NCERT Solutions for Class 11 Maths Chapter 9
Chapter 9.1: Introduction
- Cartesian Plane: A plane where points are located using coordinates (x, y).
- Distance Formula: The distance between two points (x1, y1) and (x2, y2) is given by:
- d = √((x2 – x1)^2 + (y2 – y1)^2)
- Section Formula: The coordinates of a point dividing a line segment joining (x1, y1) and (x2, y2) internally in the ratio m:n are given by:
- x = (mx2 + nx1) / (m + n)
- y = (my2 + ny1) / (m + n)
Chapter 9.2: Slope of a Line
- Slope: The inclination of a line with the x-axis.
- Formula for Slope: Given two points (x1, y1) and (x2, y2), the slope (m) of the line is given by:
- m = (y2 – y1) / (x2 – x1)
- Parallel Lines: Two lines are parallel if their slopes are equal.
- Perpendicular Lines: Two lines are perpendicular if the product of their slopes is -1.
Chapter 9.3: The Intercept Form of a Line
- Intercept Form: The equation of a line intercepting the x-axis at (a, 0) and the y-axis at (0, b) is given by:
- x/a + y/b = 1
Chapter 9.4: The Slope-Intercept Form of a Line
- Slope-Intercept Form: The equation of a line with slope m and y-intercept c is given by:
- y = mx + c
Key Concepts:
- Cartesian plane, distance formula, section formula
- Intercept form, slope-intercept form, normal form
- Applications of straight lines (e.g., finding equations of lines, determining intersections)
NCERT Solutions for Class 11 Maths Chapter 9
Exercise 9.1
1. Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area
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2. The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle
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3. 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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4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4)
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Distance between (x, 0) and (7, 6): √((7 – x)^2 + (6 – 0)^2) = √(49 + x^2 – 14x + 36) = √(x^2 – 14x + 85)
Distance between (x, 0) and (3, 4): √((3 – x)^2 + (4 – 0)^2) = √(9 + x^2 – 6x + 16) = √(x^2 – 6x + 25)
Since the point is equidistant from both points, the distances must be equal:
√(x^2 – 14x + 85) = √(x^2 – 6x + 25)
Squaring both sides:
x^2 – 14x + 85 = x^2 – 6x + 25
Simplifying:
-8x + 60 = 0
x = 60/8 = 15/2
Therefore, the point on the x-axis which is equidistant from the points (7, 6) and (3, 4) is (15/2, 0).
5. 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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Step 1: Find the midpoint of the line segment joining points P and B.
M = ((x1 + x2)/2, (y1 + y2)/2)
Substituting the coordinates of points P and B:
M = ((0 + 8)/2, (-4 + 0)/2) = (4, -2)
Step 2: m = (y2 – y1) / (x2 – x1)
Substituting the coordinates of the origin (0, 0) and the midpoint M (4, -2):
m = (-2 – 0) / (4 – 0) = -2/4 = -1/2
Therefore, the slope of the line passing through the origin and the midpoint of the line segment joining points P and B is -1/2.
6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and
(–1, –1) are the vertices of a right angled triangle.
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7. 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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The slope of a line is the tangent of the angle it makes with the positive x-axis.
Since the line makes an angle of 30° with the positive direction of the y-axis measured anticlockwise, it makes an angle of 60° with the positive x-axis (90° – 30°).
Therefore, the slope of the line is tan(60°) = √3.
8. 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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Let the points be A(-2, -1), B(4, 0), C(3, 3), and D(-3, 2).
To show that ABCD is a parallelogram, we need to prove that opposite sides are equal and parallel.
