Question

In what ratio is the line segment joining the points A(6,3) and B(-2,-5) divided by x-axis? Also, find the point of intersection (pls provide step-by-step solution if possible)​

Answer 1

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

Let assume that x - axis intersects the line segment joining the points A(6,3) and B(-2,-5) at P and divides AB internally in the ratio k : 1.

Let further assume that coordinates of point P 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:

$\sf\implies (x,y) = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)$

So, on substituting the values, we get

$\rm :\longmapsto\:(x,0) = \bigg(\dfrac{ - 2k + 6}{k + 1}, \: \dfrac{ - 5k + 3}{k + 1} \bigg)$

So, on comparing y - coordinate, we get

$\rm :\longmapsto\:\dfrac{ - 5k + 3}{k + 1} = 0$

$\rm :\longmapsto\: - 5k + 3 = 0$

$\rm :\longmapsto\: - 5k = - 3$

$\bf\implies \:k \: = \: \dfrac{3}{5}$

So, required ratio is k : 1 = 3 : 5

Now, on comparing x - coordinate, we get

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

$\rm :\longmapsto\:\dfrac{ - 2 \times \dfrac{3}{5} + 6}{\dfrac{3}{5} + 1} = x$

$\rm :\longmapsto\:\dfrac{ - \dfrac{6}{5} + 6}{\dfrac{3 + 5}{5}} = x$

$\rm :\longmapsto\:\dfrac{ \dfrac{ - 6 + 30}{5}}{\dfrac{8}{5}} = x$

$\rm :\longmapsto\:\dfrac{ \dfrac{24}{5}}{\dfrac{8}{5}} = x$

$\bf\implies \:x = 3$

Hence,

Coordinates of point of intersection, P = (3, 0)

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

Distance Formula

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane, then distance between P and Q is

$\sf\implies\boxed{\tt{ PQ = \sqrt{ {(x_{1} - x_{2}) }^{2} + {(y_{2} - y_{1})}^{2} }}}$

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:

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

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:

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

Answer 2

Answer:

3:5

Step-by-step explanation: