Exercise 7.1 answers: Coordinate Geometry

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(i) (2, 3), (4, 1)

1. Use the distance formula: distance = √[(x₂−x₁)² + (y₂−y₁)²].
2. Here x₁ = 2, y₁ = 3, x₂ = 4, y₂ = 1.
3. Distance = √[(4−2)² + (1−3)²] = √[4 + 4] = √8.
4. √8 = √(4×2) = 2√2.

Answer2√2 units

(ii) (–5, 7), (–1, 3)

1. Here x₁ = −5, y₁ = 7, x₂ = −1, y₂ = 3.
2. Distance = √[(−1−(−5))² + (3−7)²] = √[4² + (−4)²] = √[16 + 16] = √32.
3. √32 = √(16×2) = 4√2.

Answer4√2 units

(iii) (a, b), (–a, –b)

1. Here x₁ = a, y₁ = b, x₂ = −a, y₂ = −b.
2. Distance = √[(−a−a)² + (−b−b)²] = √[(−2a)² + (−2b)²] = √[4a² + 4b²].
3. Take 4 common under the root: √4 × √(a² + b²) = 2√(a² + b²).

Answer2√(a² + b²) units

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1. Use the distance formula: distance = √[(x₂−x₁)² + (y₂−y₁)²].
2. Distance between (0, 0) and (36, 15) = √[(36−0)² + (15−0)²] = √[1296 + 225] = √1521 = 39.
3. In Section 7.2, town B is 36 km east and 15 km north of town A. Put town A at the origin (0, 0) with 1 km as one unit; then town B is the point (36, 15).
4. So the distance between the two towns is this same distance: 39 km.

AnswerThe distance is 39 units, so towns A and B are 39 km apart.

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1. Let A(1, 5), B(2, 3) and C(−2, −11). Find the three distances with the distance formula.
2. AB = √[(2−1)² + (3−5)²] = √[1 + 4] = √5.
3. BC = √[(−2−2)² + (−11−3)²] = √[16 + 196] = √212 = 2√53.
4. AC = √[(−2−1)² + (−11−5)²] = √[9 + 256] = √265. This is the longest of the three.
5. If the points were on one line, the two shorter distances would add up to the longest: AB + BC = AC.
6. Square AB + BC: (√5 + √212)² = 5 + 212 + 2√1060 = 217 + 2√1060. For this to equal AC² = 265 we would need 2√1060 = 48, that is √1060 = 24, or 1060 = 576, which is false.
7. So AB + BC ≠ AC (about 16.80 against 16.28), and the three points form a triangle rather than a straight line.

AnswerNo, the points are not collinear.

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1. Let A(5, −2), B(6, 4), C(7, −2). Find all three sides with the distance formula.
2. AB = √[(6−5)² + (4−(−2))²] = √[1 + 36] = √37.
3. BC = √[(7−6)² + (−2−4)²] = √[1 + 36] = √37.
4. CA = √[(5−7)² + (−2−(−2))²] = √[4 + 0] = 2.
5. AB = BC = √37, so two sides are equal.

AnswerYes, the triangle is isosceles (AB = BC = √37).

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1. Read the four seats off the grid in Fig. 7.8: A(3, 4), B(6, 7), C(9, 4), D(6, 1).
2. Find all four sides: AB = √[3² + 3²] = √18, BC = √[3² + 3²] = √18, CD = √[3² + 3²] = √18, DA = √[3² + 3²] = √18.
3. All four sides are equal, so ABCD is at least a rhombus — check the diagonals too.
4. AC = √[6² + 0²] = 6, BD = √[0² + 6²] = 6. The diagonals are also equal.
5. Equal sides and equal diagonals together mean ABCD is a square.

AnswerChampa is correct — ABCD is a square.