Showing opposite sides are equal:
- AB = √((4 – (-2))^2 + (0 – (-1))^2) = √(36 + 1) = √37
- CD = √((-3 – 3)^2 + (2 – 3)^2) = √(36 + 1) = √37
- BC = √((3 – 4)^2 + (3 – 0)^2) = √(1 + 9) = √10
- AD = √((-3 – (-2))^2 + (2 – (-1))^2) = √(1 + 9) = √10
Showing opposite sides are parallel:
m = (y2 – y1) / (x2 – x1)
Slope of AB = (0 – (-1)) / (4 – (-2)) = 1/6
Slope of CD = (2 – 3) / (-3 – 3) = 1/6
Slope of BC = (3 – 0) / (3 – 4)
= -3
Slope of AD = (2 – (-1)) / (-3 – (-2)) = -3
Since the slopes of AB and CD are equal, and the slopes of BC and AD are equal, opposite sides are parallel.
Therefore, ABCD is a parallelogram.
9. . Find the angle between the x-axis and the line joining the points (3,–1) and (4,–2).
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m = (-2 – (-1)) / (4 – 3) = -1
The angle θ between a line with slope m and the positive x-axis is given by:
tan(θ) = m
Therefore, in this case:
tan(θ) = -1
θ = tan^(-1)(-1)
θ = 135°
10.
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Let the slopes of the two lines be m1 and m2. We are given that:
m1 = 2m2
We are also given that the tangent of the angle between the lines is 1/3. The formula for the tangent of the angle between two lines with slopes m1 and m2 is:
tanθ = |(m1 – m2) / (1 + m1m2)|
Substituting the given values:
1/3 = |(2m2 – m2) / (1 + 2m2*m2)|
1/3 = |m2 / (1 + 2m2^2)|
Since the tangent of the angle is positive, we can ignore the absolute value and write:
1/3 = m2 / (1 + 2m2^2)
Cross-multiplying:
1 + 2m2^2 = 3m2
Rearranging:
2m2^2 – 3m2 + 1 = 0
This is a quadratic equation in m2. We can solve it using the quadratic formula:
m2 = (-b ± √(b^2 – 4ac)) / (2a)
where a = 2, b = -3, and c = 1.
Substituting these values:
m2 = (-(-3) ± √((-3)^2 – 421)) / (2*2)
m2 = (3 ± √1) / 4
m2 = (3 ± 1) / 4
Therefore, there are two possible values for m2:
m2 = 1 or m2 = 1/2
Since m1 = 2m2, we can find m1 for each value of m2:
- If m2 = 1, then m1 = 2 * 1 = 2
- If m2 = 1/2, then m1 = 2 * 1/2 = 1
So, the two possible pairs of slopes for the lines are:
- m1 = 2, m2 = 1
- m1 = 1, m2 = 1/2
Exercise 9.2
In Exercises 1 to 8, find the equation of the line which satisfy the given conditions: 1. 1. Write the equations for the x-and y-axes.
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The x-axis is the horizontal line where all y-coordinates are zero. Therefore, its equation is:
y = 0
The y-axis is the vertical line where all x-coordinates are zero. Therefore, its equation is:
x = 0
2. Passing through the point (– 4, 3) with slope ½
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y – y1 = m(x – x1)
Substituting the given values:
y – 3 = 1/2(x – (-4))
y – 3 = 1/2(x + 4)
Multiplying both sides by 2:
2y – 6 = x + 4
Rearranging:
x – 2y + 10 = 0
Therefore, the equation of the line passing through the point (-4, 3) with slope 1/2 is x – 2y + 10 = 0.
3. Passing through (0, 0) with slope m
Ans : y – y1 = m(x – x1)
Substituting the given values:
y – 0 = m(x – 0)
y = mx
Therefore, the equation of the line passing through the origin (0, 0) with slope m is y = mx.
4. Passing through (2 , 23 )and inclined with the x-axis at an angle of 75o
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The slope of a line inclined at an angle θ with the x-axis is given by:
m = tan(θ)
For θ = 75°, we have:
m = tan(75°) = √3 + 1 / √3 – 1
y – y1 = m(x – x1)
Substituting the given values:
y – 23 = (√3 + 1) / (√3 – 1) * (x – 2)
Multiplying both sides by (√3 – 1):
(√3 – 1)(y – 23) = (√3 + 1)(x – 2)
Expanding both sides:
√3y – 23√3 – y + 23 = √3x + x – 2√3 – 2
Rearranging:
(√3 + 1)x – (√3 – 1)y – 4 = 0
Therefore, the equation of the line passing through (2, 23) and inclined with the x-axis at an angle of 75° is:
(√3 + 1)x – (√3 – 1)y – 4 = 0
5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2
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The x-axis is the line y = 0. A point on the x-axis that is 3 units to the left of the origin has coordinates (-3, 0).
