NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3 Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3 Straight Lines

The NCERT Solutions for Class 11 Maths Chapter 9, covering Straight Lines, are a helpful resource for students preparing for their Math board exams. The chapter covers important topics like slope, Horizontal and vertical Lines, Point-slope form, Two-point form, Slope-intercept form, Intercept – form, and Normal form. These solutions are designed to help students practice and improve their understanding using straightforward methods and formulas.

NCERT Solutions for Class 11 Maths Chapter 09

EXERCISE 9.3 PAGE NO: 167

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

Solution:

(i) x + 7y = 0

Given:

The equation is x + 7y = 0

The slope-intercept form is represented in the form ‘y = mx + c’, where m is the slope and c is the y-intercept.

So, the above equation can be expressed as

y = -1/7x + 0

∴ The above equation is of the form y = mx + c, where m = -1/7 and c = 0

(ii) 6x + 3y – 5 = 0

Given:

The equation is 6x + 3y – 5 = 0

The slope-intercept form is represented in the form ‘y = mx + c’, where m is the slope and c is the y-intercept.

So, the above equation can be expressed as

3y = -6x + 5

y = -6/3x + 5/3

= -2x + 5/3

∴ The above equation is of the form y = mx + c, where m = -2 and c = 5/3

(iii) y = 0

Given:

The equation is y = 0

The slope-intercept form is given by ‘y = mx + c’, where m is the slope and c is the y-intercept.

y = 0 × x + 0

∴ The above equation is of the form y = mx + c, where m = 0 and c = 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

Solution:

(i) 3x + 2y – 12 = 0

Given:

The equation is 3x + 2y – 12 = 0

The equation of the line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepted on the x-axis and the y-axis, respectively.

So, 3x + 2y = 12

Now, let us divide both sides by 12; we get

3x/12 + 2y/12 = 12/12

x/4 + y/6 = 1

∴ The above equation is of the form x/a + y/b = 1, where a = 4, b = 6

The intercept on the x-axis is 4.

The intercept on the y-axis is 6.

(ii) 4x – 3y = 6

Given:

The equation is 4x – 3y = 6

The equation of the line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepted on the x-axis and the y-axis, respectively.

So, 4x – 3y = 6

Now, let us divide both sides by 6; we get

4x/6 – 3y/6 = 6/6

2x/3 – y/2 = 1

x/(3/2) + y/(-2) = 1

∴ The above equation is of the form x/a + y/b = 1, where a = 3/2, b = -2

The intercept on the x-axis is 3/2.

The intercept on the y-axis is -2.

(iii) 3y + 2 = 0

Given:

The equation is 3y + 2 = 0

The equation of the line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepted on the x-axis and the y-axis, respectively.

So, 3y = -2

Now, let us divide both sides by -2; we get

3y/-2 = -2/-2

3y/-2 = 1

y/(-2/3) = 1

∴ The above equation is of the form x/a + y/b = 1, where a = 0, b = -2/3

The intercept on the x-axis is 0.

The intercept on the y-axis is -2/3.

3. Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).

Solution:

Given:

The equation of the line is 12(x + 6) = 5(y – 2).

12x + 72 = 5y – 10

12x – 5y + 82 = 0 … (1)

Now, compare equation (1) with the general equation of line Ax + By + C = 0, where A = 12, B = –5, and C = 82

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

∴ The distance is 5 units.

4. Find the points on the x-axis whose distances from the line x/3 + y/4 = 1 are 4 units.

Solution:

Given:

The equation of the line is x/3 + y/4 = 1

4x + 3y = 12

4x + 3y – 12 = 0 …. (1)

Now, compare equation (1) with the general equation of line Ax + By + C = 0, where A = 4, B = 3, and C = -12

Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.

So, the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

|4a – 12| = 4 × 5

± (4a – 12) = 20

4a – 12 = 20 or – (4a – 12) = 20

4a = 20 + 12 or 4a = -20 + 12

a = 32/4 or a = -8/4

a = 8 or a = -2

∴ The required points on the x-axis are (-2, 0) and (8, 0)

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

Solution:

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

Given:

The parallel lines are 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.

