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 \(C(x_3, y_3).\)

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 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 \((x_3, y_3) = (0, 3)\)

Area of a triangle

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

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 \(P, Q, R\)

Area of \(\Delta PQR\),

\[\begin{align}&= \frac{1}{2}\begin{bmatrix} (2 - 1)  + 1(1 - 0) \\ + 0(0 - 2)\end{bmatrix} \\ &= \frac{1}{2}(1 + 1)\\ &= 1{\text{ Square units }}\end{align}\]

By substituting the values of Points \(A, B, C\)

Area of \(\Delta ABC\),

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

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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