131
A garden is in the shape of a square. The gardener grew saplings of Ashoka tree on the boundary of the garden at the distance of $1 \text{ m}$ from each other. He wants to decorate the garden with rose plants. He chose a triangular region inside the garden to grow rose plants. In the above situation, the gardener took help from the students of class $10$. They made a chart for it which looks like the given figure.
Based on the above, answer the following questions :
(i) If A is taken as origin, what are the coordinates of the vertices of $\triangle PQR$?
(ii) (a) Find distances PQ and QR.
OR
(b) Find the coordinates of the point which divides the line segment joining points P and R in the ratio $2: 1$ internally.
(iii) Find out if $\triangle PQR$ is an isosceles triangle.
Based on the above, answer the following questions :
(i) If A is taken as origin, what are the coordinates of the vertices of $\triangle PQR$?
(ii) (a) Find distances PQ and QR.
OR
(b) Find the coordinates of the point which divides the line segment joining points P and R in the ratio $2: 1$ internally.
(iii) Find out if $\triangle PQR$ is an isosceles triangle.
Show SolutionHide Solution↓
(i) $$\begin{aligned}& P (4,6), Q (3, 2), R (6, 5) \\ & (ii) (a) PQ = \sqrt{(4-3)^2 + (6 - 2)^2} = \sqrt{17} \\ & QR = \sqrt{(3 - 6)^2 + (2 - 5)^2} = \sqrt{18} \\ & \text{OR} \\ & (b) \text{ The coordinate of required point are } \left( \frac{6\times 2+1\times 4}{3}, \frac{5\times 2+1\times 6}{3} \right) \\ & \text{i.e. } \left( \frac{16}{3}, \frac{16}{3} \right) \\ & (iii) PQ = \sqrt{(4-3)^2 + (6-2)^2} = \sqrt{17} \\ & QR = \sqrt{(3 - 6)^2 + (2 - 5)^2} = \sqrt{18} \\ & PR = \sqrt{(4 - 6)^2 + (6-5)^2} = \sqrt{5} \\ & PQ \neq QR \neq PR \\ & \triangle PQR \text{ is not isosceles}\end{aligned}$$