# Ex.7.4 Q5 Coordinate Geometry Solution - NCERT Maths Class 10

## Question

The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of \(1\; \rm m\) from each other. There is a triangular grassy lawn in the plot as shown in the following figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

(i) Taking \(A\) as origin, find the coordinates of the vertices of the triangle.

(ii) What will be the coordinates of the vertices of \(\Delta\) \(PQR\) if \(C\) is the origin? Also calculate the areas of the triangles in these cases. What do you observe?

## Text Solution

**What is known?**

Saplings of Gulmohar are planted on the boundary at a distance of \(1 \rm\,m\) from each other.

**What is unknown?**

- The coordinates of the vertices of the triangle.
- The coordinates of the vertices of \(\Delta\) \(PQR\) if \(C\) is the origin.

**Steps:**

(i) Given,

Taking \(A\) as origin, we will take \(AD\) as *\(x\)*-axis and \(AB\) as *\(y\)*- axis.

It can be observed from the figure that the coordinates of point \(P\), \(Q\), and \(R\) are \((4, 6)\), \((3, 2)\), and \((6, 5)\) respectively.

Let,

\[\begin{align}\text{P}\left( {{x_1},{y_1}} \right) = \left( {4,6} \right)\end{align}\]

Let,

\[\begin{align}\text{Q} \left( {{x_2},{y_2}} \right) = \left( {3,{\text{ }}2} \right)\end{align}\]

Let,

\[\begin{align}\text{R}\,\,\left( {{x_3},{y_3}} \right) = \left( {6,5} \right)\end{align}\]

Area of a triangle

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

By substituting the values of vertices \(P\), \(Q\), \(R\) in the Equation (1),

Area of triangle

\[\begin{align} PQR &= \frac{1}{2}\,\,\left[ {{{{x}}_1}\left( {{{{y}}_2} - {{{y}}_3}} \right) + {{{x}}_2}\left( {{{{y}}_3} - {{{y}}_1}} \right) + {{{x}}_3}\left( {{{{y}}_1} - {{{y}}_2}} \right)} \right]\\&= \frac{1}{2}\,\,[4(2 - 5) + 3(5 - 6) + 6(6 - 2)]\\&= \frac{1}{2}\,[ - 12 - 3 + 24]\\&= \frac{9}{2}\,\;\rm{}Square\;units\end{align}\]

(ii) Given,

- Taking \(C\) as origin, \(CB\) as \(x\)-axis, and \(CD\) as \(y\)-axis
- The coordinates of vertices \(P\), \(Q\), and \(R\) are \((12, 2)\), \((13, 6)\), and \((10, 3)\) respectively.
- Let \(\begin{align}\text{P}\left( {x_1,y1} \right) = \left( {12, 2} \right)\end{align}\)
- Let \(\begin{align}{\text{Q}}\left( {x_2,y2} \right) = \left( {13, 6} \right)\end{align}\)
- Let \(\begin{align}\text{R}\left( {x_3,y3} \right) = \left( {10,3} \right)\end{align}\)

Area of a triangle

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

By substituting the values of vertices \(P\), \(Q\), \(R\) in the Equation (\(1\)),

Area of triangle

\[\begin{align} PQR & \!=\! \frac{1}{2}\begin{bmatrix} {x_1}( {{y_2} \!-\! {y_3}} ) \!+\! {x_2}( {{y_3} \!-\! {y_1}} ) \!+\!\\ {x_3}( {{y_1} \!-\! {y_2} )} \end{bmatrix} \\&\!=\! \frac{1}{2}\begin{bmatrix} 12(6 \!- \!3) \!+\! 13(3 \!-\! 2)\!+\! \\ 10(2\! - 6) \end{bmatrix} \\ &\!= \!\frac{1}{2}\left[ {36 \!-\! 13 \!+\! 40} \right]\\ &\!=\! \frac{9}{2}\;\rm{}Square \;units \end{align}\]

It can be observed that the area of the triangle is same in both the cases.