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

## Question

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

## 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 \(y\)?