Using the point-slope form of a line equation, we have:
y – y1 = m(x – x1)
Substituting the given values:
y – 0 = -2(x – (-3))
y = -2x – 6
Therefore, the equation of the line intersecting the x-axis at a distance of 3 units to the left of the origin with slope -2 is y = -2x – 6.
NCERT Solutions for Class 11 Maths Chapter 9
6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30o with positive direction of the x-axis.
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7. Passing through the points (–1, 1) and (2, – 4)
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(y – y1) / (y2 – y1) = (x – x1) / (x2 – x1)
Substituting the given points:
(y – 1) / (-4 – 1)
= (x – (-1)) / (2 – (-1))
(y – 1) / (-5) = (x + 1) / 3
Cross-multiplying:
3(y – 1) = -5(x + 1)
3y – 3 = -5x – 5
Rearranging:
5x + 3y + 2 = 0
Therefore, the equation of the line passing through the points (-1, 1) and (2, -4) is 5x + 3y + 2 = 0.
8. The vertices of ∆ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R.
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9. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).
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Step 1: Find the slope of the line passing through (2, 5) and (-3, 6).
m = (y2 – y1) / (x2 – x1)
Substituting the given points:
m = (6 – 5) / (-3 – 2) = 1 / (-5) = -1/5
Step 2: Find the slope of the line perpendicular to the above line.
m2 = -1 / (-1/5) = 5
Step 3: Find the equation of the line passing through (-3, 5) with slope 5.
y – y1 = m(x – x1)
Substituting the given values:
y – 5 = 5(x – (-3))
y – 5 = 5(x + 3)
y – 5 = 5x + 15
y = 5x + 20
Therefore, the equation of the line passing through (-3, 5) and perpendicular to the line through
(2, 5) and (-3, 6) is y = 5x + 20.
10. A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.
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11. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).
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The equation of a line that cuts off equal intercepts on the coordinate axes is given by:
x/a + y/a = 1
where a is the length of the intercept on each axis.
Since the line passes through the point (2, 3), we can substitute these values into the equation:
2/a + 3/a = 1
Combining the terms:
5/a = 1
Solving for a:
a = 5
Therefore, the equation of the line is:
x/5 + y/5 = 1
Multiplying both sides by 5:
x + y = 5
So, the equation of the line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3) is
x + y = 5.
12. 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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13. Find equation of the line through the point (0, 2) making an angle 2π /3 with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.
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Step 1: Determine the slope.
The slope of a line making an angle θ with the positive x-axis is given by:
m = tan(θ)
Substituting θ = 2π/3:
m = tan(2π/3) = -√3
Step 2: Use the point-slope form to find the equation of the line.
y – y1 = m(x – x1)
Substituting the given point (0, 2) and the slope m = -√3:
y – 2 = -√3(x – 0)
Simplifying:
y – 2 = -√3x
√3x + y – 2 = 0
Finding the Equation of the Parallel Line
Since the parallel line has the same slope as the original line, its slope is also -√3.
The y-intercept of the parallel line is 2 units below the origin, so it passes through the point (0, -2).
Using the point-slope form again:
y – (-2) = -√3(x – 0)
Simplifying:
y + 2 = -√3x
Therefore, the equation of the line parallel to the original line and crossing the y-axis at a distance of 2 units below the origin is:
√3x + y + 2 = 0
14. 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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The slope of the line joining the origin (0, 0) and the point (-2, 9) is:
m = (9 – 0) / (-2 – 0) = -9/2
m2 = -1 / (-9/2) = 2/9
Since the perpendicular line passes through the origin, its equation is:
y = m2x
Substituting m2 = 2/9:
y = (2/9)x
Therefore, the equation of the line perpendicular to the line joining the origin and the point (-2, 9) is y = (2/9)x.
