Exercise 9.2 answers: Straight Lines

Open NCERT p. 163Matches NCERT’s answer

1. Every point on the x-axis has y-coordinate 0, whatever its x-coordinate is.
2. So the equation true for exactly the points on the x-axis is y = 0.
3. In the same way, every point on the y-axis has x-coordinate 0, so its equation is x = 0.

Answerx-axis: y = 0. y-axis: x = 0.

Watch the lesson Lines parallel to an axis, where one coordinate never changes

Open NCERT p. 163Matches NCERT’s answer

1. Use the point-slope form y − y₁ = m(x − x₁) with (x₁, y₁) = (−4, 3) and m = 1/2.
2. y − 3 = (1/2)(x − (−4)) = (1/2)(x + 4).
3. Multiply both sides by 2: 2y − 6 = x + 4.
4. Rearranged: x − 2y + 10 = 0.

Answerx − 2y + 10 = 0.

Watch this explained “One worked case”, 5:11 into One point and a slope, or two points: the same condition written twice

Open NCERT p. 163Matches NCERT’s answer

1. Use the point-slope form y − y₁ = m(x − x₁) with (x₁, y₁) = (0, 0).
2. y − 0 = m(x − 0), so y = mx.
3. Rearranged: mx − y = 0.

Answery = mx, i.e. mx − y = 0.

Watch this explained “One worked case”, 5:11 into One point and a slope, or two points: the same condition written twice

Open NCERT p. 163Matches NCERT’s answer

1. The slope is the tangent of the inclination: m = tan 75°.
2. Write 75° = 45° + 30° and use the tan sum formula: tan75° = (1 + 1/√3)/(1 − 1/√3) = (√3+1)/(√3−1).
3. Rationalising gives tan75° = 2 + √3.
4. Use the point-slope form with (x₁, y₁) = (2, 2√3): y − 2√3 = (2+√3)(x − 2).
5. Expand: y − 2√3 = (2+√3)x − 4 − 2√3, so y = (2+√3)x − 4.
6. Rearranged: (2+√3)x − y − 4 = 0.

Answer(2 + √3)x − y − 4 = 0.

Watch this explained “Given an angle instead of a point”, 10:47 into Steepness as the tangent of an angle, and the one line that has none

Open NCERT p. 163Matches NCERT’s answer

1. A point 3 units to the left of the origin, on the x-axis, is (−3, 0).
2. Use the point-slope form: y − 0 = −2(x − (−3)) = −2(x+3).
3. y = −2x − 6.
4. Rearranged: 2x + y + 6 = 0.

Answer2x + y + 6 = 0.

Watch this explained “The other crossing”, 2:30 into Naming a line by where it crosses the axes

Open NCERT p. 163Matches NCERT’s answer

1. Given: the line makes an angle of 30° with the positive direction of the x-axis.
2. A point 2 units above the origin, on the y-axis, is (0, 2).
3. The slope is tan 30° = 1/√3.
4. Point-slope form: y − 2 = (1/√3)(x − 0).
5. Multiply both sides by √3: √3y − 2√3 = x.
6. Rearranged: x − √3y + 2√3 = 0.

Answerx − √3y + 2√3 = 0.

Watch this explained “Given an angle instead of a point”, 10:47 into Steepness as the tangent of an angle, and the one line that has none

Open NCERT p. 164Matches NCERT’s answer

1. Slope = (−4 − 1)/(2 − (−1)) = −5/3.
2. Point-slope form with (x₁, y₁) = (−1, 1): y − 1 = (−5/3)(x − (−1)) = (−5/3)(x+1).
3. Multiply both sides by 3: 3y − 3 = −5x − 5.
4. Rearranged: 5x + 3y + 2 = 0.

Answer5x + 3y + 2 = 0.

