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?

1 Answer
Oct 5, 2016

(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