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(i) (–1, –2), (1, 0), (–1, 2), (–3, 0)

1. Let A(−1, −2), B(1, 0), C(−1, 2) and D(−3, 0), taken in order.
2. AB = √[(1+1)² + (0+2)²] = √8, BC = √[(−1−1)² + (2−0)²] = √8, CD = √[(−3+1)² + (0−2)²] = √8, DA = √[(−1+3)² + (−2−0)²] = √8.
3. All four sides are equal, so ABCD is at least a rhombus.
4. Diagonals: AC = √[(−1+1)² + (2+2)²] = √16 = 4 and BD = √[(−3−1)² + (0−0)²] = √16 = 4.
5. All four sides are equal and the two diagonals are equal, so ABCD is a square.

AnswerSquare

(ii) (–3, 5), (3, 1), (0, 3), (–1, –4)

1. Let A(−3, 5), B(3, 1), C(0, 3) and D(−1, −4).
2. AC = √[(0+3)² + (3−5)²] = √[9 + 4] = √13.
3. CB = √[(3−0)² + (1−3)²] = √[9 + 4] = √13.
4. AB = √[(3+3)² + (1−5)²] = √[36 + 16] = √52 = 2√13.
5. AC + CB = √13 + √13 = 2√13 = AB, so C lies on the segment AB: the points A, C and B are collinear.
6. Three of the four points lie on one straight line, so these four points do not form a quadrilateral.

AnswerNo quadrilateral is formed, because A, C and B are collinear.

(iii) (4, 5), (7, 6), (4, 3), (1, 2)

1. Let A(4, 5), B(7, 6), C(4, 3) and D(1, 2), taken in order.
2. AB = √[(7−4)² + (6−5)²] = √10, BC = √[(4−7)² + (3−6)²] = √18, CD = √[(1−4)² + (2−3)²] = √10, DA = √[(4−1)² + (5−2)²] = √18.
3. Opposite sides are equal (AB = CD and BC = DA), so ABCD is a parallelogram.
4. Diagonals: AC = √[(4−4)² + (3−5)²] = √4 = 2 and BD = √[(1−7)² + (2−6)²] = √52. They are not equal, so it is not a rectangle; and AB ≠ BC, so it is not a rhombus.
5. So ABCD is a parallelogram (and nothing more special).

AnswerParallelogram

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1. A point on the x-axis has the form (x, 0).
2. Its distance from (2, −5), squared, is (x−2)² + 25.
3. Its distance from (−2, 9), squared, is (x+2)² + 81.
4. Equate them: (x−2)² + 25 = (x+2)² + 81.
5. Expand: x² − 4x + 4 + 25 = x² + 4x + 4 + 81, so −4x + 29 = 4x + 85.
6. −8x = 56, so x = −7.

Answer(−7, 0)

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1. PQ² = (10−2)² + (y−(−3))² = 64 + (y+3)².
2. Set PQ² = 10² = 100, so (y+3)² = 36.
3. Take the square root of both sides: y + 3 = 6 or y + 3 = −6.
4. So y = 3 or y = −9. Both values give a real point, so keep both.

Answery = 3 or y = −9

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1. QP² = (5−0)² + (−3−1)² = 25 + 16 = 41.
2. QR² = (x−0)² + (6−1)² = x² + 25.
3. Q is equidistant from P and R, so QP² = QR²: 41 = x² + 25, giving x² = 16, so x = 4 or x = −4.
4. QR = √41 in both cases, since QR² = 41 either way.
5. For x = 4: PR = √[(4−5)² + (6−(−3))²] = √[1 + 81] = √82.
6. For x = −4: PR = √[(−4−5)² + (6−(−3))²] = √[81 + 81] = √162 = 9√2.

Answerx = 4 or x = −4; QR = √41; PR = √82 (when x = 4) or 9√2 (when x = −4).

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1. Equidistant means the two squared distances are equal: (x−3)² + (y−6)² = (x+3)² + (y−4)².
2. Expand the left side: x² − 6x + 9 + y² − 12y + 36.
3. Expand the right side: x² + 6x + 9 + y² − 8y + 16.
4. Subtract the right side from the left: −12x − 4y + 20 = 0.
5. Divide throughout by −4: 3x + y − 5 = 0.

Answer3x + y − 5 = 0 (that is, 3x + y = 5)

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