2)(1)+2(0))/(2+sqrt(2)),(sqrt(2)(1)+sqrt(2)(1)+2(0))/(2+sqrt(2)))` or

`(0,(2sqrt(2))/(2+sqrt(2))=(0,2-sqrt(2))` so the answer is A.

The graph:

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Another way is to know that the area of a triangle K can be found by K=rs where r is the radius of the incircle and s is the semiperimeter. The perimeter is `p=2+sqrt(2)==>s=1+sqrt(2)` . The area can be found easily; taking the line y=1 as the base we have a base of 2 and height 1 so the area is K=1. Then `1=r(1+sqrt(2))==>r=1/(1+sqrt(2))=sqrt(2)-1`

The radius of the incircle will be tangent to the triangle at (0,1); the center of the incircle is on the y-axis (the triangle is isosceles) and the y-coordinate will be `1-(sqrt(2)-1)=2-sqrt(2)` as above.