Question

find the value of k if the points A( 2,3), B (4,k) and C (6,-3) are collinear

Answer 1

Answer:

The value of k is 0.

Step-by-step explanation:

Given : If the points A( 2,3), B (4,k) and C (6,-3) are collinear.

To find : The value of k?

Solution :

When three points are collinear then the condition is

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

Where, $x_1 = 2, x_2 = 4, y_1 = 3, y_2 = k, x_3 = 6, y_3 = -3$

Substituting the values,

$2(k + 3) + 4(-3 - 3) + 6(3 - k) = 0$

$2k + 6 + 4(-6) + 18 - 6k = 0$

$2k + 6 - 24 + 18 - 6k = 0$

$-4k + 0 = 0$

$k=0$

Therefore, The value of k is 0.

Answer 2

Answer:

Value of k = 0

Step-by-step explanation:

Given points A(2,3) ,B(4,k) and C(6,-3).

$We\: know \:that:\\</p><p>Slope \: of \: line\: segment\\ \:joining \: points\:(x_{1},y_{1}),\:(x_{2},y_{2})\\ = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$Slope \: of \: a \:line \:AB\\=Slope \:of \:the \:line \:BC\\(Given \:A,B\:and \:are\: collinear)$

$\implies \frac{k-3}{4-2}=\frac{-3-k}{6-4}$

$\implies \frac{k-3}{2}=\frac{-3-k}{2}$

$\implies k-3 = -3-k$

$\implies k+k = -3+3$

$\implies 2k = 0$

$\implies k = \frac{0}{2}$

$\implies k = 0$

Therefore,

Value of k = 0

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