Question

28. Find the value of 'k'. If the co-ordinates of the points A(2,-2), B(-4, 2) and
C(-7, k) are collinear.
20​

Answer 1

Basic Concept Used :-

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

Given :-

• A(2, - 2)

• B(- 4, 2)

• C(- 7, k)

To Find :-

• The value of k if A, B, C are collinear.

Solution :-

Given that

• A(2, - 2)

• B(- 4, 2)

• C(- 7, k)

are collinear.

We know,

If 3 points are collinear, area of triangle is 0.

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

Here,

• • x₁ = 2

• • x₂ = - 4

• • x₃ = - 7

• • y₁ = - 2

• • y₂ = 2

• • y₃ = k

So,

On substituting all these values in above formula,

$\rm :\longmapsto\:0 = \dfrac{1}{2}\bigg(2(2 - k) - 4(k + 2) - 7( - 2 - 2)\bigg)$

$\rm :\longmapsto\:0 = \bigg(4 - 2k - 4k - 8 + 28\bigg)$

$\rm :\longmapsto\:24 - 6k = 0$

$\rm :\longmapsto\:24 = 6k$

$\bf\implies \:k = 4$

Additional Information :-

Distance Formula :-

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

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

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

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 a point on AB which divides AB in the ratio m : n, then coordinates of C is

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