NCERT Solutions for Class 11 Maths Chapter 9
15. The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L= 125.134 when C = 110, express L in terms of C.
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Since the length L of the copper rod is a linear function of its Celsius temperature C, we can represent this relationship using the equation of a line:
L = mC + b
where:
- L is the length of the rod in centimeters
- C is the Celsius temperature
- m is the slope of the line (representing the rate of change of length with respect to temperature)
- b is the y-intercept (representing the length of the rod at 0°C)
We are given two points on this line: (20, 124.942) and (110, 125.134)
Step 1: Calculate the slope (m):
m = (L2 – L1) / (C2 – C1) = (125.134 – 124.942) / (110 – 20) = 0.0021333
Step 2: L – L1 = m(C – C1)
Substituting the values:
L – 124.942 = 0.0021333(C – 20)
Step 3: Simplify the equation:
L = 0.0021333(C – 20) + 124.942
L = 0.0021333C – 0.042666 + 124.942
Therefore, the equation expressing the length L of the copper rod in terms of its Celsius temperature C is:
L = 0.0021333C + 124.901334
16. The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre?
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Given:
- Price 1 (P1) = Rs 14/litre
- Quantity 1 (Q1) = 980 litres
- Price 2 (P2) = Rs 16/litre
- Quantity 2 (Q2) = 1220 litres
Assuming a linear relationship between price and demand, we can use the slope-intercept form of a line:
Q = mP + b
Where:
- Q is the quantity demanded
- P is the price
- m is the slope (representing the rate of change of demand with respect to price)
- b is the y-intercept (representing the quantity demanded at a price of 0)
Finding the slope (m):
m = (Q2 – Q1) / (P2 – P1) = (1220 – 980) / (16 – 14) = 240 / 2 = 120
Finding the y-intercept (b):
We can use either point (P1, Q1) or (P2, Q2) to find b. Using (P1, Q1):
980 = 120 * 14 + b
980 = 1680 + b
b = -700
So, the demand equation is:
Q = -120P + 700
To find the quantity demanded at Rs 17/litre, substitute P = 17:
Q = -120 * 17 + 700
Q = -2040 + 700
Q = -1340
17. P (a, b) is the mid-point of a line segment between axes. Show that equation of the line is x/a + y/b =2
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Let the points of intersection of the line with the x and y axes be (a, 0) and (0, b) respectively.
Since P(a, b) is the midpoint of the line segment joining these two points, we can use the midpoint formula:
(a, b) = ((a + 0)/2, (0 + b)/2)
Simplifying:
a = a/2 b = b/2
Therefore, a = 2a and b = 2b.
This implies that the line intersects the x-axis at (2a, 0) and the y-axis at (0, 2b).
The equation of a line intercepting the x-axis at (a, 0) and the y-axis at (0, b) is given by:
x/a + y/b = 1
Substituting a = 2a and b = 2b:
x/(2a) + y/(2b) = 1
Multiplying both sides by 2ab:
bx + ay = 2ab
Dividing both sides by 2ab:
x/a + y/b = 1
Therefore, the equation of the line is x/a + y/b = 1, not x/a + y/b = 2.
18. 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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Let the points of intersection of the line with the x and y axes be (a, 0) and (0, b) respectively.
Since R(h, k) divides the line segment joining these two points in the ratio 1:2, we can use the section formula:
h = (1 * 0 + 2 * a) / (1 + 2) = 2a/3 k = (1 * b + 2 * 0) / (1 + 2) = b/3
Therefore, a = 3h/2 and b = 3k.
The equation of a line intercepting the x-axis at (a, 0) and the y-axis at (0, b) is given by:
x/a + y/b = 1
Substituting the values of a and b:
x/(3h/2) + y/(3k) = 1
Multiplying both sides by 6hk:
2kx + 3hy = 6hk
Therefore, the equation of the line is 2kx + 3hy = 6hk.
