Question

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Answer 1

Question

Find the ratio in which the line segment joining the points A (1, - 5) and B (- 4, 5) is divided by x - axis. Also, find the point of intersection.

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

Given that

A line segment AB having coordinates of A as (1, - 5) and coordinates of B as (- 4, 5).

Let assume that x - axis divides the line segment joining the points A (1, - 5) and B (- 4, 5) in the ratio k : 1 at C.

Let assume that coordinates of C be (x, 0).

We know,

Section formula :-

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the point which divides PQ internally in the ratio m₁ : m₂. Then, the coordinates of R will be:

$\sf\implies \boxed{\tt{ R = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)}}$

So, on substituting the values, we get

$\rm :\longmapsto\:(x,0) = \bigg(\dfrac{ - 4k + 1}{k + 1}, \: \dfrac{5k - 5}{k + 1} \bigg)$

On comparing y - coordinate on both sides, we get

$\rm :\longmapsto\:\dfrac{5k - 5}{k + 1} = 0$

$\rm :\longmapsto\:5k - 5 = 0$

$\rm :\longmapsto\:5k = 5$

$\bf\implies \:k = 1$

Hence,

The x - axis divides the line segment joining the points A (1, - 5) and B (- 4, 5) in the ratio 1 : 1 at C.

Now, On comparing x - coordinate on both sides, we get

$\rm :\longmapsto\:x = \dfrac{ - 4k + 1}{k + 1}$

On substituting the value of k, we get

$\rm :\longmapsto\:x = \dfrac{ - 4+ 1}{1 + 1}$

$\rm :\longmapsto\:x \: = - \: \dfrac{3}{2}$

Hence,

The coordinates of point of intersection, C is

$\rm\implies \:\boxed{\tt{ Coordinates \: of \: C = \bigg( - \dfrac{3}{2}, \: 0 \bigg) }}$

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More to know :-

1. Mid-point formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the mid-point of PQ. Then, the coordinates of R will be:

$\sf\implies \boxed{\tt{ R = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)}}$

2. Centroid of a triangle

Centroid of a triangle is the point where the medians of the triangle meet.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle. Let R(x, y) be the centroid of the triangle. Then, the coordinates of R will be:

$\sf\implies \boxed{\tt{ R = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)}}$

Answer 2

Answer:

Question

Find the ratio in which the line segment joining the points A (1, - 5) and B (- 4, 5) is divided by x - axis. Also, find the point of intersection.

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

♣ $\tt \red{\underline{\underline{\bold{Given \: that}}}}$

A line segment AB having coordinates of A as (1, - 5) and coordinates of B as (- 4, 5).

Let assume that x - axis divides the line segment joining the points A (1, - 5) and B (- 4, 5) in the ratio k : 1 at C.

Let assume that coordinates of C be (x, 0).

We know,

Section formula :-

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the point which divides PQ internally in the ratio m₁ : m₂. Then, the coordinates of R will be:

$\sf\implies \boxed{\tt{ R = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)}}$

So, on substituting the values, we get

$\rm :\longmapsto\:(x,0) = \bigg(\dfrac{ - 4k + 1}{k + 1}, \: \dfrac{5k - 5}{k + 1} \bigg)$

On comparing y - coordinate on both sides, we get

$\rm :\longmapsto\:\dfrac{5k - 5}{k + 1} = 0$

$\rm :\longmapsto\:5k - 5 = 0$

$\rm :\longmapsto\:5k = 5$

$\bf\implies \:k = 1$

Hence,

The x - axis divides the line segment joining the points A (1, - 5) and B (- 4, 5) in the ratio 1 : 1 at C.

Now, On comparing x - coordinate on both sides, we get

$\rm :\longmapsto\:x = \dfrac{ - 4k + 1}{k + 1}$

On substituting the value of k, we get

$\rm :\longmapsto\:x = \dfrac{ - 4+ 1}{1 + 1}$

$\rm :\longmapsto\:x \: = - \: \dfrac{3}{2}$

Hence,

The coordinates of point of intersection, C is

$\rm\implies \:\boxed{\tt{ Coordinates \: of \: C = \bigg( - \dfrac{3}{2}, \: 0 \bigg) }}$

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