If the points $( \frac{2}{5},\ \frac{1}{3}),\ ( \frac{1}{2},\ k)$ and $( \frac{4}{5},\ 0)$ are collinear, then find the value of $k$ .

Academic Mathematics NCERT Class 10

Problem Statement

If the points $( \frac{2}{5},\ \frac{1}{3}),\ ( \frac{1}{2},\ k)$ and $( \frac{4}{5},\ 0)$ are collinear, then find the value of $k$ .

Solution

Given: Points

$( \frac{2}{5},\ \frac{1}{3}),\ ( \frac{1}{2},\ k)$ and

$( \frac{4}{5},\ 0)$ are collinear

To do: To find the value of

$k$ .

Solution:

When three points A, B and C are collinear, slope of line

joining any two points (say AB) = slope of line joining any other two points (say BC).

Slope of the line passing through points

$(x_1, y_1)$ and

$(x_2,\ y_2) =\frac{y_2-y_1}{x_2-x_1}$

Slope of the line passing through

$( \frac{2}{5},\ \frac{1}{3})$ and

$( \frac{1}{2},\ k)=$ Slope of line passing through

$( \frac{1}{2},\ k)$ and

$( \frac{4}{5},\ 0)$

$\Rightarrow \frac{k-\frac{1}{3}}{\frac{1}{2}-\frac{2}{5}}=\frac{0-k}{\frac{4}{5}-\frac{1}{2}}$

$\Rightarrow \frac{\frac{3k-1}{3}}{\frac{1}{10}}=\frac{-k}{\frac{3}{10}}$

$\Rightarrow \frac{10( 3k-1)}{3}=-\frac{10k}{3}$

$\Rightarrow 30k-10=-10k$

$\Rightarrow 40k=10$

$\Rightarrow k=\frac{10}{40}$

$\Rightarrow k=\frac{1}{4}$

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Updated on: 10-Oct-2022

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