NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines Exercise 9.3
NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.3 – Straight Lines focuses on solving problems involving the general equation of a straight line and applying the concepts of slope, intercepts, and parallel/perpendicular lines. This exercise is designed to test students' understanding of how to work with the general form of the equation of a straight line.
Prepared according to the latest CBSE Class 11 Maths syllabus, Exercise 9.3 covers solving problems related to the equation of a line in general form, the relationship between the slope of two lines, and finding the angle between two lines. Students will also explore the concept of parallel lines and perpendicular lines in detail.
NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines Exercise 9.3
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The solutions are provided in a step-by-step manner, helping students develop problem-solving strategies and understand the application of formulas to find the equation of lines in different scenarios.
Question 1. Reduce the following equations into slope-intercept form and find their slopes and y-intercepts.
i. x + 7y = 0
The equation is x + 7y = 0.
We can write it as: y = −(1/7)x + 0 … (1)
This is of the form y = mx + c, where m = −1/7 and c = 0.
Therefore, the slope is −1/7 and the y-intercept is 0.
ii. 6x + 3y − 5 = 0
The given equation is 6x + 3y − 5 = 0.
We can write it as: y = (1/3)(−6x + 5) ⟹ y = −2x + 5/3 … (1)
This is of the form y = mx + c, where m = −2 and c = 5/3.
Therefore, the slope is −2 and the y-intercept is 5/3.
iii. y = 0
The given equation is y = 0.
We can write it as: y = 0·x + 0 … (1)
This is of the form y = mx + c, where m = 0 and c = 0.
Therefore, the slope is 0 and the y-intercept is 0.
Question 2. Reduce the following equations into intercept form and find their intercepts on the axes.
i. 3x + 2y − 12 = 0
The given equation is 3x + 2y − 12 = 0.
We can write it as: 3x + 2y = 12 ⟹ (3x/12) + (2y/12) = 1 ⟹ x/4 + y/6 = 1 … (1)
This is of the form x/a + y/b = 1, where a = 4 and b = 6.
Therefore, the x-intercept is 4 and the y-intercept is 6.
ii. 4x − 3y = 6
The given equation is 4x − 3y = 6.
We can write it as: (4x/6) − (3y/6) = 1 ⟹ x/(3/2) + y/(−2) = 1 … (2)
Therefore, the x-intercept is 3/2 and the y-intercept is −2.
iii. 3y + 2 = 0
The given equation is 3y + 2 = 0.
We can write it as: 3y = −2 ⟹ y/(−2/3) = 1 … (3)
This is of the form x/a + y/b = 1, where a = 0 and b = −2/3.
Therefore, the y-intercept is −2/3. It has no x-intercept.
Question 3. Find the distance of the point (−1, 1) from the line 12(x + 6) = 5(y − 2).
The equation of the line is 12(x + 6) = 5(y − 2).
⟹ 12x + 72 = 5y − 10 ⟹ 12x − 5y + 82 = 0
Comparing with Ax + By + C = 0, we get A = 12, B = −5, C = 82.
The perpendicular distance formula is: d = |Ax₁ + By₁ + C| / √(A² + B²)
Distance from (−1, 1): d = |12(−1) + (−5)(1) + 82| / √(144 + 25) = |−12 − 5 + 82| / √169 = |65| / 13
Therefore, the distance = 5 units.
Question 4. Find the points on the x-axis at distance 4 units from the line x/3 + y/4 = 1.
The given line is x/3 + y/4 = 1, or 4x + 3y − 12 = 0.
Here A = 4, B = 3, C = −12. Let (a, 0) be the point on the x-axis.
4 = |4a + 3(0) − 12| / √(16 + 9) ⟹ 4 = |4a − 12| / 5 ⟹ |4a − 12| = 20 ⟹ ±(4a − 12) = 20
Case 1: 4a − 12 = 20 ⟹ 4a = 32 ⟹ a = 8 Case 2: −(4a − 12) = 20 ⟹ 4a = −8 ⟹ a = −2
Therefore, the required points are (−2, 0) and (8, 0).