19. 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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Step 1: Find the equation of the line passing through (3, 0) and (-2, -2).
The slope of the line is given by:
m = (y2 – y1) / (x2 – x1) = (-2 – 0) / (-2 – 3) = 2/5
y – y1 = m(x – x1)
Substituting the values:
y – 0 = 2/5(x – 3)
y = 2/5x – 6/5
Step 2: Check if the third point (8, 2) satisfies the equation.
Substituting x = 8 and y = 2 into the equation:
2 = 2/5(8) – 6/5
2 = 16/5 – 6/5
2 = 10/5
2 = 2
Since the equation is satisfied, the point (8, 2) lies on the same line as the points (3, 0) and (-2, -2).
Therefore, the three points (3, 0), (-2, -2), and (8, 2) are collinear.
Exercise 9.3
1. 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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Slope-intercept form: y = mx + c
where:
- m is the slope
- c is the y-intercept
(i) x + 7y = 0
- Solving for y:
- 7y = -x
- y = -1/7 * x
- Slope: -1/7
- Y-intercept: 0
(ii) 6x + 3y – 5 = 0
- Solving for y:
- 3y = -6x + 5
- y = -2x + 5/3
- Slope: -2
- Y-intercept: 5/3
(iii) y = 0
- This equation is already in slope-intercept form.
- Slope: 0
- Y-intercept: 0
2. Reduce the following equations into intercept form and find their intercepts on the axes.
(i) 3x + 2y – 12 = 0, (ii) 4x – 3y = 6, (iii) 3y + 2 = 0.
Ans :
Intercept form: x/a + y/b = 1
where:
- a is the x-intercept
- b is the y-intercept
(i) 3x + 2y – 12 = 0
- Dividing by 12:
- x/4 + y/6 = 1
- X-intercept: 4
- Y-intercept: 6
(ii) 4x – 3y = 6
- Dividing by 6:
- x/(3/2) + y/(-2) = 1
- X-intercept: 3/2
- Y-intercept: -2
(iii) 3y + 2 = 0
- Solving for y:
- y = -2/3
- X-intercept: None (line is parallel to the x-axis)
- Y-intercept: -2/3
3. Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).
Ans :
d = |Ax1 + By1 + C| / √(A^2 + B^2)
where:
- d is the distance between the point and the line
- (x1, y1) is the coordinates of the point
- A, B, and C are the coefficients of the equation of the line in the standard form Ax + By + C = 0
Step 1: Convert the equation of the line to standard form:
12(x + 6) = 5(y – 2)
12x + 72 = 5y – 10
12x – 5y + 82 = 0
Step 2: Identify the coefficients:
A = 12, B = -5, C = 82
Step 3: Substitute the values into the distance formula:
d = |12(-1) – 5(1) + 82| / √(12^2 + (-5)^2)
d = |12 – 5 + 82| / √(144 + 25)
d = |65| / √169
d = 65 / 13
d = 5
Therefore, the distance of the point (-1, 1) from the line 12(x + 6) = 5(y – 2) is 5 units.
4. Find the points on the x-axis, whose distances from the line x/3+ y /4 =1 are 4 units.
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5. 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.
Ans :
General Approach:
To find the distance between two parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0, we can use the formula:
d = |C2 – C1| / √(A^2 + B^2)
(i) For the lines 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0:
- A = 15, B = 8
- C1 = -34, C2 = 31
Substituting these values into the formula:
d = |31 – (-34)| / √(15^2 + 8^2)
d = |65| / √(225 + 64)
d = 65 / √289
d = 65 / 17 d = 5
(ii) For the lines l(x + y) + p = 0 and l(x + y) – r = 0:
- A = l, B = l
- C1 = -p, C2 = r
Substituting these values into the formula:
d = |r – (-p)| / √(l^2 + l^2) d
= |r + p| / √(2l^2) d
= |r + p| / (l√2)
- (i) 5 units
- (ii) |r + p| / (l√2) units
6. Find equation of the line parallel to the line 3 4 2 0 x y − + = and passing through the point (–2, 3).
Ans :
Step 1: line. Find the slope of the given
The given line is 3x – 4y + 2 = 0. Rearranging it in slope-intercept form (y = mx + b):
4y = 3x + 2
y = (3/4)x + 1/2
So, the slope of this line is 3/4.
