Question

if the points (-1, 3) (2, 4) and (k, -2) are collinear. find the K
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Answer 1

Solution :-

Let the points be :-

A ( -1 , 3 )

B ( 2 , 4 )

C ( k , -2 )

Given that, these points are collinear.

That is, area of triangle ABC is zero.

$\underline{\boxed{\bf Area \ of \ triangle = \dfrac{1}{2} \times [ \ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \ ]}}$

Here :-

$\bullet \sf \ x_1 = -1 \ , \ y_1 = 3 \\\\\bullet \ x_2 = 2 \ , y_2 = 4 \\\\\bullet \ x_3 = k \ , \ y_3 = -2$

$\longrightarrow \sf \dfrac{1}{2} \times [ \ -1(4-(-2))+2(-2-3)+k(3-4) \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ -1(4+2)+2(-2-3)+k(3-4) \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ (-1 \times 6 )+( 2 \times -5)+( k \times -1) \ ] = 0 \\\\\longrightarrow \dfrac{1}{2} \times [ \ -6-10-k \ ] = 0 \\\\\longrightarrow -16-k = 0 \\\\\longrightarrow -k = 16 \\\\\longrightarrow \bf k = -16$

Answer 2

Answer :-

• Value of k $\leadsto$ $\boxed{\tt{\red{-16}}}$

Given :-

• The points (-1,3), (2,4) and (k, -2) are collinear (on a line).

To find :-

• The value of ‘k’.

Formula used :-

• Area of the triangle (∆) on the points

$\sf{(x_1,y_1),\:(x_2,y_2),\:(x_3,y_3)}$ is

$\boxed{\sf{∆=\dfrac{1}{2} |x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|}}$

Solution :-

We know that, when 3 points are collinear, there area is ‘0’. (∆ = 0)

$\sf{➪\:∆=\dfrac{1}{2} |x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|}$

$\sf{➪\:0=\dfrac{1}{2} |-1(4-(-2))+2(-2-3)+k(3-4)|}$

$\sf{➪\:0=\dfrac{1}{2} (-6-10-k)}$

$\sf{➪\:0=-6-10-k}$

$\sf{➪\:k =}$ $\bold{\red{-16}}$