Question

An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from `(7 ,1 )` to `(8 ,5 )` and the triangle's area is `27`, what are the possible coordinates of the triangle's third corner?

Answer 1

`(6.55882, 3.23529) and (8.44118, 2.76471)`

Explanation:

Let's begin by finding the length of side "a":

`a = sqrt((8-7)^2 + (5 - 1)^2)`

`a = sqrt(1^2 + 4^2)`

`a = sqrt(17)`

Use side "a" as the base of the triangle, find the height:

`Area = 1/2base xx height`

`2 = 1/2sqrt(17) xx height`

`height = (4sqrt(17))/17`

The height must bisect side "a" at the point `(7.5, 3)`

We can used the distance formula to write one equation for the two possible points:

`(4sqrt(17))/17 = sqrt((x - 7.5)^2 + (y - 3)^2)`

Square both sides:

`16/17 = (x - 7.5)^2 + (y - 3)^2" [1]"`

We need one more equation. Find the slope of side "a":

`m = (5 - 1)/(8 - 7) = 4`

Because the height is perpendicular to side "a", its slope is: `-1/4`

We can use the slope and the point to find the equation of line for the height:

`3 = -1/4(7.5) + b`

`b = 39/8`

The equation for the height is:

`y = -1/4x + 39/8" [2]"`

Solving equations [1] and [2] is very long and tedious process. I gave them to WolframAlpha. Here are the points:

`(6.55882, 3.23529) and (8.44118, 2.76471)`