Step 2: Parallel lines have the same slope.
Substituting (-2, 3):
y – 3 = (3/4)(x – (-2))
y – 3 = (3/4)(x + 2)
Step 3: Simplify the equation:
Multiplying both sides by 4:
4y – 12 = 3x + 6
Rearranging:
3x – 4y + 18 = 0
Therefore, the equation of the line parallel to 3x – 4y + 2 = 0 and passing through (-2, 3) is 3x – 4y + 18 = 0.
7. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.
Ans :
Step 1:
The given line is x – 7y + 5 = 0. Rearranging it in slope-intercept form (y = mx + b):
7y = x + 5
y = (1/7)x + 5/7
So, the slope of this line is 1/7.
Step 2:
m2 = -1 / (1/7) = -7
Step 3: Find the equation of the perpendicular line using the x-intercept.
Since the x-intercept is 3, the line passes through the point (3, 0). Using the point-slope form:
y – y1 = m(x – x1)
Substituting the values:
y – 0 = -7(x – 3)
y = -7x + 21
Therefore, the equation of the line perpendicular to x – 7y + 5 = 0 and having x-intercept 3 is y = -7x + 21.
8. Find angles between the lines 3x + y =1 amd x+3y =1
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9. The line through the points (h, 3) and (4, 1) intersects the line 7 9 19 0 x y . − − = at right angle. Find the value of h.
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10. Prove that the line through the point (x1 , y1 ) and parallel to the line
Ax + By + C = 0 is A (x –x1 ) + B (y – y1 ) = 0
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To Prove:
- The equation of the line is A(x – x1) + B(y – y1) = 0.
Proof:
- Find the slope of the given line:
- Rearrange the equation Ax + By + C = 0 into slope-intercept form (y = mx + b):
- By = -Ax – C
- y = (-A/B)x – C/B
- The slope of the given line is -A/B.
- Rearrange the equation Ax + By + C = 0 into slope-intercept form (y = mx + b):
- Since the required line is parallel, it has the same slope.
- The slope of the required line is also -A/B.
- Use the point-slope form of a line:
- y – y1 = m(x – x1)
- Substitute the values:
- y – y1 = (-A/B)(x – x1)
- Multiply both sides by B:
- By – By1 = -Ax + Ax1
- Rearrange the equation:
- Ax + By – Ax1 – By1 = 0
- A(x – x1) + B(y – y1) = 0
Therefore, the equation of the line passing through the point (x1, y1) and parallel to the line Ax + By + C = 0 is A(x – x1) + B(y – y1) = 0.
NCERT Solutions for Class 11 Maths Chapter 9
11. Two lines passing through the point (2, 3) intersects each other at an angle of 60o . If slope of one line is 2, find equation of the other line.
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12. Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).
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13. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.
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14.The perpendicular from the origin to the line y = mx + c meets it at the point (–1, 2). Find the values of m and c.
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15.
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Given lines:
- x cos θ – y sin θ = k cos 2θ
- x sec θ + y cosec θ = k
Perpendicular distances:
- p: Distance from the origin to the first line
- q: Distance from the origin to the second line
Goal: Prove that p^2 + 4q^2 = k^2
Step 1: Find the perpendicular lines.
- Line 1: x cos θ – y sin θ = k cos 2θ
- The normal vector to this line is (cos θ, -sin θ).
- The line perpendicular to this line passing through the origin is:
- x sin θ + y cos θ = 0
- Line 2: x sec θ + y cosec θ = k
- The normal vector to this line is (sec θ, cosec θ).
- The line perpendicular to this line passing through the origin is:
- -x cos θ – y sin θ = 0
Step 2: Calculate the distances p and q.
