Question

The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, –2). If the third vertex is `(7/2, y)`, find the value of y.

Answer 1

Let `A(x_1,y_1) = A(2,1), B(x_2, y_2) = B(3,-2)` and `C(x_3,y_3) = C(7/2,y)`

Now

Area(ΔABC) =`1/2|x_1(y_2 - y_3) + x_2(y_3-y_1) + x_3(y_1 - y_2)|`

`=>5 = 1/2|2(-2-y) + 3(y - 1) + 7/2(1+2)|`

`=> 10 = |-4-2y+3y - 3 + 21/2|`

`=>10 = |y + 7/2|`

`=> 10 = y + 7/2` or `-10 = y + 7/2`

`=> y = 13/2` or `y = (-27)/2`

Hence `y = 13/2 or (-27)/2`