using a scale of 2cm to 2units draw on a sheet of graph paper two perpendicular axes Ox and Oy for the intervals -10</=x</=10 and -12</=y</=12 i. draw with vertices P(2,6) , Q(6,2) and R(4,2).
ii. Draw the image P1Q1R1 of triangle PQR under an enlargement from the origin with a scale factor of 1.5 such that P->P2,Q->Q1 and R->R1. iii. draw the image P2Q2R2 of triangle PQR under a reflection in the line y=2 such that P->P2, Q->Q2 and R-> R2.
iv. find the ratio of the area of PQR to the area of P1Q1R1
To draw the axes on the graph paper:
- Mark every 2 units on the x-axis from -10 to 10.
- Mark every 2 units on the y-axis from -12 to 12.
i. Draw a triangle with vertices P(2,6), Q(6,2), and R(4,2) on the graph paper.
ii. To find the image of P1Q1R1 after an enlargement from the origin with a scale factor of 1.5:
- P(2,6) -> P1(3,9)
- Q(6,2) -> Q1(9,3)
- R(4,2) -> R1(6,3)
Draw triangle P1Q1R1 on the graph paper.
iii. To find the image of P2Q2R2 after a reflection in the line y=2:
- The line y=2 is the x axis. So, the image of each point will be mirrored across the x-axis.
- P1(3,9) -> P2(3,-5)
- Q1(9,3) -> Q2(9,-1)
- R1(6,3) -> R2(6,-1)
Draw triangle P2Q2R2 on the graph paper.
iv. To find the ratio of the area of triangle PQR to the area of triangle P1Q1R1:
- Area of PQR = 1/2 * base * height = 1/2 * 4 * 4 = 8 square units
- Area of P1Q1R1 = 1/2 * base * height = 1/2 * 6 * 6 = 18 square units
- Ratio = Area of PQR / Area of P1Q1R1 = 8 / 18 = 4 / 9.