Exercise 9.3 answers: Straight Lines

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(i) x + 7y = 0

1. x + 7y = 0, so 7y = −x, giving y = (−1/7)x + 0.
2. Slope = −1/7, y-intercept = 0.

Answerslope = −1/7, y-intercept = 0

(ii) 6x + 3y – 5 = 0

1. 6x + 3y − 5 = 0, so 3y = −6x + 5, giving y = −2x + 5/3.
2. Slope = −2, y-intercept = 5/3.

Answerslope = −2, y-intercept = 5/3

(iii) y = 0

1. y = 0 is already y = 0·x + 0.
2. Slope = 0, y-intercept = 0.

Answerslope = 0, y-intercept = 0

Watch this explained “Going backwards”, 9:59 into Naming a line by where it crosses the axes

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(i) 3x + 2y – 12 = 0

1. 3x + 2y = 12. Divide by 12: x/4 + y/6 = 1.
2. x-intercept = 4, y-intercept = 6.

Answerx-intercept = 4, y-intercept = 6

(ii) 4x – 3y = 6

1. 4x − 3y = 6. Divide by 6: x/(3/2) + y/(−2) = 1.
2. x-intercept = 3/2, y-intercept = −2.

Answerx-intercept = 3/2, y-intercept = −2

(iii) 3y + 2 = 0

1. 3y + 2 = 0 gives y = −2/3, a line parallel to the x-axis.
2. It never crosses the x-axis, so there is no x-intercept.
3. y-intercept = −2/3.

Answerno x-intercept; y-intercept = −2/3

Watch this explained “Going backwards”, 9:59 into Naming a line by where it crosses the axes

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1. Expand the line: 12x + 72 = 5y − 10.
2. Bring everything to one side: 12x − 5y + 82 = 0.
3. Distance = |12(−1) − 5(1) + 82| ÷ √(12² + (−5)²).
4. = |−12 − 5 + 82| ÷ √(144 + 25) = |65| ÷ 13 = 5.

Answer5 units

Watch this explained “Put it in shape first”, 13:11 into The shortest route from a point to a line, read off its coefficients

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1. Clear fractions: multiply x/3 + y/4 = 1 by 12 to get 4x + 3y − 12 = 0.
2. A point on the x-axis is (x, 0). Its distance is |4x − 12| ÷ √(4² + 3²) = |4x − 12| ÷ 5.
3. Set this equal to 4: |4x − 12| = 20.
4. 4x − 12 = 20 gives x = 8. 4x − 12 = −20 gives x = −2.
5. The points are (8, 0) and (−2, 0).

Answer(8, 0) and (−2, 0)

Watch this explained “Two answers, and a fold”, 11:46 into The shortest route from a point to a line, read off its coefficients

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(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

1. Both lines have the same A = 15 and B = 8, so they are parallel. Here C₁ = −34 and C₂ = 31.
2. Distance = |C₁ − C₂| / √(A² + B²) = |−34 − 31| / √(225 + 64) = 65/√289 = 65/17.

Answer65/17 units

(ii) l (x + y) + p = 0 and l (x + y) – r = 0

1. Expand: lx + ly + p = 0 and lx + ly − r = 0. Both have A = l and B = l (with l ≠ 0), so they are parallel. Here C₁ = p and C₂ = −r.
2. Distance = |C₁ − C₂| / √(A² + B²) = |p − (−r)| / √(l² + l²) = |p + r| / √(2l²).
3. √(2l²) = √2 · |l|, because a square root is never negative.
4. So the distance is |p + r| / (√2 |l|). When l > 0 and p + r > 0, this is (p + r)/(√2 l).

Answer|p + r| / (√2 |l|) units

Watch this explained “Using it”, 10:16 into The gap between two parallel lines as one distance measured once

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1. A line parallel to 3x − 4y + 2 = 0 has the same leading terms: 3x − 4y + C = 0 for some C.
2. Put in the point (−2, 3): 3(−2) − 4(3) + C = 0, so −6 − 12 + C = 0, giving C = 18.
3. The line is 3x − 4y + 18 = 0.

