Question

in what ratio does x axis divides the line joining the points (6,2) (3,-4)​

Answer 1

Answer:

1:2

Step-by-step explanation:

Find your answer in image

Answer 2

Answer:

x-axis divides the join of points (6, 2) and (3, - 4) internally in the ratio 1 : 2

Step-by-step explanation:

Let assume that x axis divides the line joining the points (6,2) and (3,-4) in the ratio k : 1 at (x, 0).

We know,

Section Formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the cartesian plane and C(x, y) be the point which divides AB internally in the ratio m₁ : m₂, then the coordinates of C is given by

$\begin{gathered} \boxed{\tt{ (x, y) = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)}} \\ \end{gathered}$

So, on substituting the values in above formula, we get

$\sf \: (x,0) = \bigg(\dfrac{3 \times k + 6 \times 1}{k + 1}, \: \dfrac{- 4 \times k + 2 \times 1}{k + 1} \bigg)$

$\sf \: (x,0) = \bigg(\dfrac{3k + 6}{k + 1}, \: \dfrac{- 4k + 2}{k + 1} \bigg)$

On comparing y- coordinate, we get

$\sf \: \dfrac{- 4k + 2}{k + 1} = 0$

$\sf \: {- 4k + 23} = 0$

$\sf \: {4k} = 2$

$\implies\sf \: k = \dfrac{1}{2}$

Hence, x-axis divides the join of points (6, 2) and (3, - 4) internally in the ratio 1 : 2

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${{ \sf{Additional\:Information}}}$

1. Distance Formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the cartesian plane, then distance between A and B is given by

$\begin{gathered}\boxed{\tt{ AB \: = \sqrt{ {(x_{2} - x_{1}) }^{2} + {(y_{2} - y_{1})}^{2} }}} \\ \end{gathered}$

2. Section formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the cartesian plane and C(x, y) be the point which divides AB internally in the ratio m₁ : m₂, then the coordinates of C is given by

$\begin{gathered} \boxed{\tt{ (x, y) = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)}} \\ \end{gathered}$

3. Mid-point formula

Let A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane and C(x, y) be the mid-point of AB, then the coordinates of C is given by

$\begin{gathered}\boxed{\tt{ (x,y) = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)}} \\ \end{gathered}$

4. Centroid of a triangle

Centroid of a triangle is defined as the point at which the medians of the triangle meet and is represented by the symbol G.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle and G(x, y) be the centroid of the triangle, then the coordinates of G is given by

$\begin{gathered}\boxed{\tt{ (x, y) = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)}} \\ \end{gathered}$

5. Area of a triangle

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle, then the area of triangle is given by

$\begin{gathered}\boxed{\tt{ Area =\dfrac{1}{2}\bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|}} \\ \end{gathered}$