NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 covers the various forms of equations of straight lines. Students learn to write equations using slope-point form, two-point form, slope-intercept form, intercept form, and normal form. This exercise also includes real-life application problems involving linear relationships. The step-by-step solutions help students build a strong understanding of coordinate geometry concepts required for board exams.
Question 1. Write the equation for the x and y-axes.
The y-coordinate of every point on the x-axis is 0. Therefore, the equation of the x-axis is y = 0.
NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 – Straight Lines
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Find the distance between P (x1, y1) and Q (x2, y2) when : (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.
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The owner of a milk store finds that, he can sell 980 litres of milk each week at ₹14/litre and 1220 litres of milk each week at ₹16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹17/litre?
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Find the distance between parallel lines
(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0
(ii) l (x + y) + p = 0 and l (x + y) – r = 0.
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Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).
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Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts.
(i) x + 7y = 0,
(ii) 6x + 3y – 5 = 0,
(iii) y = 0.
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By using the concept of equation of a line, prove that the three points (3, 0), (– 2, – 2) and (8, 2) are collinear.
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Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.
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The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.
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Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P(0, – 4) and B(8, 0).
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Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
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Consider the following population and year graph (Fig 10.10), find the slope of the line AB and using it, find what will be the population in the year 2010?
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Find the angle between the x-axis and the line joining the points (3,–1) and (4,–2).
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Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.
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Find the value of x for which the points (x, – 1), (2, 1) and (4, 5) are collinear.
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Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
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Find the equation of the line through the intersection of lines x + 2y – 3 = 0 and 4x – y + 7 = 0 and which is parallel to 5x + 4y – 20 = 0
NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 – Straight Lines
The x-coordinate of every point on the y-axis is 0. Therefore, the equation of the y-axis is x = 0.
Question 2. Find the equation of the line which passes through the point (−4, 3) with slope 1/2.
We know that the equation of the line passing through point (x₀, y₀) with slope m is: y − y₀ = m(x − x₀)
The equation of the line passing through (−4, 3) with slope 1/2: y − 3 = (1/2)(x + 4) ⟹ 2(y − 3) = (x + 4) ⟹ 2y − 6 = x + 4
Therefore, the equation of the line is x − 2y + 10 = 0.
Question 3. Find the equation of the line which passes through (0, 0) with slope m.
We know that the equation of the line passing through (x₀, y₀) with slope m is: y − y₀ = m(x − x₀)
The equation of the line passing through (0, 0) with slope m: y − 0 = m(x − 0)
Therefore, the equation of the line passing through origin is y = mx.
Question 4. Find the equation of the line which passes through (2, 2√3) and is inclined with the x-axis at an angle of 75°.
The slope of the line inclined at 75° with the x-axis: m = tan 75° = tan(45° + 30°)
Using the addition formula: m = (tan 45° + tan 30°) / (1 − tan 45° · tan 30°) = (1 + 1/√3) / (1 − 1/√3) = (√3 + 1) / (√3 − 1)
Rationalising: = (√3 + 1)² / (√3 − 1)(√3 + 1) = (3 + 1 + 2√3) / (3 − 1) = (4 + 2√3) / 2 = 2 + √3
Equation through (2, 2√3) with slope (2 + √3): y − 2√3 = (2 + √3)(x − 2) y − 2√3 = 2x − 4 + √3x − 2√3 y = 2x − 4 + √3x
Therefore, the equation of the line is (2 + √3)x − y − 4 = 0.
Question 5. Find the equation of the line which intersects the x-axis at a distance of 3 units to the left of origin with slope −2.
If a line with slope m makes x-intercept d, then the equation is: y = m(x − d)
Here, d = −3 (3 units to the left) and m = −2.
y = −2[x − (−3)] ⟹ y = −2x − 6
Therefore, the equation of the line is 2x + y + 6 = 0.
Question 6. Find the equation of the line which intersects the y-axis at a distance of 2 units above the origin and makes an angle of 30° with the positive direction of the x-axis.
If a line with slope m makes y-intercept c, then: y = mx + c
Here, c = 2 and m = tan 30° = 1/√3.
y = (1/√3)x + 2 ⟹ √3y = x + 2√3 ⟹ √3y − x − 2√3 = 0
Therefore, the equation of the line is x − √3y + 2√3 = 0.
Question 7. Find the slope of the line which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
If a line makes an angle of 30° with the positive y-axis, then it makes an angle of (90° + 30°) = 120° with the positive x-axis.
Slope = tan 120° = tan(180° − 60°) = −tan 60° = −√3
Therefore, the slope of the line is −√3.
Question 8. Find the equation of the line which is at a perpendicular distance of 5 units from the origin and the angle made by the perpendicular with the positive x-axis is 30°.