By using the formula,

The distance (d) between parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is given by

∴ The distance between parallel lines is 65/17

(ii) l(x + y) + p = 0 and l (x + y) – r = 0

Given:

The parallel lines are l (x + y) + p = 0 and l (x + y) – r = 0

lx + ly + p = 0 and lx + ly – r = 0

By using the formula,

The distance (d) between parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is given by

∴ The distance between parallel lines is |p+r|/l2

6. Find the equation of the line parallel to the line 3x − 4y + 2 = 0 and passing through the point (–2, 3).

Solution:

Given:

The line is 3x – 4y + 2 = 0

So, y = 3x/4 + 2/4

= 3x/4 + ½

Which is of the form y = mx + c, where m is the slope of the given line.

The slope of the given line is 3/4

We know that parallel lines have the same slope.

∴ Slope of other line = m = 3/4

The equation of line having slope m and passing through (x1, y1) is given by

y – y1 = m (x – x1)

∴ The equation of the line having slope 3/4 and passing through (-2, 3) is

y – 3 = ¾ (x – (-2))

4y – 3 × 4 = 3x + 3 × 2

3x – 4y = 18

∴ The equation is 3x – 4y = 18

7. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.

Solution:

Given:

The equation of line is x – 7y + 5 = 0

So, y = 1/7x + 5/7 [which is of the form y = mx + c, where m is the slope of the given line.]

The slope of the given line is 1/7

The slope of the line perpendicular to the line having slope m is -1/m

The slope of the line perpendicular to the line having a slope of 1/7 is -1/(1/7) = -7

So, the equation of the line with slope -7 and the x-intercept 3 is given by y = m(x – d)

y = -7 (x – 3)

y = -7x + 21

7x + y = 21

∴ The equation is 7x + y = 21

8. Find angles between the lines √3x + y = 1 and x + √3y = 1.

Solution: Given:

The lines are √3x + y = 1 and x + √3y = 1

So, y = -√3x + 1 … (1) and

y = -1/√3x + 1/√3 …. (2)

The slope of the line (1) is m1 = -√3, while the slope of the line (2) is m2 = -1/√3

Let θ be the angle between two lines.

So,

θ = 30°

∴ The angle between the given lines is either 30° or 180°- 30° = 150°

9. The line through the points (h, 3) and (4, 1) intersects the line 7x − 9y −19 = 0. At the right angle. Find the value of h.

Solution:

Let the slope of the line passing through (h, 3) and (4, 1) be m1

Then, m1 = (1-3)/(4-h) = -2/(4-h)

Let the slope of line 7x – 9y – 19 = 0 be m2

7x – 9y – 19 = 0

So, y = 7/9x – 19/9

m2 = 7/9

Since the given lines are perpendicular,

m1 × m2 = -1

-2/(4-h) × 7/9 = -1

-14/(36-9h) = -1

-14 = -1 × (36 – 9h)

36 – 9h = 14

9h = 36 – 14

h = 22/9

∴ The value of h is 22/9

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.

Solution:

Let the slope of line Ax + By + C = 0 be m

Ax + By + C = 0

So, y = -A/Bx – C/B

m = -A/B

By using the formula,

Equation of the line passing through point (x1, y1) and having slope m = -A/B is

y – y1 = m (x – x1)

y – y1= -A/B (x – x1)

B (y – y1) = -A (x – x1)

∴ A(x – x1) + B(y – y1) = 0

So, the line through point (x1, y1) and parallel to the line Ax + By + C = 0 is A (x – x1) + B (y – y1) = 0

Hence, proved.

11. Two lines passing through point (2, 3) intersects each other at an angle of 60o. If the slope of one line is 2, find the equation of the other line.

Solution:

Given: m1 = 2

Let the slope of the first line be m1

And let the slope of the other line be m2.

The angle between the two lines is 60°.

So,

12. Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).

Solution:

Given:

The right bisector of a line segment bisects the line segment at 90°.

End-points of the line segment AB are given as A (3, 4) and B (–1, 2).

