The points A(0, 3), B(-2, a) and C(-1, 4) are the vertices of a Δ ABC right –angledat A. Find the value of a.

Given, the vertices of a Δ ABC are right –angled at A.

Using Pythagoras theorem
⇒ AB ² + AC ² = BC ²
AB ² = (- 2 - 0) ² + (a - 3) ² = 4 + (a – 3) ²
BC ² = (- 1 + 2) ² + (4 - a) ² = 1 + (4 – a) ²
AC ² = (- 1 - 0) ² + (4 - 3) ² = 1 + 1 = 2
Since, AB ² + AC ² = BC ²
⇒ 4 + (a – 3) ² + 2 = 1 + (4 – a) ²
⇒ 4 + a ² + 9 – 6a + 2 = 1 + 16 + a ² – 8a
⇒ 2a = 2
∴ a = 1.