Miscellaneous Exercise answers: Straight Lines

Open NCERT p. 172Checked by computer

(a) Parallel to the x-axis

1. A line parallel to the x-axis has no x in its equation, so the coefficient of x must be 0.
2. Coefficient of x is (k − 3), so k − 3 = 0, i.e. k = 3.
3. Check: at k=3 the coefficient of y is −(4−9) = 5, which is not 0, so this is really a line.

Answerk = 3

(b) Parallel to the y-axis

1. A line parallel to the y-axis has no y in its equation, so the coefficient of y must be 0.
2. Coefficient of y is −(4 − k²), so 4 − k² = 0, i.e. k = 2 or k = −2.
3. Check: for both values the coefficient of x, (k−3), is not 0, so both give genuine lines.

Answerk = 2 or k = −2

(c) Passing through the origin

1. A line passes through the origin when x=0, y=0 satisfies the equation.
2. This needs the constant term to be 0: k² − 7k + 6 = 0.
3. Factorise: (k − 1)(k − 6) = 0, so k = 1 or k = 6.

Answerk = 1 or k = 6

Watch this explained “One equation with a letter in it”, 9:31 into Naming Ax + By + C = 0, the one form the distance formula ahead will take

Open NCERT p. 172Matches NCERT’s answer

1. Let the x-intercept be a and the y-intercept be b, so a + b = 1 and ab = −6.
2. a and b are roots of t² − t − 6 = 0, i.e. (t−3)(t+2) = 0, so {a, b} = {3, −2}.
3. Case 1 (a=3, b=−2): x/3 + y/(−2) = 1, which clears to 2x − 3y − 6 = 0.
4. Case 2 (a=−2, b=3): x/(−2) + y/3 = 1, which clears to 3x − 2y + 6 = 0.

Answer2x − 3y − 6 = 0 or 3x − 2y + 6 = 0

Watch this explained “Sums and products”, 12:18 into Naming a line by where it crosses the axes

Open NCERT p. 173Matches NCERT’s answer

1. Clear the intercept form: x/3 + y/4 = 1 becomes 4x + 3y − 12 = 0.
2. A point on the y-axis has the form (0, y).
3. Distance from (0, y) to the line = |4(0)+3y−12| ÷ √(4²+3²) = |3y−12| ÷ 5.
4. Set this equal to 4: |3y − 12| = 20.
5. So 3y−12 = 20 (giving y = 32/3) or 3y−12 = −20 (giving y = −8/3).

Answer(0, 32/3) and (0, −8/3)

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

Open NCERT p. 173Matches NCERT’s answer

1. Two-point form through (cos θ, sin θ) and (cos φ, sin φ): (y − sin θ)(cos φ − cos θ) = (x − cos θ)(sin φ − sin θ).
2. Gather terms: (sin φ − sin θ)x − (cos φ − cos θ)y + (sin θ cos φ − cos θ sin φ) = 0, and sin θ cos φ − cos θ sin φ = sin(θ − φ).
3. Distance from the origin: d = |sin(θ − φ)| ÷ √((sin φ − sin θ)² + (cos φ − cos θ)²).
4. Under the root: sin²φ + sin²θ + cos²φ + cos²θ − 2(sin φ sin θ + cos φ cos θ) = 2 − 2cos(θ − φ) = 4 sin²((θ − φ)/2), so the root is 2|sin((θ − φ)/2)|.
5. On top: |sin(θ − φ)| = 2|sin((θ − φ)/2)| × |cos((θ − φ)/2)|.
6. The two points are different, so sin((θ − φ)/2) ≠ 0 and it cancels: d = |cos((θ − φ)/2)|.

Answer|cos((θ − φ)/2)|

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

Open NCERT p. 173Matches NCERT’s answer

1. Solve x − 7y + 5 = 0 and 3x + y = 0 together to find where they cross.
2. From the second equation, y = −3x. Substitute: x − 7(−3x) + 5 = 0 → 22x + 5 = 0 → x = −5/22.
3. A line parallel to the y-axis has the form x = constant.
4. So the required line is x = −5/22, i.e. 22x + 5 = 0.

