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Mathematics

Three friends Govind, Rishi and Kanika are participating in a Treasure Hunt organized in their school playground. The playground is mapped using a coordinate grid where each square represents 1 meter.

Use a graph paper for this question taking 1 cm = 1 unit along both x and y axes. Reflection, RSA Mathematics Solutions ICSE Class 10.

At a cerlain point in the game, they each stand at different spots waiting for their next clue. Their positions are recorded on the grid as points:

  • Govind is at point P

  • Rishi is at point Q

  • Kanika is at point R

The coordinate map is shown alongside.

Based on the above information, answer the following questions:

(i) Is Q the midpoint of segment PR? Justify your answer.

(ii) A new clue directs them to reach point M, which divides segment PQ in the ratio 2 : 3. Find the coordinates of M.

Section Formula

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Answer

(i) From graph,

Coordinates of P = (4, 13), Q = (7, 8) and R = (11, 2)

By mid-point formula,

Mid-point = (x1+x22,y1+y22)\Big(\dfrac{x1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

Substituting values we get :

Mid-point of PR =(4+112,13+22)=(152,152)=(7.5,7.5).\text{Mid-point of PR } = \Big(\dfrac{4 + 11}{2}, \dfrac{13 + 2}{2}\Big) \\[1em] = \Big(\dfrac{15}{2}, \dfrac{15}{2}\Big) \\[1em] = (7.5, 7.5).

Hence, Q is not the midpoint of segment PR.

(ii) Given,

M divides PQ in the ratio 2 : 3.

By section formula,

M = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

Substituting values we get :

M=(2×7+3×42+3,2×8+3×132+3)=(14+125,16+395)=(265,555)=(265,11).\Rightarrow M = \Big(\dfrac{2 \times 7 + 3 \times 4}{2 + 3}, \dfrac{2 \times 8 + 3 \times 13}{2 + 3}\Big) \\[1em] = \Big(\dfrac{14 + 12}{5}, \dfrac{16 + 39}{5}\Big) \\[1em] = \Big(\dfrac{26}{5}, \dfrac{55}{5}\Big) \\[1em] = \Big(\dfrac{26}{5}, 11\Big).

Hence, M = (265,11)\Big(\dfrac{26}{5}, 11\Big).

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