# Ex.7.3 Q3 Coordinate Geometry Solution - NCERT Maths Class 10

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## Question

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are $$(0, -1), (2, 1)$$ and $$(0, 3).$$ Find the ratio of this area to the area of the given triangle.

## Text Solution

Reasoning:

Let ABC be any triangle whose vertices are  A $$(x_1, y_1),$$ B $$(x_2, y_2)$$ and C $$(x_3, y_3).$$

\begin{align}\text{Area of a triangle}=\frac{{{1}}}{{{2}}}\left\{ {{{{x}}_{{1}}}\left( {{{{y}}_{{2}}} - {{{y}}_{{3}}}} \right){{ + }}{{{x}}_{{2}}}\left( {{{{y}}_{{3}}} - {{{y}}_{{1}}}} \right){{ + }}{{{x}}_{{3}}}\left( {{{{y}}_{{1}}} - {{{y}}_{{2}}}} \right)} \right\} \end{align}

What is Known:

The $$x$$ and $$y$$ co-ordinates of the vertices of the triangle.

What is Unknown?

The ratio of this area to the area of the given triangle.

Solution:

From the given figure,

Given,

• Let A $$(x_1, y_1) = (0, -1)$$
• Let B $$(x_2, y_2) = (2 , 1)$$
• Let C $$(x_3, y_3) = (0, 3)$$

\begin{align}\text{Area of a triangle}= \frac{{{1}}}{{{2}}}\left\{ {{{{x}}_{{1}}}\left( {{{{y}}_{{2}}} - {{{y}}_{{3}}}} \right){{ + }}{{{x}}_{{2}}}\left( {{{{y}}_{{3}}} - {{{y}}_{{1}}}} \right){{ + }}{{{x}}_{{3}}}\left( {{{{y}}_{{1}}} - {{{y}}_{{2}}}} \right)} \right\} \\...\text{{Equation}}\,\left( {{1}} \right)\end{align}

By substituting the values of vertices, $$A, B, C$$ in (1),

Let $$P, Q, R$$ be the mid-points of the sides of this triangle.

Coordinates of $$P, Q,$$ and $$R$$ are given by

\begin{align}{P = \left( {\frac{{0 + 2}}{2},\frac{{ - 1 + 1}}{2}} \right) = (1,0)}\\{Q = \left( {\frac{{0 + 0}}{2},\frac{{3 - 1}}{2}} \right) = (0,1)}\\{R = \left( {\frac{{2 + 0}}{2},\frac{{1 + 3}}{2}} \right) = (1,2)}\end{align}

By substituting the values of Points $$P, Q, R$$

\begin{align}{\text{ Area of }}\Delta PQR&= \frac{1}{2}\{ (2 - 1)\, + \,1(1 - 0) + 0(0 - 2)\} \\ &= \frac{1}{2}(1 + 1)\\ &= 1{\text{ Square units }}\end{align}

By substituting the values of Points $$A, B, C$$

\begin{align}\text{ Area if }\Delta ABC &= \frac{1}{2}[0(1 - 3) + 2\{ 3 - ( - 1)\} + 0( - 1 - 1)]\\ &= \frac{1}{2}\{ 8\} \\ &= 4{\text{ Square units }}\end{align}

Therefore, Ratio of this area $$\Delta$$PQR to the area of the triangle \begin{align}\Delta {\text{ABC = 1: 4}}\end{align}

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