A(1,k) B(3,5) C(6,2) are vertices of triangle ABC if angle abc = 90 degree find the value of k please someone give me answer very urgent
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
Rearranging and solving, we get, K = 3 (Answer)
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