Question

find the ratio in which zx plane divides the join of the points (2,4,5) and (3,-6,8)​

Answer 1

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

Given that,

• zx plane divides the line segment joining the points (2,4,5) and (3,-6,8).

Let Assume that

• zx plane divides the line segment joining the points (2,4,5) and (3,-6,8) in the ratio k : 1.

• Let the point of intersection of zx plane and give line segment joining the points (2,4,5) and (3,-6,8) be (x, 0, z).

We know

Section Formula

Let P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) be two points in the coordinate plane and R(x, y, z) be the point which divides PQ internally in the ratio m₁ : m₂. Then, the coordinates of R will be:

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

So, on substituting the values, we get

$\rm :\longmapsto\:(x,0,z) = \bigg(\dfrac{3k + 2}{k + 1}, \dfrac{ - 6k + 4}{k + 1}, \dfrac{5k + 8}{k + 1} \bigg)$

So, on comparing, we get

$\rm :\longmapsto\:\dfrac{ - 6k + 4}{k + 1} = 0$

$\rm :\longmapsto\: - 6k + 4 = 0$

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

$\rm :\longmapsto\:k = \dfrac{2}{3}$

So, Required ratio is 2 : 3

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More to know

1. Mid-point formula

Let P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) be two points in the coordinate plane and R(x, y) be the mid-point of PQ. Then, the coordinates of R will be:

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

2. Centroid of a triangle

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

Let A(x₁, y₁, z₁) and B(x₂, y₂, z₂) and C(x₃, y₃, z₃) be the vertices of a triangle. Let R(x, y) be the centroid of the triangle. Then, the coordinates of R will be:

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