The normal form of a line is: x cosω + y sinω = p
Here, p = 5 and ω = 30°.
x cos 30° + y sin 30° = 5 x(√3/2) + y(1/2) = 5 √3x + y = 10
Therefore, the equation of the line is √3x + y − 10 = 0.
Question 9. The vertices of △PQR are P(2, 1), Q(−2, 3) and R(4, 5). Find equation of the median through the vertex R.
Let RL be the median through R. Then L is the midpoint of PQ.
By midpoint formula: L = ((2 + (−2))/2, (1 + 3)/2) = (0, 2)
Equation of line through R(4, 5) and L(0, 2): y − 5 = ((2 − 5)/(0 − 4))(x − 4) y − 5 = (−3/−4)(x − 4) y − 5 = (3/4)(x − 4) 4(y − 5) = 3(x − 4) 4y − 20 = 3x − 12
Therefore, the equation of the median through R is 3x − 4y + 8 = 0.
Question 10. Find the equation of the line passing through (−3, 5) and perpendicular to the line through the points (2, 5) and (−3, 6).
Slope of the line through (2, 5) and (−3, 6): m = (6 − 5)/(−3 − 2) = 1/(−5) = −1/5
Since two perpendicular lines have slopes that are negative reciprocals: Slope of required line = −1/(−1/5) = 5
Equation through (−3, 5) with slope 5: y − 5 = 5(x + 3) y − 5 = 5x + 15
Therefore, the equation of the line is 5x − y + 20 = 0.
Question 11. A line perpendicular to the line segment joining (1, 0) and (2, 3) divides it in the ratio 1 : n. Find the equation of the line.
By section formula, the point dividing (1, 0) and (2, 3) in ratio 1 : n is: ((n·1 + 1·2)/(1 + n), (n·0 + 1·3)/(1 + n)) = ((n + 2)/(n + 1), 3/(n + 1))
Slope of line joining (1, 0) and (2, 3) = (3 − 0)/(2 − 1) = 3
Slope of perpendicular line = −1/3
Equation through ((n + 2)/(n + 1), 3/(n + 1)) with slope −1/3: y − 3/(n + 1) = (−1/3)(x − (n + 2)/(n + 1)) 3[(n + 1)y − 3] = −[x(n + 1) − (n + 2)] 3(n + 1)y − 9 = −(n + 1)x + (n + 2) (n + 1)x + 3(n + 1)y = n + 2 + 9
Therefore, the equation of the line is (1 + n)x + 3(1 + n)y = n + 11.
Question 12. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through (2, 3).
The intercept form of a line is: x/a + y/b = 1
Since equal intercepts: a = b ⟹ x/a + y/a = 1 ⟹ x + y = a
Since the line passes through (2, 3): 2 + 3 = a ⟹ a = 5
Therefore, the equation of the line is x + y = 5.
Question 13. Find the equation of the line passing through (2, 2) and cutting off intercepts on the axes whose sum is 9.
The intercept form: x/a + y/b = 1, where a + b = 9, so b = 9 − a.
x/a + y/(9 − a) = 1
Since the line passes through (2, 2): 2/a + 2/(9 − a) = 1 ⟹ 2(9 − a + a) / (a(9 − a)) = 1 ⟹ 18 = 9a − a² ⟹ a² − 9a + 18 = 0 ⟹ (a − 6)(a − 3) = 0 ⟹ a = 6 or a = 3
If a = 6, then b = 3: x/6 + y/3 = 1 ⟹ x + 2y − 6 = 0
If a = 3, then b = 6: x/3 + y/6 = 1 ⟹ 2x + y − 6 = 0
Therefore, the equations of the line are x + 2y − 6 = 0 or 2x + y − 6 = 0.
Question 14. Find equation of the line through (0, 2) making an angle 2π/3 with the positive x-axis. Also find the equation of the line parallel to it and crossing the y-axis at a distance of 2 units below the origin.
Slope of line making angle 2π/3 with positive x-axis: m = tan(2π/3) = −√3
Equation through (0, 2) with slope −√3: y − 2 = −√3(x − 0) ⟹ √3x + y − 2 = 0
For the parallel line (slope = −√3), crossing y-axis 2 units below origin means it passes through (0, −2): y − (−2) = −√3(x − 0) y + 2 = −√3x ⟹ √3x + y + 2 = 0
Therefore, the equations are √3x + y − 2 = 0 and √3x + y + 2 = 0.
Question 15. The perpendicular from the origin to a line meets it at the point (−2, 9). Find the equation of the line.
