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Question
The points A (3, 0), B (a, -2) and C (4, -1) are the vertices of triangle ABC right angled at vertex A. Find the value of a.
Solution
AB = `sqrt(("a" - 3)^2 + (-2 - 0)^2)`
= `sqrt("a"^2 + 9 - 6"a" + 4)`
= `sqrt("a"^2 - 6"a" + 13)`
BC = `sqrt((4 - "a")^2 + (-1+2)^2)`
= `sqrt("a"^2 + 16 - 8"a" + 1)`
= `sqrt("a"^2 - 8"a" + 17)`
CA = `sqrt((3 - 4)^2 + (0 + 1)^2)`
= `sqrt(1+1)`
= `sqrt(2)`
Since, triangle ABC is a right-angled at A, we have:
AB2 + AC2 = BC2
⇒ a2 - 6a + 13 + 2 = a2 - 8a + 17
⇒ 2a = 2
⇒ a = 1
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