# 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\(\begin{align} =\sqrt{\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)^{2}}\end{align}\)

**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,

\(\begin{align}\sqrt {{{\left( {{{\text{x}}_1} - {{\text{x}}_2}} \right)}^2} + {{\left( {{{\text{y}}_1} - {{\text{y}}_2}} \right)}^2}} \qquad \qquad ...\;{\text{Equation}}\;(1)\end{align}\)

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

Therefore,

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

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