if A(9,-9), (1,-3) are the vertices of right angled isosceles triangle, then find the third vertex
Answers
Correct Question :- if A(9,-9), C(1,-3) are the end points of hypotenuse of a right angled isosceles triangle, then the third vertex ?
Solution :-
Let us assume that, the co-ordinates of third vertex B are (x,y).
since it is given that, ∆ABC is a right angled isosceles triangle and AC is hypotenuse of the ∆ .
so,
- AB = BC .
- AB² + BC² = AC² .
- slope of AB * slope of BC = (-1) { AB ⟂ BC . }
then,
→ AB = BC
→ (9-x)² + (-9-y)² = (1 - x)² + (-3 - y)²
→ 81 + x² - 18x + 81 + y² + 18y = 1 + x² - 2x + 9 + y² + 6y
→ 162 - 18x + 18y = 1 - 2x + 9 + 6y
→ 162 - 10 = - 2x + 18x + 6y - 18y
→ 152 = 16x - 12y
→ 152 = 4(4x - 3y)
→ 38 = (4x - 3y)
→ 4x = 38 + 3y
→ x = (38 + 3y)/4 -------- Eqn.(1)
Now ,
→ Slope of AB * Slope of BC = (-1)
→ {(y + 9)/(x - 9)} * {(-3-y)/(1 - x)} = (-1)
→ (y + 9)(y + 3)(-1) = (-1)(x - 9)(1 - x)
→ (y + 9)(y + 3) = (x - 9)(1 - x)
Putting value of x from Eqn.(1) , we get,
→ (y² + 12y + 27) = [{(38 + 3y)/4 - 9} * {1 - (38 + 3y)/4}]
→ (y² + 12y + 27) = [{(38 + 3y - 36)}/4} * {4 - (38 + 3y)/4}]
→ (y² + 12y + 27) = [(3y + 2/4) * (-3y - 34)/4]
→ (y² + 12y + 27) = [(-9y² - 102y - 6y - 68)/16)
→ (16y² + 192y + 432 + 9y² + 108y + 68) = 0
→ 25y² + 300y + 500 = 0
→ 25(y² + 12y + 20) = 0
→ y² + 12y + 20 = 0
→ y² + 10y + 2y + 20 = 0
→ y(y + 10) + 2(y + 10) = 0
→ (y + 10)(y + 2) = 0
→ y = (-2) or, (-10) .
Putting value of y in Eqn.(1) we get,
→ If y = (-10)
→ x = (38 - 30)/4 = 8/4 = 2 .
and,
→ If y = (-2)
→ x = (38 - 6)/4 = 32/4 = 8 .
Hence, Coordinates of third vertex are (2, -10) or (8 , -2) .
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