Question

The area of triangle is 5. two of its vertices are (2,1) and (3,-2). find the third vertex if it lies on the line y=x+3.

Answer 1

The expression for area formed by (x1,y1),(x2,y2) and (x3,y3) is

(1/2)*(x1(y2-y3)+x2(y3-y1)+x3(y1-y2))

In our case , the third vertex is on line y=x+3, meaning it is (x,x+3). The area is given as 5.

Hence 5 = (1/2)*(2*(-2-(x+3))+3*((x+3)-1)+x(1+2)))

5=(1/2)*(2*(-2-x-3)+3(x+2)+x*3))=(1/2)*(2*(-x-5)+3x+6+3x)=(1/2)*(-2x-10+3x+6+3x)

5=(1/2)*(4x-4)=2x-2 i.e. x=7/2=3.5

hence y= 6.5

The third vertex is (3.5,6.5)