Exercise 7.2 answers: Coordinate Geometry

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1. Use the section formula: a point dividing A(x₁,y₁) and B(x₂,y₂) in the ratio m₁:m₂ is ((m₁x₂+m₂x₁)/(m₁+m₂), (m₁y₂+m₂y₁)/(m₁+m₂)).
2. Here A(−1, 7), B(4, −3), m₁ = 2, m₂ = 3.
3. x = (2×4 + 3×(−1))/(2+3) = (8−3)/5 = 1.
4. y = (2×(−3) + 3×7)/(2+3) = (−6+21)/5 = 3.

Answer(1, 3)

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1. The two points of trisection cut the segment into three equal parts, so the nearer one divides it in the ratio 1:2 and the farther one in the ratio 2:1.
2. For the ratio 1:2 from A(4,−1) to B(−2,−3): x = (1×(−2)+2×4)/3 = 6/3 = 2, y = (1×(−3)+2×(−1))/3 = −5/3.
3. For the ratio 2:1: x = (2×(−2)+1×4)/3 = 0, y = (2×(−3)+1×(−1))/3 = −7/3.

Answer(2, −5/3) and (0, −7/3)

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1. Take A as the origin, with the numbered chalk lines along the x-axis and the 100 flower-pot positions along AD as the y-axis, so AD = 100 m.
2. Niharika stands on the 2nd line, ¼ of the way along AD: her green flag is at (2, ¼×100) = (2, 25).
3. Preet stands on the 8th line, ⅕ of the way along AD: her red flag is at (8, ⅕×100) = (8, 20).
4. Distance between the flags = √[(8−2)² + (20−25)²] = √[36 + 25] = √61.
5. Rashmi's blue flag is at the midpoint: ((2+8)/2, (25+20)/2) = (5, 45/2).

AnswerThe flags are √61 m apart; Rashmi should post the blue flag at (5, 22.5).

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1. Let the point (−1, 6) divide the segment joining (−3, 10) and (6, −8) in the ratio k : 1.
2. By the section formula, the x-coordinate of this point is (6k + (−3))/(k + 1), and it must equal −1.
3. So 6k − 3 = −(k + 1) = −k − 1, which gives 7k = 2, so k = 2/7.
4. Check with the y-coordinate: (−8k + 10)/(k + 1) = (−16/7 + 10)/(9/7) = (54/7)/(9/7) = 6, which is the given y-coordinate. So (−1, 6) does lie on the segment.
5. The ratio is 2/7 : 1, which is the same as 2 : 7.

AnswerThe point divides the segment in the ratio 2 : 7.

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1. Let the x-axis divide AB in the ratio k : 1. On the x-axis, y = 0.
2. By the section formula: 0 = (5k + (−5))/(k+1), so 5k − 5 = 0, giving k = 1.
3. So the ratio is 1 : 1 — the x-axis bisects AB.
4. The x-coordinate: x = (1×(−4) + 1×1)/(1+1) = −3/2.

AnswerRatio 1 : 1; the point of division is (−3/2, 0).

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1. Let A(1,2), B(4,y), C(x,6), D(3,5). In a parallelogram, the diagonals AC and BD bisect each other, so their midpoints are equal.
2. Midpoint of AC = ((1+x)/2, (2+6)/2) = ((1+x)/2, 4).
3. Midpoint of BD = ((4+3)/2, (y+5)/2) = (7/2, (y+5)/2).
4. Equate x-coordinates: (1+x)/2 = 7/2, so x = 6.
5. Equate y-coordinates: 4 = (y+5)/2, so y = 3.

Answerx = 6, y = 3

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1. The centre of a circle is the midpoint of any diameter, so the centre is the midpoint of A and B.
2. Let A = (x, y). Then (2, −3) = ((x+1)/2, (y+4)/2).
3. From the first coordinate: (x+1)/2 = 2, so x = 3.
4. From the second coordinate: (y+4)/2 = −3, so y = −10.

AnswerA = (3, −10)

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1. AP is 3/7 of AB, so the remaining part PB is 4/7 of AB — a fraction of the whole is not the same as a ratio of the two pieces.
2. So AP : PB = 3 : 4.
3. By the section formula with A(−2,−2), B(2,−4), m₁=3, m₂=4: x = (3×2 + 4×(−2))/7 = −2/7.
4. y = (3×(−4) + 4×(−2))/7 = −20/7.

AnswerP = (−2/7, −20/7)

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1. Four equal parts need three dividing points, at ratios 1:3, 1:1 (the midpoint) and 3:1 from A to B.
2. At 1:3: x = (1×2+3×(−2))/4 = −1, y = (1×8+3×2)/4 = 7/2.
3. At 1:1 (midpoint): x = (−2+2)/2 = 0, y = (2+8)/2 = 5.
4. At 3:1: x = (3×2+1×(−2))/4 = 1, y = (3×8+1×2)/4 = 13/2.

Answer(−1, 7/2), (0, 5), (1, 13/2)

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1. Let A(3,0), B(4,5), C(−1,4), D(−2,−1). The diagonals are AC and BD.
2. AC = √[(3−(−1))² + (0−4)²] = √[16+16] = √32 = 4√2.
3. BD = √[(4−(−2))² + (5−(−1))²] = √[36+36] = √72 = 6√2.
4. Area = ½ × AC × BD = ½ × 4√2 × 6√2 = ½ × 48 = 24.

Answer24 square units

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