Question

Find the ratio in which P(4,m) divides the line segment joining the points A(2,3) and B(6,-3). Hence find m.

Answer 1

Answer:

m = + 4

Step-by-step explanation:

Answer 2

Formula Used :-

Section Formula :-

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

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

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

Let's solve the problem now!!

$\large\underline{\sf{Solution-}}$

Let P(4, m) divides the line segment joining the points A(2, 3) and B(6, -3) in the ratio k : 1.

So,

By Section Formula, we have

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

Here,

• x = 4

• y = m

• m = k

• n = 1

• x₁ = 2

• x₂ = 6

• y₁ = 3

• y₂ = - 3

So,

on substituting all these values in above formula, we get

$\bf \:( 4, m) = \bigg(\dfrac{6 \times k + 1 \times 2}{k + 1} , \dfrac{ - 3 \times k + 1 \times 3}{k + 1} \bigg)$

$\bf \:( 4, m) = \bigg(\dfrac{6k + 2}{k + 1} , \dfrac{ - 3k + 3}{k + 1} \bigg)$

On comparing x - coordinate, we get

$\rm :\longmapsto\:4 = \dfrac{6k + 2}{k + 1}$

$\rm :\longmapsto\:4k + 4 = 6k + 2$

$\rm :\longmapsto\:6k - 4k = 4 - 2$

$\rm :\longmapsto\:2k = 2$

$\bf\implies \:k = 1$

Hence, the required ratio is 1 : 1.

This means, P is the midpoint of line segment joining AB.

Now,

On comparing y - coordinate, we have

$\rm :\longmapsto\:m = \dfrac{ - 3k + 3}{k + 1}$

On substituting the value of k = 1, we get

$\rm :\longmapsto\:m = \dfrac{ - 3 \times 1 + 3}{1 + 1}$

$\rm :\longmapsto\:m = \dfrac{ - 3 + 3}{1 + 1}$

$\bf\implies \:m \: = \: 0$

Additional Information :-

Distance Formula :-

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

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

Midpoint Formula :-

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

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

Area of triangle :-

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