Answer22x + 5 = 0

Watch this explained “A crossing, and one coordinate of it”, 8:32 into Lines parallel to an axis, where one coordinate never changes

Open NCERT p. 173Matches NCERT’s answer

1. Find where x/4 + y/6 = 1 meets the y-axis: put x = 0, giving y = 6, so the point is (0, 6).
2. Clear the given line: x/4 + y/6 = 1 becomes 3x + 2y − 12 = 0, with slope −3/2.
3. A perpendicular line has slope = −1 ÷ (−3/2) = 2/3.
4. Equation through (0, 6): y − 6 = (2/3)(x − 0).
5. Clear the fraction: 3y − 18 = 2x, i.e. 2x − 3y + 18 = 0.

Answer2x − 3y + 18 = 0

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. 173Matches NCERT’s answer

1. The three lines are y = x, y = −x, and x = k.
2. y=x meets y=−x at the origin (0, 0).
3. y=x meets x=k at (k, k). y=−x meets x=k at (k, −k).
4. So the triangle's corners are (0,0), (k,k), (k,−k).
5. Area = 1/2 |x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)| = 1/2 |0(2k) + k(−k) + k(−k)| = 1/2 × 2k² = k².

Answerk² square units

Watch this explained “An area that is really a test”, 0:40 into Steepness as the tangent of an angle, and the one line that has none

Open NCERT p. 173Matches NCERT’s answer

1. Two of the lines don't have p in them: 3x + y − 2 = 0 and 2x − y − 3 = 0. Find where these two cross first.
2. Adding the two equations cancels y: 5x − 5 = 0, so x = 1. Then y = 2 − 3x = −1.
3. For all three lines to meet at one point, the third line px + 2y − 3 = 0 must also pass through (1, −1).
4. Put x = 1, y = −1 into it: p(1) + 2(−1) − 3 = 0, which gives p − 5 = 0.
5. So p = 5.

Answerp = 5

Watch this explained “Three lines, one point”, 10:51 into Naming Ax + By + C = 0, the one form the distance formula ahead will take

Open NCERT p. 173One way to think about it

1. Concurrent means all three lines pass through one common point. Call that point (x₀, y₀).
2. Since (x₀, y₀) lies on each line: y₀ − m₁x₀ = c₁, y₀ − m₂x₀ = c₂, and y₀ − m₃x₀ = c₃.
3. Use these to write each bracket on the left of the claim in terms of x₀ alone: c₂ − c₃ = (m₃ − m₂)x₀, c₃ − c₁ = (m₁ − m₃)x₀, and c₁ − c₂ = (m₂ − m₁)x₀.
4. Multiply the first by m₁, the second by m₂, the third by m₃, and add: m₁(c₂ − c₃) + m₂(c₃ − c₁) + m₃(c₁ − c₂) = x₀[ m₁(m₃ − m₂) + m₂(m₁ − m₃) + m₃(m₂ − m₁) ].
5. Expand the bracket: m₁m₃ − m₁m₂ + m₁m₂ − m₂m₃ + m₂m₃ − m₁m₃. Every term cancels with another, so the bracket is 0.
6. So the whole right-hand side is x₀ × 0 = 0, which proves m₁(c₂ − c₃) + m₂(c₃ − c₁) + m₃(c₁ − c₂) = 0.

In shortm₁(c₂ − c₃) + m₂(c₃ − c₁) + m₃(c₁ − c₂) = 0, proved.

Watch this explained “Three lines, one point”, 10:51 into Naming Ax + By + C = 0, the one form the distance formula ahead will take

Open NCERT p. 173Matches NCERT’s answer

1. Given: the angle is 45° and the line is x – 2y = 3.
2. Write the given line as y = x/2 − 3/2, so its slope is m₁ = 1/2.
3. Let the slope of the line we want be m. The angle between two lines uses tan θ = |(m − m₁)/(1 + m·m₁)|.
4. Put θ = 45°, so tan θ = 1: (m − 1/2)/(1 + m/2) = 1 or (m − 1/2)/(1 + m/2) = −1.
5. First case: m − 1/2 = 1 + m/2, so m/2 = 3/2, giving m = 3.
6. Second case: m − 1/2 = −1 − m/2, so (3/2)m = −1/2, giving m = −1/3.
7. Line through (3, 2) with slope 3: y − 2 = 3(x − 3), which tidies to 3x − y − 7 = 0.
8. Line through (3, 2) with slope −1/3: y − 2 = −(1/3)(x − 3), which tidies to x + 3y − 9 = 0.