Question 5. Find the distance between the following parallel lines.
i. 15x + 8y − 34 = 0 and 15x + 8y + 31 = 0
Here A = 15, B = 8, C₁ = −34, C₂ = 31.
d = |C₁ − C₂| / √(A² + B²) = |−34 − 31| / √(225 + 64) = |−65| / √289 = 65/17
Therefore, the distance between the parallel lines = 65/17 units.
ii. l(x + y) + p = 0 and l(x + y) − r = 0
The lines can be written as lx + ly + p = 0 and lx + ly − r = 0.
Here A = l, B = l, C₁ = p, C₂ = −r.
d = |C₁ − C₂| / √(A² + B²) = |p − (−r)| / √(l² + l²) = |p + r| / (l√2)
Therefore, d = (1/√2) · |p + r| / l units.
Question 6. Find the equation of the line parallel to 3x − 4y + 2 = 0 and passing through (−2, 3).
The slope of 3x − 4y + 2 = 0 is m = 3/4.
Since parallel lines have the same slope, the slope of the required line = 3/4.
Equation through (−2, 3) with slope 3/4: y − 3 = (3/4)(x + 2) ⟹ 4y − 12 = 3x + 6
Therefore, the equation of the required line is 3x − 4y + 18 = 0.
Question 7. Find the equation of the line perpendicular to x − 7y + 5 = 0 and having x-intercept 3.
The slope of x − 7y + 5 = 0 is 1/7.
Slope of the perpendicular line = −1/(1/7) = −7.
Equation with slope −7 and passing through (3, 0): y − 0 = −7(x − 3) ⟹ y = −7x + 21
Therefore, the equation of the required line is 7x + y = 21.
Question 8. Find the angle between the lines √3x + y = 1 and x + √3y = 1.
From the equations: y = −√3x + 1 ⟹ m₁ = −√3 y = −(1/√3)x + 1/√3 ⟹ m₂ = −1/√3
Using the angle formula: tan θ = |(m₁ − m₂) / (1 + m₁m₂)| = |(−√3 + 1/√3) / (1 + (−√3)(−1/√3))| = |(−3+1)/√3 / (1+1)| = |(−2/√3)/2| = 1/√3
⟹ θ = 30°
Therefore, the angle between the lines is 30° or 150°.
Question 9. The line through (h, 3) and (4, 1) intersects 7x − 9y − 19 = 0 at right angle. Find h.
Slope of line through (h, 3) and (4, 1): m₁ = (1 − 3)/(4 − h) = −2/(4 − h)
Slope of 7x − 9y − 19 = 0: m₂ = 7/9
Since the lines are perpendicular: m₁ × m₂ = −1
(−2/(4 − h)) × (7/9) = −1 ⟹ −14/(36 − 9h) = −1 ⟹ 14 = 36 − 9h ⟹ 9h = 22
Therefore, h = 22/9.
Question 10. Prove that the line through (x₁, y₁) parallel to Ax + By + C = 0 is A(x − x₁) + B(y − y₁) = 0.
Slope of Ax + By + C = 0 is m = −A/B.
Parallel lines have the same slope, so slope of required line = −A/B.
Equation through (x₁, y₁): y − y₁ = (−A/B)(x − x₁) ⟹ B(y − y₁) = −A(x − x₁) ⟹ A(x − x₁) + B(y − y₁) = 0
Hence proved.
Question 11. Two lines passing through (2, 3) intersect at an angle of 60°. If slope of one line is 2, find the equation of the other line.
Given m₁ = 2. Let slope of the other line be m₂.
Using tan 60° = |(m₁ − m₂)/(1 + m₁m₂)|: √3 = |(2 − m₂)/(1 + 2m₂)|
This gives: m₂ = (2 − √3)/(2√3 + 1) or m₂ = −(2 + √3)/(2√3 − 1)
Case 1: Equation through (2, 3): (√3 − 2)x + (2√3 + 1)y = −1 + 8√3
Case 2: Equation through (2, 3): (2 + √3)x + (2√3 − 1)y = 1 + 8√3
Question 12. Find the equation of the right bisector of the line segment joining (3, 4) and (−1, 2).
Midpoint of AB = ((3−1)/2, (4+2)/2) = (1, 3)
Slope of AB = (2 − 4)/(−1 − 3) = −2/−4 = 1/2
Slope of perpendicular bisector = −1/(1/2) = −2
Equation through (1, 3) with slope −2: y − 3 = −2(x − 1) ⟹ y − 3 = −2x + 2 ⟹ 2x + y = 5
Therefore, the equation of the right bisector is 2x + y = 5.
