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

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## 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.
• \begin{align}\text{Let P}\left( {{x_1},{\text{ }}{y_1}} \right){\text{ }} = {\text{ }}\left( {4,{\text{ }}6} \right)\end{align}
• \begin{align}\text{Let Q} \left( {{x_2},{\text{ }}{y_2}} \right){\text{ }} = {\text{ }}\left( {3{\text{ }},{\text{ }}2} \right)\end{align}
• \begin{align}\text{Let R}\,\,\left( {{x_3},{\text{ }}{y_3}} \right){\text{ }} = {\text{ }}\left( {6,{\text{ }}5} \right)\end{align}

Area of a triangle\begin{align}=\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)\qquad \ldots . . \end{align} Equation(1)

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

Area of triangle\begin{align} PQR &= \frac{1}{2}\,\,\left[ {{{\text{x}}_1}\left( {{{\text{y}}_2} - {{\text{y}}_3}} \right) + {{\text{x}}_2}\left( {{{\text{y}}_3} - {{\text{y}}_1}} \right) + {{\text{x}}_3}\left( {{{\text{y}}_1} - {{\text{y}}_2}} \right)} \right] \end{align}

\begin{align} &= \frac{1}{2}\,\,[4(2 - 5) + 3(5 - 6) + 6(6 - 2)] \end{align}

\begin{align} &= \frac{1}{2}\,[ - 12 - 3 + 24] \end{align}

\begin{align} &= \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.
• \begin{align}\text{Let P}\left( {x1,{\text{ }}y1} \right) = \left( {12,{\text{ }}2} \right)\end{align}
• \begin{align}{\text{ Let Q}}\left( {x2,{\text{ }}y2} \right) = \left( {13,{\text{ }}6} \right)\end{align}
• \begin{align}\text{Let R}\left( {x3,{\text{ }}y3} \right) = \left( {10,3} \right)\end{align}

Area of a triangle \begin{align}=\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) \quad \ldots \ldots \end{align}Equation (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] \end{align}

\begin{align} &= \frac{1}{2}\,\,\,\left[ {12\,(6 - 3) + 13(3 - 2) + 10(2 - 6)} \right] \end{align}

\begin{align} &= \frac{1}{2}\,\,\left[ {36 - 13 + 40} \right] \end{align}

\begin{align} &= \frac{9}{2}\,\,\;\rm{}Square \;units \end{align}

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

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