# Without using distance formula, show that the points $A\left( {4,2} \right),B\left( { - 4,4} \right)\& C\left( {10,6} \right)$ are the vertices of a right angled triangle.

Last updated date: 25th Mar 2023

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Hint: In this question, let’s use the concept of slope of a line. Find it with the help of coordinates of the points. If the product of slopes of 2 lines is -1 then the lines will be perpendicular to each other.

Given points are $A\left( {4,2} \right),B\left( { - 4,4} \right)\& C\left( {10,6} \right)$

We will use the concept of slopes of line to solve the question.

We know that for a line joining point $p\left( {{x_1},{y_1}} \right)\& q\left( {{x_2},{y_2}} \right)$. Slope of the line $pq$ is:

$\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

So the slope of line $AB$ is:

$

= \dfrac{{4 - \left( { - 2} \right)}}{{ - 4 - 4}} \\

= \dfrac{{4 + 2}}{{ - 8}} = \dfrac{6}{{ - 8}} = \dfrac{{ - 3}}{4} \\

$

Slope of line $BC$ is:

$

= \dfrac{{6 - 4}}{{10 - \left( { - 4} \right)}} \\

= \dfrac{2}{{10 + 4}} = \dfrac{2}{{14}} = \dfrac{1}{7} \\

$

Also slope of line $AC$ is:

$

= \dfrac{{6 - \left( { - 2} \right)}}{{10 - 4}} \\

= \dfrac{{6 + 2}}{6} = \dfrac{8}{6} = \dfrac{4}{3} \\

$

After multiplying slopes of line $AB$ and $AC$ we get

$ \Rightarrow \dfrac{{ - 3}}{4} \times \dfrac{4}{3} = - 1$

Since the product of the slopes of these two lines is -1. So these lines are perpendicular.

Hence points ABC makes a right angled triangle.

Note: The concept of slope used above in order to prove the triangle as right angled are from geometry of lines and are very handy in solving such questions of coordinate geometry. So these concepts must be remembered. This problem can also be solved by finding the distance between the points and using Pythagoras theorem but as mentioned in the question it should not be solved that way.

Given points are $A\left( {4,2} \right),B\left( { - 4,4} \right)\& C\left( {10,6} \right)$

We will use the concept of slopes of line to solve the question.

We know that for a line joining point $p\left( {{x_1},{y_1}} \right)\& q\left( {{x_2},{y_2}} \right)$. Slope of the line $pq$ is:

$\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

So the slope of line $AB$ is:

$

= \dfrac{{4 - \left( { - 2} \right)}}{{ - 4 - 4}} \\

= \dfrac{{4 + 2}}{{ - 8}} = \dfrac{6}{{ - 8}} = \dfrac{{ - 3}}{4} \\

$

Slope of line $BC$ is:

$

= \dfrac{{6 - 4}}{{10 - \left( { - 4} \right)}} \\

= \dfrac{2}{{10 + 4}} = \dfrac{2}{{14}} = \dfrac{1}{7} \\

$

Also slope of line $AC$ is:

$

= \dfrac{{6 - \left( { - 2} \right)}}{{10 - 4}} \\

= \dfrac{{6 + 2}}{6} = \dfrac{8}{6} = \dfrac{4}{3} \\

$

After multiplying slopes of line $AB$ and $AC$ we get

$ \Rightarrow \dfrac{{ - 3}}{4} \times \dfrac{4}{3} = - 1$

Since the product of the slopes of these two lines is -1. So these lines are perpendicular.

Hence points ABC makes a right angled triangle.

Note: The concept of slope used above in order to prove the triangle as right angled are from geometry of lines and are very handy in solving such questions of coordinate geometry. So these concepts must be remembered. This problem can also be solved by finding the distance between the points and using Pythagoras theorem but as mentioned in the question it should not be solved that way.

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