Question 13. Find the coordinates of the foot of the perpendicular from (−1, 3) to the line 3x − 4y − 16 = 0.
Let (a, b) be the foot of the perpendicular.
Slope of line joining (−1, 3) and (a, b): m₁ = (b − 3)/(a + 1) Slope of 3x − 4y − 16 = 0: m₂ = 3/4
Since perpendicular: m₁ × m₂ = −1 ⟹ (b − 3)/(a + 1) × 3/4 = −1 ⟹ 3b − 9 = −4a − 4 ⟹ 4a + 3b = 5 … (1)
Since (a, b) lies on 3x − 4y = 16: 3a − 4b = 16 … (2)
Solving (1) and (2): a = 68/25, b = −49/25
Therefore, the foot of perpendicular = (68/25, −49/25).
Question 14. The perpendicular from the origin to y = mx + c meets it at (−1, 2). Find m and c.
Slope of line joining (0, 0) and (−1, 2) = 2/(−1) = −2
Since this line is perpendicular to y = mx + c: m × (−2) = −1 ⟹ m = 1/2
Since (−1, 2) lies on y = mx + c: 2 = (1/2)(−1) + c ⟹ c = 2 + 1/2 = 5/2
Therefore, m = 1/2 and c = 5/2.
Question 15. If p and q are lengths of perpendiculars from the origin to the lines x cosθ − y sinθ = k cos2θ and x secθ + y cosecθ = k respectively, prove that p² + 4q² = k².
For line (1): x cosθ − y sinθ = k cos2θ
p = |−k cos2θ| / √(cos²θ + sin²θ) = |k cos2θ| … (3)
For line (2): x secθ + y cosecθ = k
q = |−k| / √(sec²θ + cosec²θ) = k sinθ cosθ … (4)
From (3) and (4): p² + 4q² = k²cos²2θ + 4k²sin²θcos²θ = k²cos²2θ + k²(2sinθcosθ)² = k²cos²2θ + k²sin²2θ = k²(cos²2θ + sin²2θ) = k²
Hence proved that p² + 4q² = k².
Question 16. In △ABC with vertices A(2, 3), B(4, −1) and C(1, 2), find the equation and length of the altitude from vertex A.
Let AD be the altitude from A. Since AD ⊥ BC:
Slope of BC = (2 − (−1))/(1 − 4) = 3/(−3) = −1
Slope of AD = −1/(−1) = 1
Equation of AD through A(2, 3): y − 3 = 1(x − 2) ⟹ x − y + 1 = 0 ⟹ y − x = 1
Equation of BC: x + y − 3 = 0
Length of AD = |1(2) + 1(3) − 3| / √(1² + 1²) = |2| / √2 = √2
Therefore, the equation of the altitude is y − x = 1 and its length is √2 units.
Question 17. If p is the length of the perpendicular from the origin to the line whose intercepts on the axes are a and b, show that 1/p² = 1/a² + 1/b².
The equation of the line with intercepts a and b is x/a + y/b = 1, or bx + ay − ab = 0.
p = |b(0) + a(0) − ab| / √(b² + a²) = ab / √(a² + b²)
Squaring both sides: p² = (ab)² / (a² + b²)
⟹ (a² + b²) / (a²b²) = 1/p²
⟹ 1/a² + 1/b² = 1/p²
Hence proved that 1/p² = 1/a² + 1/b².
FAQs – Class 11 Maths Chapter 9 Exercise 9.3 Straight Lines
Q1. What is the focus of Exercise 9.3?
Exercise 9.3 focuses on solving problems related to the general equation of a straight line, the slope of lines, and calculating the angle between two lines. It also includes questions on finding equations for parallel and perpendicular lines.
Q2. Why is Exercise 9.3 important for exams?
This exercise is important because it consolidates the knowledge of the general form of a straight line and its applications, which is essential for solving complex problems in coordinate geometry and further topics in mathematics.
Q3. How can students prepare effectively for Exercise 9.3?
Students should:
- Understand the general equation of a straight line and its properties.
- Practice calculating the slope of lines in different forms.
- Solve problems involving parallel and perpendicular lines.
- Work on problems involving the angle between two lines.