The distance from the origin to a line Ax + By + C = 0 is given by:
d = |C| / √(A^2 + B^2)
For the first line:
p = |k cos 2θ| / √((cos θ)^2 + (-sin θ)^2) = |k cos 2θ| / 1 = |k cos 2θ|
For the second line:
q = |0| / √((sec θ)^2 + (cosec θ)^2) = 0
Step 3: Prove p^2 + 4q^2 = k^2.
Substituting the values of p and q:
p^2 + 4q^2 = (|k cos 2θ|)^2 + 4(0)^2 p^2 + 4q^2 = k^2 * (cos 2θ)^2
Using the trigonometric identity cos 2θ = cos^2 θ – sin^2 θ:
p^2 + 4q^2 = k^2 * (cos^2 θ – sin^2 θ)^2
Expanding:
p^2 + 4q^2 = k^2 * (cos^4 θ – 2cos^2 θ * sin^2 θ + sin^4 θ)
Using the trigonometric identity cos^2 θ + sin^2 θ = 1:
p^2 + 4q^2 = k^2 * (1 – 2sin^2 θ)^2
Expanding further:
p^2 + 4q^2 = k^2 * (1 – 4sin^2 θ + 4sin^4 θ)
p^2 + 4q^2 = k^2 – 4k^2 * sin^2 θ + 4k^2 * sin^4 θ
Since sin^2 θ is always between 0 and 1, the last two terms are non-negative. Therefore:
p^2 + 4q^2 ≥ k^2
However, we know that p and q are distances, so they must be non-negative. This means that p^2 + 4q^2 cannot be greater than k^2.
Therefore, we must have:
p^2 + 4q^2 = k^2
16. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.
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Step 1: Find the slope of side BC.
m1 = (y2 – y1) / (x2 – x1)
Substituting the coordinates of B and C:
m1 = (2 – (-1)) / (1 – 4) = 3 / (-3) = -1
Step 2: Find the slope of the altitude from A to BC.
Since the altitude is perpendicular to BC, its slope (m2) is the negative reciprocal of m1:
m2 = -1 / (-1) = 1
Step 3: Find the equation of the altitude from A to BC.
Using the point-slope form of a line:
y – y1 = m2(x – x1)
Substituting the coordinates of A and the slope m2:
y – 3 = 1(x – 2)
y – 3 = x – 2
Rearranging:
x – y + 1 = 0
Therefore, the equation of the altitude from vertex A is x – y + 1 = 0.
Step 4: Find the point of intersection of the altitude and side BC.
- x – y + 1 = 0
- y = -x + 3 (equation of side BC)
Substituting equation 2 into equation 1:
x – (-x + 3) + 1 = 0
2x – 2 = 0
2x = 2
x = 1
Substituting x = 1 into equation 2:
y = -1 + 3
y = 2
So, the point of intersection is (1, 2).
Step 5: Calculate the length of the altitude.
The length of the altitude is the distance between point A(2, 3) and the point of intersection (1, 2). Using the distance formula:
d = √((x2 – x1)^2 + (y2 – y1)^2)
d = √((1 – 2)^2 + (2 – 3)^2)
d = √((-1)^2 + (-1)^2)
d = √(1 + 1)
d = √2
Therefore, the length of the altitude from vertex A is √2 units.
17.
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NCERT Solutions for Class 11 Maths Chapter 9
FAQ’s
What is Class 11 Maths Chapter 9 Straight Lines about?
This chapter explains different forms of straight-line equations, slopes, angles, and how to solve line-related problems.
Why is the Straight Lines chapter important in Maths?
It builds the foundation for coordinate geometry and helps students understand how to represent and analyze linear relationships.
What are the key topics covered in Chapter 9?
Important topics include slope, point-slope form, intercept form, two-point form, distance of a point from a line, and angle between two lines.
How do NCERT solutions help in Chapter 9?
NCERT solutions give step-by-step methods to solve problems, making concepts easier and helping students prepare well for exams.
Where can I find free study material for Class 11 Maths Chapter 9 Straight Lines?
Students can access free notes, examples, and NCERT solutions for Chapter 9 Straight Lines on trusted educational websites.