Answer3x − 4y + 18 = 0

Watch this explained “Using it”, 10:16 into The gap between two parallel lines as one distance measured once

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1. Write the given line as y = x/7 + 5/7, so its slope is 1/7.
2. A line perpendicular to it has slope = −1 ÷ (1/7) = −7.
3. "x intercept 3" means the line passes through the point (3, 0).
4. Point-slope form: y − 0 = −7(x − 3).
5. Simplify: y = −7x + 21, i.e. 7x + y − 21 = 0.

Answer7x + y − 21 = 0

Watch this explained “Where the minus sign comes from”, 7:34 into Equal slopes mean parallel; slopes multiplying to minus one mean perpendicular

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1. Line 1: √3x + y = 1 has slope m₁ = −√3.
2. Line 2: x + √3y = 1 has slope m₂ = −1/√3.
3. tan θ = |(m₁ − m₂) ÷ (1 + m₁m₂)| = |(−√3 + 1/√3) ÷ 2| = 1/√3.
4. So θ = 30°.
5. The two angles made at the crossing add up to 180°, so the other angle is 150°.

Answer30° and 150°

Watch this explained “Angles, in the plural”, 12:33 into Recovering the angle between two lines from their two slopes

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1. Slope of the line through (h, 3) and (4, 1) is (1−3) ÷ (4−h) = −2/(4−h).
2. Slope of 7x − 9y − 19 = 0 is 7/9 (from y = 7x/9 − 19/9).
3. The two lines are perpendicular, so the product of their slopes is −1.
4. [−2/(4−h)] × (7/9) = −1.
5. Solve: −14 = −9(4−h) → −14 = −36 + 9h → 9h = 22 → h = 22/9.

Answerh = 22/9

Watch this explained “Four points, one missing number”, 10:55 into Equal slopes mean parallel; slopes multiplying to minus one mean perpendicular

Open NCERT p. 168One way to think about it

1. Case 1: B ≠ 0. The given line Ax + By + C = 0 has slope −A/B.
2. A parallel line has the same slope, so the line through (x₁, y₁) is y − y₁ = (−A/B)(x − x₁).
3. Multiply both sides by B: B(y − y₁) = −A(x − x₁), that is, A(x − x₁) + B(y − y₁) = 0.
4. Case 2: B = 0. Then A ≠ 0, and the given line is Ax + C = 0, i.e. x = −C/A, a line parallel to the y-axis.
5. The line parallel to it through (x₁, y₁) is x = x₁, i.e. A(x − x₁) = 0, which is A(x − x₁) + B(y − y₁) = 0 with B = 0.
6. So in every case the required line is A(x − x₁) + B(y − y₁) = 0.

In shortShown: the line through (x₁, y₁) parallel to Ax + By + C = 0 is A(x − x₁) + B(y − y₁) = 0.

Watch this explained “The same line, or not”, 7:08 into Naming Ax + By + C = 0, the one form the distance formula ahead will take

Open NCERT p. 168Checked by computerAnswers can differ: one example

1. Let the slope of the other line be m. The known line has slope 2.
2. The angle between them is 60°, so |(m − 2) ÷ (1 + 2m)| = tan 60° = √3.
3. Removing the modulus gives two cases: (m − 2) ÷ (1 + 2m) = √3 or (m − 2) ÷ (1 + 2m) = −√3.
4. Case 1: m − 2 = √3 + 2√3m, so m(1 − 2√3) = 2 + √3 and m = (2 + √3) ÷ (1 − 2√3).
5. Line through (2, 3): (2√3 − 1)(y − 3) = −(2 + √3)(x − 2), which simplifies to (2 + √3)x + (2√3 − 1)y − 1 − 8√3 = 0.
6. Case 2: m − 2 = −√3 − 2√3m, so m(1 + 2√3) = 2 − √3 and m = (2 − √3) ÷ (1 + 2√3).
7. Line through (2, 3): (1 + 2√3)(y − 3) = (2 − √3)(x − 2), which simplifies to (√3 − 2)x + (2√3 + 1)y + 1 − 8√3 = 0.
8. The 60° can open on either side of the known line, so both lines are correct answers.

