Question

find the value of k for which the point A(-2,3),B(1,2),C(k,0) are collinear solve distance formula

Answer 1

Points are collinear means slope of AB,BC and AC.
i.e, Slope of AB = Slope of BC
(2-3)/(1-(-2))=(0-2)/(k-1)
-1/3 = -2/k-1
-k+1 = -6
-k = -7
k = 7

Answer 2

The value of k is 7.

Step-by-step explanation:

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

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{y_3-y_2}{x_3-x_2}$

The given points are A(-2,3),B(1,2) and C(k,0).

If these points are collinear then

$\dfrac{2-3}{1-(-2)}=\dfrac{0-2}{k-1}$

$\dfrac{-1}{3}=\dfrac{-2}{k-1}$

$\dfrac{1}{3}=\dfrac{2}{k-1}$

On cross multiplication we get

$k-1=6$

Add 1 on both sides.

$k-1+1=6+1$

$k=7$

Therefore, the value of k is 7.

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