Watch this explained “Two points, and no slope at all”, 5:51 into One point and a slope, or two points: the same condition written twice

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1. A median joins a vertex to the midpoint of the opposite side.
2. Midpoint of PQ = ((2 + (−2))/2, (1 + 3)/2) = (0, 2).
3. The median through R joins R (4, 5) and (0, 2).
4. Slope = (5 − 2) / (4 − 0) = 3/4.
5. Using point (0, 2): y − 2 = (3/4)(x − 0).
6. Multiply by 4 and rearrange: 3x − 4y + 8 = 0.

Answer3x − 4y + 8 = 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. Slope of the line through (2, 5) and (−3, 6) = (6 − 5)/(−3 − 2) = −1/5.
2. A perpendicular line has slope = −1 ÷ (−1/5) = 5.
3. Use point (−3, 5) with slope 5: y − 5 = 5(x + 3).
4. Expand and rearrange: 5x − y + 20 = 0.

Answer5x − y + 20 = 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

Open NCERT p. 164Matches NCERT’s answer

1. Slope of the segment from (1, 0) to (2, 3) = (3 − 0)/(2 − 1) = 3.
2. A perpendicular line has slope = −1/3.
3. The dividing point, using ratio 1 : n from (1, 0) to (2, 3), is ((n + 2)/(n + 1), 3/(n + 1)).
4. Write y − 3/(n + 1) = (−1/3)(x − (n + 2)/(n + 1)).
5. Clear the fractions and collect terms: (n + 1)x + 3(n + 1)y − (n + 11) = 0.

Answer(n + 1)x + 3(n + 1)y − (n + 11) = 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. Let both intercepts be a (a ≠ 0). The intercept form x/a + y/b = 1 becomes x/a + y/a = 1, that is x + y = a.
2. The line passes through (2, 3): 2 + 3 = a, so a = 5.
3. The line is x + y = 5, that is x + y − 5 = 0.
4. Note: the line through the origin and (2, 3), which is 3x − 2y = 0, meets both axes at the origin, so both its intercepts are 0. Taking 'cuts off equal intercepts' to mean non-zero intercepts, as the intercept form needs, the answer is the single line x + y − 5 = 0.

Answerx + y − 5 = 0

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

Open NCERT p. 164Checked by computerAnswers can differ: one example

1. Let the intercepts be a and b, so the line is x/a + y/b = 1, with a + b = 9.
2. Put in the point (2, 2): 2/a + 2/b = 1.
3. Substitute b = 9 − a and clear fractions: a² − 9a + 18 = 0.
4. Factor: (a − 3)(a − 6) = 0, so a = 3 or a = 6.
5. a = 3 gives b = 6: x/3 + y/6 = 1, that is 2x + y − 6 = 0.
6. a = 6 gives b = 3: x/6 + y/3 = 1, that is x + 2y − 6 = 0.
7. Both lines meet all the conditions, so both are valid answers.

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

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

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1. Slope = tan(2π/3) = −√3.
2. The point (0, 2) is on the y-axis, so it is the y-intercept: y = −√3 x + 2.
3. Rearranged: √3 x + y − 2 = 0.
4. A parallel line has the same slope, −√3.
5. "2 units below the origin" on the y-axis means y-intercept = −2: y = −√3 x − 2.
6. Rearranged: √3 x + y + 2 = 0.

Answer√3 x + y − 2 = 0; the parallel line is √3 x + y + 2 = 0

Watch this explained “Given an angle instead of a point”, 10:47 into Steepness as the tangent of an angle, and the one line that has none

Open NCERT p. 164Matches NCERT’s answer

1. The segment from the origin to (−2, 9) is perpendicular to the line, since (−2, 9) is the foot of the perpendicular.
2. Slope of that segment = 9/(−2) = −9/2.
3. So the line's own slope = −1 ÷ (−9/2) = 2/9.
4. Use point (−2, 9) with slope 2/9: y − 9 = (2/9)(x + 2).
5. Multiply by 9 and rearrange: 2x − 9y + 85 = 0.

