Question

10. A(1, k), B(3,5) and C(6, 2) are vertices of Triangle ABC. If Angle ABC = 90°, find the value of k.​

Answer 1

Answer:

Using distance formula d = sqrt((x2 - x1)^2 + (y2 - y1)^2) to find distance between two points A(x1, x2) and B(y1, y2), we can find the distance between A and B, B and C and A and C respectively.

We have, AB = sqrt((3-1)^2 + (5-k)^2)

= sqrt(2^2 + (5-k)^2

= sqrt(4 + (5-k)^2) (Equation 1)

BC = sqrt((6-3)^2 + (2-5)^2)

= sqrt(3^2 + (-3)^2)

= sqrt (18) (Equation 2)

AC = sqrt((6-1)^2 + (2-k)^2)

= sqrt(5^2 + (2-k)^2)

= sqrt((25 + (2-k)^2) (Equation 2)

Using Pythagoras theorem, we have AC^2 = AB^2 + BC^2

Substituting the values of AB, BC and AC from equation 1, 2 and 3 respectively, we have:

(sqrt((25 + (2-k)^2)) ^2 = (sqrt(4 + (5-k)^2)) ^2 + (sqrt (18))^2

= 25 + (2-k)^2 = 4 + (5-k)^2 + 18

= 25 + 4 + k^2 – 4k = 4 + 25 + k^2 -10k + 18