

InterviewSolution
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Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3. |
Answer» Let the third vertex be C (x, y) And, given A (2, 0) & B (2, 5) We have, Length of AB = √[(2 – 2)2 + (5 – 0)2] = √[(0)2 + (5)2] = √[0 + 25] = 5 units Length of BC = √[(x – 2)2 + (y – 5)2] = √[x2 – 4x + 4 + y2 – 10y + 25] = √[ x2 – 4x + y2 – 10y + 29] units Length of AC = √[(x – 2)2 + (y – 0)2] = √[x2 – 4x + 4 + y2] units Given that, AC = BC = 3 So, AC2 = BC2 = 9 x2 – 4x + 4 + y2 = x2 – 4x + y2 – 10y + 29 10y = 25 y = 25/10 = 2.5 And, AC2 = 9 x2 – 4x + 4 + y2 = 9 x2 – 4x + 4 + (2.5)2 = 9 x2 – 4x + 4 + 6.25 = 9 x2 – 4x + 1.25 = 0 D = (-4)2 – 4 x 1 x 1.25 = 16 – 5 = 11 So, the roots are x = -(-4) + √11/ 2 = (4 + 3.31)/ 2 = 3.65 And, x = -(-4) – √11/ 2 = (4 – 3.31)/ 2 = 0.35 Therefore, the third vertex can be C (3.65, 2.5) or (0.35, 2.5). |
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