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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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