# Ex.7.1 Q8 Coordinate Geometry Solution - NCERT Maths Class 10

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## Question

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

Video Solution
Coordinate Geometry
Ex 7.1 | Question 8

## Text Solution

Reasoning:

The distance between the two points can be measured using the Distance Formula which is given by:

Distance Formula

\begin{align} =\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}\end{align}

What is Known?

The $$x$$ and $$y$$ co-ordinates of the points which is at a distance of $$10$$ units.

What is Unknown?

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

Steps:

Given,

Distance between points $$A \;(2, -3)$$ and $$B \;(10, y)$$ is $$10$$ units.

We know that the distance between the two points is given by the Distance Formula,

\begin{align}&\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}}\right)}^2}}\;\;\dots(1)\end{align}

By Substituting the values of points $$A \;(2, -3)$$ and $$B \;(10, y)$$ in Equation (1)

Therefore,

\begin{align}\sqrt {{{(2 - 10)}^2} + {{( - 3 - y)}^2}} &= 10\\\sqrt {{{( - 8)}^2} + {{(3 + {\text{y}})}^2}} &= 10\\{\text{Squaring on both sides}}\\64 + {(y + 3)^2} &= 100\\{(y + 3)^2} &= 36\\y + 3 &= \pm 36\\y + 3 &= \pm 6\\y + 3 &= 6\\{\text{or}}\\{\text{y}} + 3 &= - 6\end{align}

Therefore, \begin{align}y = 3\quad{\text{or}}\quad -y= 9\end{align} are the possible values for $$y$$?

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