Question

Find the value of k for which the points A(6,1),B (k,-6),C(0,-7)are collinear

Answer 1

given points are collinear so ∆ABC s area is 0
X1(y2-y3)+X2(y3-y1)+X3(y1-y2)=0
6(-6+7)+k(-7-1)+0(1+6)=0
6+(-8k)+0=0
6-8k+0=0
-8k=0-0-6
k=-6/-8
k=3/4

Answer 2

The value of k is 3/4

Solution:

If three points $(x_1, y_1), (x_2, y_2) , (x_3, y_3)$ are collinear then

$x_1(y_2 - y_3) + x_2( y_3 - y_1)+ x_3(y_1 - y_2) = 0$

From given,

$(x_1, y_1) = (6, 1)\\\\(x_2, y_2) = (k, -6)\\\\(x_3, y_3) = (0, -7)$

Substituting we get,

$6(-6+7) + k(-7-1) + 0(1+6) = 0\\\\6 \times 1 + k \tmes -8 + 0 = 0\\\\6 -8k = 0\\\\8k = 6\\\\4k = 3\\\\k = \frac{3}{4}$

Thus value of k is 3/4

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