An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(4 ,9 )# to #(1 ,5 )# and the triangle's area is #32 #, what are the possible coordinates of the triangle's third corner?

1 Answer
Feb 4, 2018

Coordinates of vertex A #color(brown)(12.74, -0.68)#

Explanation:

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Given : B (4,9), C (1,5), #A_t = 32#, AB = AC, #AN (h) _|_# bisector of BC

To find A coordinates.

Coordinates of N (4+1)/2, (9+5)/2 = N (5/2, 7)

#BC = a = sqrt((1-4)^2 + (5-9)^2) = 5#

#A_t = 32 = (1/2) a * h = (1/2) * 5 * h#

#h = 12.8#

Slope of BC #m_a = (5-9) / (1 - 4) = 4/3#

Slope of #AN _|_ BC = m_h = -1 / m_a = -3/4#

Equation of AN

#y - 7 = -(3/4) (x - (5/2))#

#y - 7 = -(3/8)(2x - 5)#

#8y - 56 = -6x + 15#

#8y + 6x = 71# Eqn (1)

AN = h = 12.8 = sqrt((x-(5/2))^2 + (y-7)^2)#

#(y-7)^2 + (x-(5/2))^2 = 12.8^2# Eqn (2)

Solving equations (1), (2), we get the coordinates of vertex A

#A (12.74, -0.68)#

Verification :

#AB = c = sqrt((4-12.74)^2 + (9+0.68)^2) = 13.0419#

#AC = b = sqrt((1-12.74)^2 + (5+0.68)^2) = 13.0419#

#b = c = 13.0419# since it is an isosceles triangle