Question

Find the value of k for which (8,1), (k,-1),(2,-5)are collinear.

Answer 1

Answer:

here is the answer of your question.

assume that ,

A(8,1), B(k,-1), C(2,-5)

Answer 2

AnswEr:

Let us Consider that the Given points be A(8,1), B(k,-1) & C(2, -5).

If the Above Given points are Collinear so, they will lie on the same plane & there area will be zero.

So, Now we can say that

Area of Triangle ∆ABC = 0.

We know that,

$\large{\underline{\boxed{\sf{\red{ \dfrac{1}{2} [x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3(y_1 - y_2)}}}}}$

Here,

• x1 = 8, x2 = k & x3 = 2
• y1 = 1, y2 = -1 & y3 = -5

$\dag\bold{\underline{\sf{\red{Now,\; Putting\; Values}}}}$

$:\implies\sf\; \dfrac{1}{2} [ 8(-1 +5) + k(-5 -1) + 2(1 +1)] = 0$

$:\implies\sf\; 8(4) + 6k + 2(2) = 0 \times \; 2$

$:\implies\sf\; 32 + 6k + 4 = 0$

$:\implies\sf\; 6k + 36 = 0$

$:\implies\sf\; \dfrac{36}{6}$

$:\implies\large\boxed{\sf{\red{k = 6}}}$

$\rule{150}3$