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# Ex.7.3 Q1 Coordinate Geometry Solution - NCERT Maths Class 10

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## Question

Find the area of the triangle whose vertices are:

(i) $$(2, 3), (-1, 0), (2, -4)$$

(ii) $$(-5, -1), (3, -5), (5, 2)$$

Video Solution
Coordinate Geometry
Ex 7.3 | Question 1

## Text Solution

Reasoning:

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

Area of a triangle

\begin{align} \! = \! \frac{1}{2} \begin{bmatrix} x_1 \left( y_2 \! - \! y_3 \right) \! + \! x_2 \left( y_3 \! - \! y_1 \right) + \\ x_3 \left( y_1 \! - \! y_2 \right) \end{bmatrix} \end{align}

What is the known:

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

What is the Unknown:

The area of the triangle.

Solution:

(i) Given,

• Let  $$A\,(x_1, y_1) = (2, 3)$$
• Let  $$B\,(x_2, y_2) = (-1 , 0)$$
• Let  $$C\,(x_3, y_3) = (2, -4)$$

Area of a triangle

\begin{align} \! = \! \frac{1}{2} \begin{bmatrix} x_1 \left( y_2 \! - \! y_3 \right) \! + \! x_2 \left( y_3 \! - \! y_1 \right) \! + \! \\ x_3 \left( y_1 \! - \! y_2 \right) \end{bmatrix} \cdots (1) \end{align}

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

Area of the given triangle

\begin{align}&= \! \frac{1}{2} \! \begin{bmatrix}2 ( 0 \! - \! ( - 4)) \! + \! ( - 1)(( - 4) \! - \! (3)) \\ + 2(3 \! - \! 0) \end{bmatrix} \\&= \! \frac{1}{2}(8 + 7 + 6)\\ &= \! \frac{{{\text{21}}}}{{\text{2}}}{\text{ Square units }}\end{align}

(ii) Given,

• Let $$A(x_1, y_1) = (-5, -1)$$
• Let $$B(x_2, y_2) = (3, -5)$$
• Let $$C(x_3, y_3) = (5, 2)$$

Area of a triangle

\begin{align}= \! \frac{1}{2} \! \begin{bmatrix} x_1 \left( y_2 \! - \! y_3 \right) \! + \! x_2 \left( y_3 \! - \! y_1 \right) + \\x_3 \left( y_1 \! - \! y_2 \right) \end{bmatrix} \;\; \cdots (1) \end{align}

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

Area of the given triangle

\begin{align} &= \! \frac{1}{2} \begin{bmatrix} ( - 5) (- 5) \! - \! ( - 2) \! + \! 3(2 \! - \! ( - 1)) + \\5 \! - \! 1 \! - \! ( - 5) \end{bmatrix} \\&= \frac{1}{2}(35 \! + \! 9 \! + \! 20) \\ &= \text{32 Square units } \end{align}

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