Tharunya was thrilled to know that the football tournament is fixed with a monthly timeframe from 20th July to 20th August 2023 and for the first time in the FIFA Women's World Cup's history, two nations host in 10 venues. Her father felt that the game can be better understood if the position of players is represented as points on a coordinate plane. (i) At an instance, the midfielders and forward formed a parallelogram. Find the position of the central midfielder (D) if the position of other players who formed the parallelogram are: A(1,2), B(4,3) and C(6,6). (ii) Check if the Goal keeper G(-3,5), Sweeper H(3,1) and Wing-back K(0,3) fall on a same straight line. (ii) Check if the Full-back J(5,-3) and centre-back I(-4,6) are equidistant from forward C(0,1) and if C is the mid-point of IJ. (iii) If Defensive midfielder A(1,4), Attacking midfielder B(2,-3) and Striker E(a,b) lie on the same straight line and B is equidistant from A and E, find the position of E.

MathsCBSE Class 10 · Standard · Sample paper, session 2023-24 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard sample paper 2023-24: question 37

From the official CBSE sample question paper for 2023-24, section E, question 37, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q37 · 4 marks · Case-basedCompetency-basedCoordinate Geometry

Tharunya was thrilled to know that the football tournament is fixed with a monthly timeframe from 20th July to 20th August 2023 and for the first time in the FIFA Women's World Cup's history, two nations host in 10 venues. Her father felt that the game can be better understood if the position of players is represented as points on a coordinate plane.

(i)
At an instance, the midfielders and forward formed a parallelogram. Find the position of the central midfielder (D) if the position of other players who formed the parallelogram are: A(1,2), B(4,3) and C(6,6).
1 mark
(ii)
Check if the Goal keeper G(-3,5), Sweeper H(3,1) and Wing-back K(0,3) fall on a same straight line.
2 marks
OR
(ii)
Check if the Full-back J(5,-3) and centre-back I(-4,6) are equidistant from forward C(0,1) and if C is the mid-point of IJ.
2 marks
(iii)
If Defensive midfielder A(1,4), Attacking midfielder B(2,-3) and Striker E(a,b) lie on the same straight line and B is equidistant from A and E, find the position of E.
1 mark

Figure in the paper: Football pitch diagram with player positions labelled: J Full-back, I Centre-back, G Goalkeeper, H Sweeper, K Wing-back, A Defensive midfielder, D Central midfielder, B Attacking midfielder, C Forward, E Striker, F Winger. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

Open one hint at a time and try again before the next.

Hint 1

In a parallelogram the diagonals bisect each other.

Hint 2

For collinearity, check whether the two shorter distances add up to the longest one.

Hint 3

If B is on line AE and equidistant from A and E, B must be the midpoint of AE.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Let $D = (a, b)$ . The diagonals of a parallelogram bisect each other, so midpoint of AC = midpoint of BD:
$(\frac{1 + 6}{2}, \frac{2 + 6}{2}) = (\frac{4 + a}{2}, \frac{3 + b}{2})$
½ mark
(i) $4 + a = 7 \Rightarrow a = 3$ and $3 + b = 8 \Rightarrow b = 5$ .
The central midfielder is at D(3, 5).
½ mark
(ii)(a) $G H = (3 + 3)^{2} + (1 - 5)^{2} = 36 + 16 = 52 = 213$
½ mark
(ii)(a) $G K = (0 + 3)^{2} + (3 - 5)^{2} = 9 + 4 = 13$
½ mark
(ii)(a) $H K = (3 - 0)^{2} + (1 - 3)^{2} = 13$
½ mark
(ii)(a) $G K + H K = 13 + 13 = 213 = G H$ .
So G, H and K lie on the same straight line, with K between G and H.
½ mark
(ii)(b) $C J = (5 - 0)^{2} + (- 3 - 1)^{2} = 25 + 16 = 41$
½ mark
(ii)(b) $C I = (- 4 - 0)^{2} + (6 - 1)^{2} = 16 + 25 = 41$ .
So $C J = C I$ : J and I are equidistant from C.
½ mark
(ii)(b) Midpoint of IJ $= (\frac{5 - 4}{2}, \frac{- 3 + 6}{2}) = (\frac{1}{2}, \frac{3}{2})$
½ mark
(ii)(b) $(\frac{1}{2}, \frac{3}{2}) \neq = (0, 1)$ , so C is not the midpoint of IJ.
½ mark
(iii) A, B, E are collinear and $A B = B E$ , so B is the midpoint of AE:
$(\frac{1 + a}{2}, \frac{4 + b}{2}) = (2, - 3)$
½ mark
(iii) $1 + a = 4 \Rightarrow a = 3$ and $4 + b = - 6 \Rightarrow b = - 10$ .
So E is (3, −10).
½ mark

Answer: (i) D(3, 5); (ii)(a) Yes, collinear since GK + HK = GH; (ii)(b) Equidistant (both √41), but C is not the midpoint of IJ (midpoint is (1/2, 3/2)); (iii) E(3, −10)

Where the marks are

(i) Equates the midpoints of the diagonals (½ mark)
(i) Gets D(3,5) (½ mark)
(ii)(a) GH = 2√13 (½ mark)
(ii)(a) GK = √13 (½ mark)
(ii)(a) HK = √13 (½ mark)
(ii)(a) GK + HK = GH, so collinear (½ mark)
(ii)(b) CJ = √41 (½ mark)
(ii)(b) CI = √41, so equidistant (½ mark)
(ii)(b) Midpoint of IJ = (1/2, 3/2) (½ mark)
(ii)(b) Concludes C is not the midpoint (½ mark)
(iii) B is the midpoint of AE; sets up the equations (½ mark)
(iii) Gets E(3, −10) (½ mark)

Revise the chapter

Chapter: Coordinate Geometry (notes, formulas, where marks are lost and every past question)
Topic: Finding the Ratio and Unknowns with Section Formula

Practise this chapterBoard-style questions from Coordinate Geometry, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2023-24 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard sample paper 2023-24: question 37

Hints

Step-by-step solution

Revise the chapter

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