Question

Two corners of an isosceles triangle are at `(8 ,3 )` and `(5 ,4 )`. If the triangle's area is `4`, what are the lengths of the triangle's sides?

Answer 1

The length of the sides are `sqrt 10, sqrt 10, sqrt 8` and the points are `(8,3), (5,4) and (6,1)`

Explanation:

Let the points of the triangle be `(x_1,y_1),( x_2, y_2),( x_3, y_3).`

Area of triangle is A = `((x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2))/2)`

Given `A = 4, (x_1,y_1) = (8,3), (x_2, y_2) = (5,4)`

Substituting we have the below Area equation:

`((8(4 – y_3) + 5(y_3 – 3) + x_3(3 – 4))/2 )= 4`

`((8(4 – y_3) + 5(y_3 – 3) + x_3(3 – 4)) = 8`

`(32 – 8y_3) + (5y_3 – 15) + (-1x_3) = 8`

`17 – 3y_3 -x_3 = 8`

`– 3y_3 -x_3 = (8-17)`

`– 3y_3 -x_3 = -9`

`3y_3 + x_3 = 9` ----> Equation 1

Distance between points `(8,3), (5,4)` using distance formula is

`sqrt ( (8-5)^2+(3-4)^2)` = `sqrt ( 3^2+(-1)^2)` = `sqrt 10`

Distance between points `(x_3,y_3), (5,4)` using distance formula is
`sqrt ((x_3 -5)^2 + (y_3 - 4)^2)` = `sqrt 10`

Squaring both sides and subsituting `x_3 = 9 - 3y_3` from equation 1, we get a quadratic equation.

`(9-3y_3 - 5)^2 + (y_3 - 4 )^ 2 = 0`

`(4-3y_3)^2 + (y_3 - 4 )^ 2 = 0`

Factorizing this, we get `(y-1)(10y-22) = 0`

y = 1 or y = 2.2 . y=2.2 can be discarded. Hence, the third point has to be (6,1).

By calculating the distances for points `(8,3), (5,4) and (6,1)`, we get `sqrt 8` for the length of the base.