Question

If the points A(2,3) , B(4,k) , C(6,-3) are collinear , find the value of k.​

Answer 1

Basic Concept Used :-

• Three points A, B, C are collinear, if and only if Area of triangle ABC =0.

Given :-

• Three points A(2, 3), B(4, k) and C(6, - 3) are collinear

To Find :-

• The value of 'k'.

Solution :-

Given that

• A(2, 3), B(4, k) and C(6, - 3) are collinear

We know,

• If three points are collinear, then area of triangle is 0.

Area of triangle is given by

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

Here,

• • x₁ = 2

• • x₂ = 4

• • x₃ = 6

• • y₁ = 3

• • y₂ = k

• • y₃ = - 3

$\rm :\longmapsto\:0 = \dfrac{1}{2}\bigg(2(k + 3) + 4( - 3 - 3) + 6(3 - k)\bigg)$

$\rm :\longmapsto\:0 = \bigg(2k + 6 + 4( - 6) + 18 - 6k\bigg)$

$\rm :\longmapsto\:0 = \bigg(2k + 6 -24 + 18 - 6k\bigg)$

$\rm :\longmapsto\:0 = - 4k + 24 - 24$

$\rm :\longmapsto\:0 = - 4k$

$\bf\implies \:k \: = \: 0$

Additional Information :-

Distance Formula :-

Let us consider a line segment joining the points A and B, then distance between A and B is given by

$\sf\implies \:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

$\sf \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: B(x_2,y_2)$

Midpoint Formula :-

Let us consider a line segment joining the points A and B and let C (x, y) be the midpoint of AB, then coordinates of C is

$\boxed{ \quad\sf \:( x, y) = \bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} \bigg) \quad}$

$\sf \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: B(x_2,y_2)$

Section Formula :-

Let us consider a line segment joining the points A and B and let C (x, y) be the point on AB which divides AB in the ratio m : n internally, then coordinates of C is

$\boxed{ \quad\sf \:( x, y) = \bigg(\dfrac{nx_1+mx_2}{m + n} , \dfrac{ny_1+my_2}{m + n} \bigg) \quad}$

$\sf \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: B(x_2,y_2)$