# Ex.7.4 Q4 Coordinate Geometry Solution - NCERT Maths Class 10

## Question

The two opposite vertices of a square are \(\begin{align}\left( { - 1,\,2} \right){\text{ }}{\text{ and}}\left( {3,\,2} \right).\end{align}\) Find the coordinates of the other two vertices.

## Text Solution

**Reasoning:**

**What is known?**

The \(x\) and \(y\) co-ordinates of the two opposite vertices of a square.

**What is unknown?**

The coordinates of the other two vertices.

**Steps:**

From the Figure,

Given,

- Let \(ABCD\) be a square having known vertices \(\begin{align}{A}\,\,\left( { - 1,{\text{ }}2} \right){\text{ and}}{\text{ }}{C}\,\,\left( {3,{\text{ }}2} \right)\end{align}\)as vertices \(A\) and \(C\) respectively.
- Let \(\begin{align}\text{B}\left( {x, y} \right)\end{align}\) be one unknown vertex

We know that the sides of a square are equal to each other.

\(\begin{align}∴ {{AB = BC}}\end{align}\)

By Using Distance formula to find distance between points \(AB\; {\rm and } \;BC ,\)

\(\begin{align} \sqrt {{{(x_1 - 1)}^2} + {{(y_1 - 2)}^2}} &= \sqrt {{{(x_1 - 3)}^2} + {{(y_1 - 2)}^2}} \end{align}\)

\(\begin{align} {x_1^2} + 2x + 1 + {y_1^2} - 4y + 4 \end{align}\) \(\begin{align} &= {x_1^2} + 9 - 6x + {y_1^2} + 4 - 4y _1\qquad \end{align}\) (By Simplifying & Transposing)

\(\begin{align} 8x_1 &= 8 \end{align}\)

\(\begin{align} x_1 &= 1\end{align}\)

We know that in a square, all interior angles are of \(\begin{align}{90^\circ }.\end{align}\)

In \(\begin{align}\Delta ABC\end{align}\)

\(\begin{align}{{AB}^{2}+{BC}^{2}={AC}^{2}} \quad \text { [By Pythagorastheorem }] \end {align}\)

Distance formula is used to find distance between \(AB\), \(BC\) and \(AC\)

\(\begin{align} {\left( {\sqrt {{{(1 + 1)}^2} + {{(y_1 - 2)}^2}} } \right)^2} + {\left( {\sqrt {{{(1 - 3)}^2} + {{(y_1 - 2)}^2}} } \right)^2}\end{align}\) \(\begin{align} &= {\left( {\sqrt {{{(3 + 1)}^2} + {{(2 - 2)}^2}} } \right)^2} \end{align}\)

\(\begin{align} 4 + {y_1^2} + 4 - 4y_1 + 4 + {y_1^2} - 4y_1 + 4 &= 16 \end{align}\)

\(\begin{align} 2{y_1^2} + 16 - 8y &= 16 \end{align}\)

\(\begin{align} 2{y_1^2} - 8y &= 0 \end{align}\)

\(\begin{align} y_1(y_1 - 4) &= 0 \end{align}\)

\(\begin{align}y_1 &= 0\,\,or\,\,4\end{align}\)

Hence the required vertices are \(\begin{align}B{\text{ }}\left( {1,\,\,0} \right){\text{ }}{\text{ and}}\;D{\text{ }}\left( {1,{\text{ }}4} \right)\end{align}\)