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

Go back to  'Ex.7.3'

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 \(\begin{align}\Delta {\text{ABC = 1: 4}}\end{align}\)

  
Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school