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 ,5 )` to `(8 ,1 )` and the triangle's area is `15`, what are the possible coordinates of the triangle's third corner?

Answer 1

`(15/34, 42/34)` and `(495/34, 162/34)`

Explanation:

We can find the length of 'a' by finding the distance between the two points:

`a = sqrt((7 - 8)² + (5 - 1)²)`

`a = sqrt((-1)² + (4)²)`

`a = sqrt(1 + 16)`

`a = sqrt(17)`

Let side 'a' be the base of the triangle.

Using the area, we can compute the height:

`A = (1/2)bh = (1/2)ah`

`15 = (1/2)(sqrt17)h`

`h = 30sqrt17/17`

The height must lie on the line that is the perpendicular bisector of side 'a'. Let's find the equation of that line:

Side 'a' goes from left to right 1 unit and down 4 units (For later use, remember this is slope, -4), therefore, the midpoint goes from left to right `1/2` unit and down 2 units.

The midpoint is `(15/2, 3)`

A perpendicular line will have a slope that is the negative reciprocal of -4:

`-1/-4 = 1/4`

Using the point-slope form of the equation of a line, `y-y_1 = m(x - x_1)`, we write an equation of a line upon which both possible vertices must lie:

`y - 3 = 1/4(x - 15/2)`

`y = 1/4x + 9/8`

Using the equation for a circle we write an equation where the radius is `r = h = 30sqrt17/17` and the center is the midpoint `(15/2, 3)`:

`(30sqrt17/17)² = (x - 15/2)² + (y - 3)²`

`900/17 = x² - 15x + 225/4 + y² -6y + 9`

Substitute `1/4x + 9/8` for y:

`900/17 = x² - 15x + 225/4 + (1/4x + 9/8)² -6(1/4x + 9/8) + 9`

I used Wolframalpha to solve this:

`x = 15/34` and `x = 495/34`

The corresponding y coordinates are:

`y = 42/34` and `y = 162/34`