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) \((-5, -1), (3, -5), (5, 2)\)

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
\[\begin{align}=\frac{{{1}}}{{{2}}}\left\{ {{{{x}}_{{1}}}\left( {{{{y}}_{{2}}} - {{{y}}_{{3}}}} \right){{ + }}{{{x}}_{{2}}}\left( {{{{y}}_{{3}}} - {{{y}}_{{1}}}} \right){{ + }}{{{x}}_{{3}}}\left( {{{{y}}_{{1}}} - {{{y}}_{{2}}}} \right)} \right\} & & \end{align}\]

What is 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
\[\begin{align} = \frac{{{1}}}{{{2}}}\left\{ {{{{x}}_{{1}}}\left( {{{{y}}_{{2}}} - {{{y}}_{{3}}}} \right){{ + }}{{{x}}_{{2}}}\left( {{{{y}}_{{3}}} - {{{y}}_{{1}}}} \right){{ + }}{{{x}}_{{3}}}\left( {{{{y}}_{{1}}} - {{{y}}_{{2}}}} \right)} \right\}\\\qquad ...\,{\text{Equation}}\,\left( {{1}} \right)\end{align}\]

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

Area of the given triangle
\[\begin{align}&= \frac{{\text{1}}}{{\text{2}}}{\text{[2\{ 0} - {\text{(}} - {\text{4)\} + (}} - {\text{1)\{ (}} - {\text{4)}} - {\text{(3)\} + 2(3}} - {\text{0)]}}}\\&= \frac{{\text{1}}}{{\text{2}}}{\text{\{ 8 + 7 + 6\} }}\\ &= \frac{{{\text{21}}}}{{\text{2}}}{\text{ Square units }}\end{align}\]

(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
\[\begin{align} = \frac{{{1}}}{{{2}}}\left\{ {{{{x}}_{{1}}}\left( {{{{y}}_{{2}}} - {{{y}}_{{3}}}} \right){{ + }}{{{x}}_{{2}}}\left( {{{{y}}_{{3}}} - {{{y}}_{{1}}}} \right){{ + }}{{{x}}_{{3}}}\left( {{{{y}}_{{1}}} - {{\text{y}}_{{2}}}} \right)} \right\} \\  ...\,{\rm{Equation}}\,\left( {{1}} \right)\end{align}\]
By substituting the values of vertices, A, B, C in the Equation (1),

Area of the given triangle

\[\begin{align} &= \frac{1}{2}[( - 5)\ (- 5) - ( - 2) + 3(2 - ( - 1))+ 5  - 1 - ( - 5)\ ]\\&= {\frac{{{1}}}{{{2}}}{{[35 + 9 + 20] }}}\\ &= {{\text{32 square units }}}\end{align}\]

  
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