Answer3x − y − 7 = 0 and x + 3y − 9 = 0

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

Open NCERT p. 173Matches NCERT’s answer

1. Solve the two given lines together to find where they cross: 4x + 7y − 3 = 0 and 2x − 3y + 1 = 0 meet at (1/13, 5/13).
2. A line with equal, non-zero intercepts a on both axes can be written as x + y = a, i.e. x + y − a = 0.
3. This line must pass through (1/13, 5/13), so a = 1/13 + 5/13 = 6/13.
4. So the line is x + y − 6/13 = 0. Multiplying through by 13 clears the fraction: 13x + 13y − 6 = 0.

Answer13x + 13y − 6 = 0

Watch this explained “Equal crossings”, 11:00 into Naming a line by where it crosses the axes

Open NCERT p. 173One way to think about it

1. Show: the line through the origin making an angle θ with y = mx + c is y/x = (m ± tan θ)/(1 ∓ m tan θ).
2. Let the line we want have slope k. Since it passes through the origin, its equation is y = kx, i.e. y/x = k.
3. The angle between two lines of slopes k and m satisfies tan θ = (k − m)/(1 + km) or tan θ = (m − k)/(1 + km), depending on which line is measured from which.
4. First case: tan θ (1 + km) = k − m, so k − k·m tanθ = m + tanθ, giving k(1 − m tanθ) = m + tanθ, so k = (m + tanθ)/(1 − m tanθ).
5. Second case: tan θ (1 + km) = m − k, so m − tanθ = k + k·m tanθ, giving k(1 + m tanθ) = m − tanθ, so k = (m − tanθ)/(1 + m tanθ).
6. Together, k = (m ± tanθ)/(1 ∓ m tanθ), and since y/x = k, this proves y/x = (m ± tanθ)/(1 ∓ m tanθ).

In shorty/x = (m ± tanθ)/(1 ∓ m tanθ), proved.

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

Open NCERT p. 173Matches NCERT’s answer

1. Suppose the line x + y = 4 cuts the join of (−1, 1) and (5, 7) in the ratio k : 1.
2. By the section formula, the cutting point is ( (5k − 1)/(k+1), (7k + 1)/(k+1) ).
3. This point lies on x + y = 4, so (5k − 1)/(k+1) + (7k + 1)/(k+1) = 4.
4. Combine the fractions: (12k)/(k+1) = 4, so 12k = 4k + 4, giving 8k = 4, so k = 1/2.
5. A ratio of k : 1 = 1/2 : 1 is the same as 1 : 2.

Answer1 : 2

Open NCERT p. 173Matches NCERT’s answer

1. "Along the line 2x − y = 0" means: move from (1, 2) in the direction of that line (slope 2) until you hit 4x + 7y + 5 = 0, and measure that path.
2. Points on the path through (1, 2) with slope 2 can be written as (1 + t, 2 + 2t) for some number t.
3. Put this into 4x + 7y + 5 = 0: 4(1 + t) + 7(2 + 2t) + 5 = 0, which gives 18t + 23 = 0, so t = −23/18.
4. The meeting point is (1 − 23/18, 2 − 46/18) = (−5/18, −5/9).
5. The distance from (1, 2) to this point is √[(23/18)² + (23·2/18)²] = (23/18)√(1 + 4) = 23√5/18.

Answer23√5 / 18 units

Watch this explained “Measured some other way”, 14:53 into The shortest route from a point to a line, read off its coefficients

Open NCERT p. 173Checked by computer

1. The point where the new line meets x + y = 4 lies on that line, so call it (x, 4 − x).
2. Its distance from (−1, 2) must be 3: (x + 1)² + (4 − x − 2)² = 9, that is (x + 1)² + (2 − x)² = 9.
3. Expand: x² + 2x + 1 + 4 − 4x + x² = 9, so 2x² − 2x − 4 = 0, i.e. x² − x − 2 = 0, which factors as (x − 2)(x + 1) = 0.
4. So x = 2, giving the meeting point (2, 2), or x = −1, giving the meeting point (−1, 5).
5. The line from (−1, 2) to (2, 2) keeps y = 2, so it is parallel to the x-axis (angle 0°). The line from (−1, 2) to (−1, 5) keeps x = −1, so it is parallel to the y-axis (angle 90°).
6. Both meeting points are exactly 3 units from (−1, 2), so both directions work.