Answer(2 + √3)x + (2√3 − 1)y − 1 − 8√3 = 0, or the other possible line (√3 − 2)x + (2√3 + 1)y + 1 − 8√3 = 0

Watch this explained “One angle, one slope, two lines”, 9:35 into Recovering the angle between two lines from their two slopes

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1. Midpoint of (3, 4) and (−1, 2) = ((3−1)/2, (4+2)/2) = (1, 3).
2. Slope of the segment = (2−4) ÷ (−1−3) = 1/2.
3. The right bisector is perpendicular to the segment, so its slope = −1 ÷ (1/2) = −2.
4. It passes through the midpoint (1, 3): y − 3 = −2(x − 1).
5. Simplify: y = −2x + 5, i.e. 2x + y − 5 = 0.

Answer2x + y − 5 = 0

Watch this explained “One equation, a great deal of work”, 11:27 into One point and a slope, or two points: the same condition written twice

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1. The line 3x − 4y − 16 = 0 has slope 3/4, so a line perpendicular to it has slope −4/3.
2. This perpendicular line passes through (−1, 3): y − 3 = −4/3 (x + 1).
3. Clear fractions: 3y − 9 = −4x − 4, i.e. 4x + 3y − 5 = 0.
4. Solve 3x − 4y − 16 = 0 and 4x + 3y − 5 = 0 together.
5. The solution is x = 68/25, y = −49/25 — this crossing point is the foot of the perpendicular.

Answer(68/25, −49/25)

Watch the lesson The shortest route from a point to a line, read off its coefficients

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1. (−1, 2) lies on the line, so 2 = m(−1) + c, i.e. c − m = 2.
2. The segment from the origin to (−1, 2) has slope 2 ÷ (−1) = −2.
3. This segment is perpendicular to the line, so (−2) × m = −1.
4. Solve: m = 1/2.
5. From c − m = 2: c = 2 + 1/2 = 5/2.

Answerm = 1/2, c = 5/2

Watch this explained “Where the minus sign comes from”, 7:34 into Equal slopes mean parallel; slopes multiplying to minus one mean perpendicular

Open NCERT p. 168One way to think about it

1. First line: x cosθ − y sinθ − k cos2θ = 0. Since cos²θ+sin²θ=1, distance from origin p = |k cos2θ|.
2. Second line: x secθ + y cosecθ − k = 0. Distance from origin q = |k| ÷ √(sec²θ + cosec²θ).
3. So q² = k² ÷ (sec²θ + cosec²θ) = k² sin²θ cos²θ, using sec²θ+cosec²θ = 1/(sin²θcos²θ).
4. Since sin2θ = 2 sinθ cosθ, we get 4q² = 4k² sin²θ cos²θ = k² sin²2θ.
5. Also p² = k² cos²2θ. Add: p² + 4q² = k²(cos²2θ + sin²2θ) = k².

In shortShown: p² + 4q² = k².

Watch this explained “What is left after the division”, 5:21 into The shortest route from a point to a line, read off its coefficients

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1. Slope of BC = (2 − (−1)) ÷ (1 − 4) = 3 ÷ (−3) = −1.
2. The altitude from A is perpendicular to BC, so its slope = −1 ÷ (−1) = 1.
3. It passes through A(2, 3): y − 3 = 1 × (x − 2), i.e. x − y + 1 = 0.
4. Equation of BC: y − (−1) = −1(x − 4), i.e. x + y − 3 = 0.
5. Length of altitude = distance from A(2, 3) to line x+y−3=0 = |2+3−3| ÷ √(1²+1²) = 2/√2 = √2.

AnswerEquation: x − y + 1 = 0; Length: √2 units

Watch this explained “One equation, a great deal of work”, 11:27 into One point and a slope, or two points: the same condition written twice

Open NCERT p. 168One way to think about it

1. The line cuts the x-axis at (a, 0) and the y-axis at (0, b), with a ≠ 0 and b ≠ 0, so its equation is x/a + y/b = 1.
2. Multiply by ab: bx + ay − ab = 0.
3. Distance from the origin: p = |b × 0 + a × 0 − ab| ÷ √(b² + a²) = |ab| ÷ √(a² + b²).
4. Square both sides: p² = a²b² ÷ (a² + b²).
5. Take reciprocals: 1/p² = (a² + b²) ÷ (a²b²) = a² ÷ (a²b²) + b² ÷ (a²b²) = 1/b² + 1/a².
6. So 1/p² = 1/a² + 1/b².

In shortShown: 1/p² = 1/a² + 1/b².

Watch this explained “A point and two crossings”, 1:40 into The shortest route from a point to a line, read off its coefficients