An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (2 ,5 ) to (8 ,1 ) and the triangle's area is 15 , what are the possible coordinates of the triangle's third corner?

1 Answer
Dec 19, 2017

Coordinates of the third vertex is (187/9), (80/3)

Explanation:

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Area = 15

a = sqrt ( (8-2)^2 + (1-5)^2) = sqrt 52 = 2sqrt 13

Area = (1/2) a * h

15 = (1/2) * 2 sqrt 13 * h

h = 15/ ( sqrt 13)

tan theta = h / (a/2) = (15/sqrt13) / ((2 sqrt 13)/2) = 15/13

Slope of line AB is tan theta = 15/13
Equation of line AB is given by
y - 5 = (15/13) * ( x - 2)
13y - 65 = 15x - 30
15x - 13y = —35. Eqn (1)

Slope of base m = (1-5)/(8-2) = -2/3
Slope of altitude m_1 = -(1/m) = 3/2

Midpoint of base a is (8+2)/2, (1+5)/2= (5, 3)

Equation of altitude and passing through point A and having slope 3/2 is
y - 3 = (3/2) * (x - 5)
2y - 6 = 3x - 15
3x - 2y = 9 Eqn (2)

Solving Eqns (1) & (2), we get coordinates of point A.

x = (187/9), y =( 80/3)

Coordinates of the third corner A is (187/9, 80/3)