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.

Answer 1

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