Slope of line joining origin (0, 0) and (−2, 9): m₁ = (9 − 0)/(−2 − 0) = −9/2
Since the required line is perpendicular to this: m₂ = −1/m₁ = −1/(−9/2) = 2/9
Equation through (−2, 9) with slope 2/9: y − 9 = (2/9)(x + 2) 9y − 81 = 2x + 4
Therefore, the equation of the line is 2x − 9y + 85 = 0.
Question 16. The length L (in cm) of a copper rod is a linear function of its Celsius temperature C. If L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.
The two points satisfying the linear relation are (20, 124.942) and (110, 125.134).
Using the two-point form: (L − 124.942) = ((125.134 − 124.942)/(110 − 20))(C − 20) (L − 124.942) = (0.192/90)(C − 20)
Therefore, L = (0.192/90)(C − 20) + 124.942.
Question 17. The owner of a milk store finds he can sell 980 litres at Rs 14/litre and 1220 litres at Rs 16/litre. Assuming a linear relationship, how many litres could he sell weekly at Rs 17/litre?
The two points are (14, 980) and (16, 1220).
Equation of the line: y − 980 = ((1220 − 980)/(16 − 14))(x − 14) y − 980 = (240/2)(x − 14) y − 980 = 120(x − 14) y = 120(x − 14) + 980
When x = 17: y = 120(17 − 14) + 980 = 120 × 3 + 980 = 360 + 980 = 1340
Therefore, the owner could sell 1340 litres of milk weekly at Rs 17/litre.
Question 18. P(a, b) is the mid-point of a line segment between axes. Show that the equation of the line is x/a + y/b = 2.
Let A(0, y) and B(x, 0) be the endpoints of the line segment on the axes.
Since P(a, b) is the midpoint of AB: (0 + x)/2 = a ⟹ x = 2a (y + 0)/2 = b ⟹ y = 2b
So A = (0, 2b) and B = (2a, 0).
Equation of the line through (0, 2b) and (2a, 0): (y − 2b) = ((0 − 2b)/(2a − 0))(x − 0) y − 2b = (−b/a)x ay − 2ab = −bx bx + ay = 2ab
Dividing both sides by ab: x/a + y/b = 2
Hence proved that x/a + y/b = 2.
Question 19. Point R(h, k) divides a line segment between the axes in the ratio 1 : 2. Find equation of the line.
Let A(x, 0) and B(0, y) be the endpoints on the axes.
Since R(h, k) divides AB in ratio 1 : 2, by the section formula: (1·0 + 2·x)/(1 + 2) = h ⟹ 2x/3 = h ⟹ x = 3h/2 (1·y + 2·0)/(1 + 2) = k ⟹ y/3 = k ⟹ y = 3k
So A = (3h/2, 0) and B = (0, 3k).
Equation of line through (3h/2, 0) and (0, 3k): y − 0 = ((3k − 0)/(0 − 3h/2))(x − 3h/2) y = (−2k/h)(x − 3h/2) hy = −2kx + 3hk
Therefore, the equation of the line is 2kx + hy = 3hk.
Question 20. By using the concept of equation of a line, prove that the three points (3, 0), (−2, −2) and (8, 2) are collinear.
To prove collinearity, it is sufficient to show that the line passing through (3, 0) and (−2, −2) also passes through (8, 2).
Equation of line through (3, 0) and (−2, −2): y − 0 = ((−2 − 0)/(−2 − 3))(x − 3) y = (−2/−5)(x − 3) 5y = 2(x − 3) 5y = 2x − 6 i.e., 2x − 5y = 6
Checking point (8, 2): LHS = 2(8) − 5(2) = 16 − 10 = 6 = RHS ✓
Therefore, the line passing through (3, 0) and (−2, −2) also passes through (8, 2).
Hence, the points (3, 0), (−2, −2) and (8, 2) are collinear.
Frequently Asked Questions (FAQs)
Q1. What is the slope-point form of the equation of a line? The slope-point form is y − y₁ = m(x − x₁), where (x₁, y₁) is a point on the line and m is its slope. It is used when the slope and one point on the line are known.
Q2. What is the normal form of a line? The normal form is x cosω + y sinω = p, where p is the perpendicular distance from the origin to the line and ω is the angle the perpendicular makes with the positive x-axis.
Q3. How do you find the equation of a median of a triangle? First, find the midpoint of the opposite side using the midpoint formula. Then write the equation of the line joining the vertex and that midpoint using the two-point form.
Q4. What does it mean for three points to be collinear? Three points are collinear if they all lie on the same straight line. To verify, find the equation of the line through any two points and check whether the third point satisfies that equation.
Q5. How many questions are in NCERT Class 11 Maths Exercise 9.2? Exercise 9.2 contains 20 questions covering various forms of equations of straight lines including slope-point form, two-point form, intercept form, normal form, and application-based problems.