Let the mid-point of AB be (x, y).

x = (3-1)/2= 2/2 = 1

y = (4+2)/2 = 6/2 = 3

(x, y) = (1, 3)

Let the slope of line AB be m1

m1 = (2 – 4)/(-1 – 3)

= -2/(-4)

= 1/2

And let the slope of the line perpendicular to AB be m2

m2 = -1/(1/2)

= -2

The equation of the line passing through (1, 3) and having a slope of –2 is

(y – 3) = -2 (x – 1)

y – 3 = – 2x + 2

2x + y = 5

∴ The required equation of the line is 2x + y = 5

13. Find the coordinates of the foot of the perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.

Solution:

Let us consider the coordinates of the foot of the perpendicular from (-1, 3) to the line 3x – 4y – 16 = 0 be (a, b)

So, let the slope of the line joining (-1, 3) and (a, b) be m1

m= (b-3)/(a+1)

And let the slope of the line 3x – 4y – 16 = 0 be m2

y = 3/4x – 4

m2 = 3/4

Since these two lines are perpendicular, m1 × m2 = -1

(b-3)/(a+1) × (3/4) = -1

(3b-9)/(4a+4) = -1

3b – 9 = -4a – 4

4a + 3b = 5 …….(1)

Point (a, b) lies on the line 3x – 4y = 16

3a – 4b = 16 ……..(2)

Solving equations (1) and (2), we get

a = 68/25 and b = -49/25

∴ The coordinates of the foot of perpendicular are (68/25, -49/25)

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.

Solution:

Given:

The perpendicular from the origin meets the given line at (–1, 2).

The equation of the line is y = mx + c

The line joining the points (0, 0) and (–1, 2) is perpendicular to the given line.

So, the slope of the line joining (0, 0) and (–1, 2) = 2/(-1) = -2

The slope of the given line is m.

m × (-2) = -1

m = 1/2

Since point (-1, 2) lies on the given line,

y = mx + c

2 = 1/2 × (-1) + c

c = 2 + 1/2 = 5/2

∴ The values of m and c are 1/2 and 5/2, respectively.

15. If p and q are the lengths of perpendiculars from the origin to the lines x cos θ − y sin θ = k cos 2θ and x sec θ + y cosec θ = k, respectively, prove that p2 + 4q2 = k2

Solution:

Given:

The equations of the given lines are

x cos θ – y sin θ = k cos 2θ …………………… (1)

x sec θ + y cosec θ = k ……………….… (2)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

q = k cos θ sin θ

Multiply both sides by 2, and we get

2q = 2k cos θ sin θ = k × 2sin θ cos θ

2q = k sin 2θ

Squaring both sides, we get

4q2 = k2 sin22θ …………………(4)

Now add (3) and (4); we get

p2 + 4q2 = k2 cos2 2θ + k2 sin2 2θ

p2 + 4q2 = k2 (cos2 2θ + sin2 2θ) [Since, cos2 2θ + sin2 2θ = 1]

∴ p2 + 4q2 = k2

Hence proved.

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 vertex A.

Solution:

Let AD be the altitude of triangle ABC from vertex A.

So, AD is perpendicular to BC.

Given:

Vertices A (2, 3), B (4, –1) and C (1, 2)

Let the slope of the line BC = m1

m1 = (- 1 – 2)/(4 – 1)

m1 = -1

Let the slope of the line AD be m2

m1 × m2 = -1

-1 × m2 = -1

m2 = 1

The equation of the line passing through the point (2, 3) and having a slope of 1 is

y – 3 = 1 × (x – 2)

y – 3 = x – 2

y – x = 1

Equation of the altitude from vertex A = y – x = 1

Length of AD = Length of the perpendicular from A (2, 3) to BC

The equation of BC is

y + 1 = -1 × (x – 4)

y + 1 = -x + 4

x + y – 3 = 0 …………………(1)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

Now, compare equation (1) to the general equation of the line, i.e., Ax + By + C = 0; we get

[where, A = 1, B = 1 and C = -3]

∴ The equation and the length of the altitude from vertex A are y – x = 1 and

√2 units, respectively.

17. If p is the length of the perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that 1/p2 = 1/a2 + 1/b2

Solution:

The equation of a line whose intercepts on the axes are a and b is x/a + y/b = 1

bx + ay = ab

bx + ay – ab = 0 ………………..(1)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

Now, square on both sides; we get

∴ 1/p2 = 1/a2 + 1/b2

Hence, proved.

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