Find the values of $y$ for which the distance between the points $P (2, -3)$ and $Q (10, y)$ is 10 units.

Given: 

The distance between the points $P( 2,\ -3)$ and $Q( 10,\ y)$ is $10$ units.

To do: 

We have to find the values of $y$.

Solution:

We know that,

The distance between two points $( x_1,\ y_1)$ and $(x_2,\ y_2)=\sqrt{( x_2-x_1)^2+( y_2-y_1)^2}$

$\Rightarrow PQ=10$

$\Rightarrow \sqrt{( 10-2)^2+( y-( -3))^2}=10$

$\Rightarrow \sqrt{8^2+( y+3)^2}=10$

Squaring on both sides, we get,

$\Rightarrow 8^2+( y+3)^2=(10)^2$

$\Rightarrow ( y+3)^2=100-64$

$\Rightarrow ( y+3)^2=36$

$\Rightarrow y+3=\pm 6$

If $y+3=6\ \Rightarrow y=6-3=3$

If $y+3=-6\ \Rightarrow y=-6-3=-9$ 

$\Rightarrow y=3$ or $y=-9$

The values of $y$ are $-9$ and $3$.

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Updated on: 10-Oct-2022

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