AnswerThe line must be drawn parallel to the x-axis (angle 0°) or parallel to the y-axis (angle 90°).

Watch this explained “Measured some other way”, 14:53 into The shortest route from a point to a line, read off its coefficients

Open NCERT p. 173Checked by computer

1. The legs are parallel to the axes, so one leg is level (y stays fixed) and the other is upright (x stays fixed). They meet at the right-angle corner.
2. Each end of the hypotenuse lies on one leg, so the corner takes its x from one end and its y from the other. That gives two possible corners: (−4, 3) or (1, 1).
3. Corner at (−4, 3): the level leg joins (1, 3) and (−4, 3), so it is y = 3, i.e. y − 3 = 0. The upright leg joins (−4, 1) and (−4, 3), so it is x = −4, i.e. x + 4 = 0.
4. Corner at (1, 1): the level leg joins (−4, 1) and (1, 1), so it is y − 1 = 0. The upright leg joins (1, 3) and (1, 1), so it is x − 1 = 0.
5. Both triangles have the given hypotenuse and legs parallel to the axes, so both pairs are correct. The first pair is y − 3 = 0 and x + 4 = 0.

Answery − 3 = 0 (parallel to the x-axis) and x + 4 = 0 (parallel to the y-axis); the other possible triangle gives y − 1 = 0 and x − 1 = 0.

Watch this explained “Two answers where the question expects one”, 9:18 into Lines parallel to an axis, where one coordinate never changes

Open NCERT p. 173Matches NCERT’s answer

1. Write the mirror x + 3y = 7 as y = −x/3 + 7/3, so its slope is −1/3.
2. The ray from (3, 8) to its image meets the mirror at a right angle, so it has slope 3 (because 3 × (−1/3) = −1).
3. That line through (3, 8) is y − 8 = 3(x − 3), i.e. y = 3x − 1.
4. It meets the mirror where x + 3(3x − 1) = 7, so 10x = 10, x = 1 and y = 2. This is the foot of the perpendicular, (1, 2).
5. The foot is the midpoint of (3, 8) and its image (h, k): (3 + h)/2 = 1 and (8 + k)/2 = 2, so h = −1 and k = −4.

AnswerThe image is (−1, −4).

Open NCERT p. 173Checked by computerAnswers can differ: one example

1. The two given lines have slopes 3 and 1/2 (from 2y = x + 3, i.e. y = x/2 + 3/2).
2. "Equally inclined" to the line of slope m means the two angles made with it have the same tangent size: |(3 − m)/(1 + 3m)| = |(1/2 − m)/(1 + m/2)|.
3. Equal sizes with opposite signs (equal, same sign, forces the two given lines to have the same slope, which they don't) means (3 − m)/(1 + 3m) = −(1/2 − m)/(1 + m/2).
4. Cross-multiply and tidy: this becomes a quadratic in m, 7m² − 2m − 7 = 0.
5. Using the quadratic formula: m = [2 ± √(4 + 4·7·7)]/(2·7) = [2 ± √200]/14 = [2 ± 10√2]/14 = (1 ± 5√2)/7.

Answerm = (1 + 5√2)/7 or m = (1 − 5√2)/7

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

Open NCERT p. 173One way to think about it

1. The distance from P(x, y) to x + y − 5 = 0 is |x + y − 5|/√2, and to 3x − 2y + 7 = 0 it is |3x − 2y + 7|/√13.
2. The condition is |x + y − 5|/√2 + |3x − 2y + 7|/√13 = 10.
3. The expression x + y − 5 is positive on one side of its line and negative on the other, and the same is true of 3x − 2y + 7. So while P stays on one side of each line, each expression keeps one sign and its bars can be replaced by a plain + or −.
4. Take P on the side where both expressions are positive: (x + y − 5)/√2 + (3x − 2y + 7)/√13 = 10.
5. Multiply by √26 = √2 × √13: √13(x + y − 5) + √2(3x − 2y + 7) = 10√26, that is (√13 + 3√2)x + (√13 − 2√2)y + (7√2 − 5√13 − 10√26) = 0.
6. The coefficient of x, √13 + 3√2, is not 0, so this is a first-degree equation: a straight line. Any other choice of signs changes only signs in the same working, and the x-coefficient √13 ± 3√2 is again not 0, so again a line. Hence P must move on a line.