Answer2x − 9y + 85 = 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. L is a linear function of C, so treat (C, L) as points (20, 124.942) and (110, 125.134).
2. Slope = (125.134 − 124.942)/(110 − 20) = 0.192/90 = 4/1875.
3. Using point (20, 124.942): L − 124.942 = (4/1875)(C − 20).
4. Simplify to get L in terms of C: L = 4C/1875 + 187349/1500 (≈ 4C/1875 + 124.899).

AnswerL = (4/1875)C + 187349/1500

Watch this explained “Two measurements, one straight law”, 12:38 into One point and a slope, or two points: the same condition written twice

Open NCERT p. 164Matches NCERT’s answer

1. Treat (price, litres) as two points: (14, 980) and (16, 1220).
2. Slope = (1220 − 980)/(16 − 14) = 240/2 = 120 litres for every rupee rise in price.
3. Using point (14, 980): litres = 980 + 120 × (price − 14).
4. At price 17: litres = 980 + 120 × 3 = 980 + 360 = 1340.

Answer1340 litres

Watch this explained “Two measurements, one straight law”, 12:38 into One point and a slope, or two points: the same condition written twice

Open NCERT p. 164One way to think about it

1. Show: x/a + y/b = 2.
2. Let the line meet the x-axis at (p, 0) and the y-axis at (0, q).
3. P (a, b) is the midpoint of these two points: a = p/2 and b = q/2, so p = 2a and q = 2b.
4. Write the intercept form of the line using these intercepts: x/(2a) + y/(2b) = 1.
5. Multiply both sides by 2: x/a + y/b = 2, which is the required equation.

In shortShown: x/a + y/b = 2.

Watch this explained “Both crossings at once”, 4:14 into Naming a line by where it crosses the axes

Open NCERT p. 164Checked by computerReads two ways: both answers shown

1. Let the line meet the x-axis at A(a, 0) and the y-axis at B(0, b). Its equation is x/a + y/b = 1.
2. The book does not say from which end the ratio 1 : 2 is measured. NCERT's answer key measures it from the x-axis end, AR : RB = 1 : 2, so that reading leads.
3. AR : RB = 1 : 2: by the section formula, R = ((1 × 0 + 2 × a)/3, (1 × b + 2 × 0)/3) = (2a/3, b/3).
4. So h = 2a/3 and k = b/3, giving a = 3h/2 and b = 3k.
5. The line is 2x/(3h) + y/(3k) = 1. Multiply by 3hk: 2kx + hy = 3hk.
6. BR : RA = 1 : 2 (measured from the y-axis end): R = (a/3, 2b/3), so a = 3h and b = 3k/2.
7. The line is x/(3h) + 2y/(3k) = 1. Multiply by 3hk: kx + 2hy = 3hk.

AnswerRead as AR : RB = 1 : 2 measured from the x-axis end (NCERT's answer key): 2kx + hy = 3hk. Read as measured from the y-axis end: kx + 2hy = 3hk.

Watch this explained “Both crossings at once”, 4:14 into Naming a line by where it crosses the axes

Open NCERT p. 164One way to think about it

1. Find the equation of the line through the first two points, (3, 0) and (−2, −2).
2. Slope = (−2 − 0)/(−2 − 3) = 2/5.
3. Using point (3, 0): y − 0 = (2/5)(x − 3).
4. Multiply by 5 and rearrange: 2x − 5y − 6 = 0.
5. Check the third point (8, 2): 2(8) − 5(2) − 6 = 16 − 10 − 6 = 0.
6. Since (8, 2) satisfies the same equation, all three points lie on one line — they are collinear.

In shortShown: (8, 2) lies on 2x − 5y − 6 = 0, the line through (3, 0) and (−2, −2), so the three points are collinear.

Watch this explained “Two points, and no slope at all”, 5:51 into One point and a slope, or two points: the same condition written twice