In shortP moves on the line (√13 + 3√2)x + (√13 − 2√2)y + (7√2 − 5√13 − 10√26) = 0 (taking P where both expressions are positive), so P moves on a line.

Watch this explained “The sign thrown away”, 10:44 into The shortest route from a point to a line, read off its coefficients

Open NCERT p. 174Matches NCERT’s answer

1. Divide the first line by 3 so both lines share the same left-hand coefficients: 3x + 2y − 7/3 = 0 and 3x + 2y + 6 = 0.
2. A line midway between two parallel lines 3x + 2y + c₁ = 0 and 3x + 2y + c₂ = 0 is 3x + 2y + (c₁+c₂)/2 = 0.
3. Here (c₁ + c₂)/2 = (−7/3 + 6)/2 = (11/3)/2 = 11/6.
4. So the middle line is 3x + 2y + 11/6 = 0. Multiplying by 6 clears the fraction: 18x + 12y + 11 = 0.

Answer18x + 12y + 11 = 0

Watch this explained “The line down the middle”, 8:56 into The gap between two parallel lines as one distance measured once

Open NCERT p. 174Matches NCERT’s answer

1. When a ray reflects off the x-axis, the reflected ray behaves as if it came from the mirror image of the source, reflected across the x-axis.
2. The mirror image of (1, 2) in the x-axis is (1, −2).
3. So A must lie on the straight line joining (1, −2) to the point the reflected ray reaches, (5, 3), and A must also lie on the x-axis (y = 0).
4. The line through (1, −2) and (5, 3) has slope (3 − (−2))/(5 − 1) = 5/4, giving y + 2 = (5/4)(x − 1).
5. Set y = 0: 2 = (5/4)(x − 1), so x − 1 = 8/5, giving x = 13/5.
6. So A = (13/5, 0).

AnswerA = (13/5, 0)

Open NCERT p. 174One way to think about it

1. Write the line as (cosθ/a)x + (sinθ/b)y − 1 = 0. Let X = √(a² − b²) for short, and let D = √[(cosθ/a)² + (sinθ/b)²].
2. Distance from (X, 0): d₁ = |X cosθ/a − 1| / D. Distance from (−X, 0): d₂ = |−X cosθ/a − 1| / D.
3. Multiply the insides of the bars: (X cosθ/a − 1)(−X cosθ/a − 1) = 1 − X²cos²θ/a², so d₁d₂ = |1 − X²cos²θ/a²| / D².
4. Since X² = a² − b², 1 − X²cos²θ/a² = 1 − cos²θ + b²cos²θ/a² = sin²θ + b²cos²θ/a² = (a²sin²θ + b²cos²θ)/a². This is positive, so the bars can be dropped.
5. Also D² = cos²θ/a² + sin²θ/b² = (b²cos²θ + a²sin²θ)/(a²b²).
6. Divide: d₁d₂ = [(a²sin²θ + b²cos²θ)/a²] ÷ [(a²sin²θ + b²cos²θ)/(a²b²)] = b².

In shortThe product of the two perpendicular lengths is b², proved.

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

Open NCERT p. 174Matches NCERT’s answer

1. First find the junction: solve 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0 together. This gives x = −1/17, y = 22/17.
2. The quickest route from a point to a line is always the perpendicular from that point onto the line — any other path is longer.
3. The target path is 6x − 7y + 8 = 0, with slope 6/7, so the perpendicular path must have slope −7/6.
4. The path through (−1/17, 22/17) with slope −7/6 is y − 22/17 = −(7/6)(x + 1/17).
5. Clearing fractions and tidying this gives 119x + 102y − 125 = 0.

Answer119x + 102y − 125 = 0

Watch this explained “And it stands square”, 0:54 into The shortest route from a point to a line, read off its coefficients