NCERT Solutions Class 11 – Chapter 9 Straight Lines – Exercise 9.3

Question 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 that,

The equation is x + 7y = 0

Slope – intercept form is represented as y = mx + c, where m is the slope and c is the y intercept

Therefore, the above equation can be represented as,

y = -1/7x + 0

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

(ii) 6x + 3y – 5 = 0

Given that,

The equation is 6x + 3y – 5 = 0

Slope – intercept form is represented as y = mx + c, where m is the slope and c is the y intercept

Therefore, the above equation can be represented as,

3y = -6x + 5

y = -6/3x + 5/3 = -2x + 5/3

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

(iii) y = 0

Given that,

The equation is y = 0

Slope – intercept form is represented as y = mx + c, where m is the slope and c is the y intercept

Therefore, the above equation can be represented as,

y = 0 × x + 0

Hence, the above equation is in the form of y = mx + c, where m = 0 and c = 0.

Question 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 that,

The equation is 3x + 2y – 12 = 0

As we know that equation of line in intercept form is x/a + y/b = 1, where a and b are intercepts on x axis and y axis respectively.

Therefore, 3x + 2y = 12

Let’s divide both sides by 12, and we will get

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

x/4 + y/6 = 1

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

Intercept on x axis = 4

Intercept on y axis = 6

(ii) 4x – 3y = 6

Given that,

The equation is 4x – 3y = 6

As we know that equation of line in intercept form is x/a + y/b = 1, where a and b are intercepts on x axis and y axis respectively.

Therefore, 4x – 3y = 6

let’s divide both sides by 6, and we will get

4x/6 – 3y/6 = 6/6

2x/3 – y/2 = 1

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

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

Intercept on x axis = 3/2

Intercept on y axis = -2

(iii) 3y + 2 = 0

Given that,

The equation is 3y + 2 = 0

As we know that equation of line in intercept form is x/a + y/b = 1, where a and b are intercepts on x axis and y axis respectively.

Therefore, 3y = -2

Let’s divide both sides by -2, and we will get 

3y/-2 = -2/-2

3y/-2 = 1

y/(-2/3) = 1

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

Intercept on x axis = 0

Intercept on y axis = -2/3

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

Solution: 

Given that,

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

12x + 72 = 5y – 10

12x – 5y + 82 = 0      ——–(i)

Now, after comparing equation (i) with general equation of line Ax + By + C = 0, we get 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,

d = |Ax1 + By1 + C| / √A² + B²

Given points (x1, y1) are (-1, 1)

Distance of point (-1, 1) from the given point is 

d = |12 x (-1) + (-5) + 82| / √12² + (-5)² = 65 / 13 units

= 5 units.

Hence, the distance is 5 units.

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

Solution: 

Given that,

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

4x + 3y = 12

4x + 3y – 12 = 0      ——(i)

Now, after comparing equation (i) with general equation of line Ax + By + C = 0, we get A = 4, B = 3, and C = -12

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

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

d = |Ax1 + By1 + C| / √A² + B²

4 = |4a + 3 × 0 – 12| / √4² + 3²

4 = |4a – 12| / 5

|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

Hence, the required points on the x – axis are (-2, 0) and (8, 0)

Question 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 that,

The parallel lines are 15x + 8y – 34 = 0 & 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,

d = |C₁ – C₂| / √A² + B²

From given equation we get, A = 15, B = 8, C₁ = -34, C₂ = 31

Now apply the formula and calculate distance between parallel lines,

d = |-34 – 31| / √15² + 8² = |-65| / √225 + 64

= 65 / 17 

Hence, the distance between parallel lines is 65/17.

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

Given that,

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,

By using the formula, the distance d between parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 is given by,

d = |C₁ – C₂| / √A² + B²

From given equation we get, A = l, B = l, C₁ = p, C₂ = -r

Now apply the formula and calculate distance between parallel lines,

d = |p – (-r)| / √l² + l² = |p+ r| / √2 l  = |p+r|/l√2

Hence, the distance between parallel lines is |p+r|/l√2

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

Solution: 

Given that,

The line is 3x – 4y + 2 = 0

Therefore, y = 3x/4 + 2/4 = 3x/4 + ½

The above equation is in the form of y = mx + c, where m is the slope of the given line.

Therefore, slope of the given line is 3/4

As we know that parallel line have same slope.

Therefore, slope of other line = m = 3/4

Equation of line having slope m and passing through (x₁, y₁) is given by, y – y₁ = m (x – x₁)

Now put the value of slope 3/4 and points (-2, 3) in above formula, and we get,

y – 3 = ¾ (x – (-2))

4y – 3 × 4 = 3x + 3 × 2

3x – 4y = 18

Hence, the equation is 3x – 4y = 18

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

Solution: 

Given that,

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

Therefore, y = 1/7x + 5/7 

The above equation is in the form of y = mx + c, where m is the slope of the given line.

Therefore, slope of the given line is 1/7

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

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

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

y = -7 (x – 3)

y = -7x + 21

7x + y = 21

Hence, the equation is 7x + y = 21

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

Solution: 

Given that,

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

Therefore, y = -√3x + 1     ———(i)   &

y = -1/√3x + 1/√3    ——–(ii)

Slope of line (i) is -√3, while the slope of line (ii) is -1/√3

Let’s assume that θ be the angle between two lines,

As we know that,

tanθ = |m1 – m2| / |1 + m₁m₂|

Put the values of m₁ and m₂ formula, and we will get,

= |-√3 – (-1/√3)| / |1 + (-√3)(-1/√3)| = 1 / √3

θ = 30°

Hence, the angle between the given lines is either 30° or 180°- 30° = 150°

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

Solution: 

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

Therefore, m1 = (1-3)/(4-h) = -2/(4-h)

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

7x – 9y – 19 = 0

Therefore, y = 7/9x – 19/9

m₂ = 7/9

Given that, 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

Hence, the value of h is 22/9.

Question 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’s assume that the slope of line Ax + By + C = 0 be m,

Ax + By + C = 0

Therefore, y = -A/B x – C/B

m = -A / B

By using the formula,

As we know that the 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)

Hence, A(x – x1) + B(y – y1) = 0

Therefore, 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.

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

Solution:

Given that, m1 = 2

Let’s assume that the slope of the first line be m1 and

The slope of the other line be m2.

Angle between the two lines is 60° (Given)

Therefore,

tanθ = |m1 – m2| / |1 + m1m2|

tan 60^o = |2 – m2| / |1 + 2m2|

√3 = ±((2 – m2) / (1 + 2m2))

After rationalization, we got,

m2 = (2 – √3) / (2√3 + 1) and m2 = -(2 – √3) / (2√3 + 1)

Case 1: When m2 = (2 – √3) / (2√3 + 1)

Therefore, the equation of line passing through point (2, 3) and having slope m2 = (2 – √3) / (2√3 + 1) is :

y – 2 = ((2 – √3) / (2√3 + 1)) x (x – 2)

After solving above equation we got,

(√3 – 2)x + (2√3 + 1) y = 8√3 – 1

Equation of line is (√3 – 2)x + (2√3 + 1) y = 8√3 – 1.

Case 2: When m2 = -(2 – √3) / (2√3 + 1)

Therefore, the equation of line passing through point (2, 3) and having slope m2 = -(2 – √3) / (2√3 + 1) is :

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

After solving above equation we got,

(√3 + 2)x + (2√3 – 1) y = 8√3 + 1

Equation of line is (√3 + 2)x + (2√3 – 1) y = 8√3 + 1.

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

Solution:

Given that,

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

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

Let’s assume that 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’s the slope of line AB be m1

m1 = (2 – 4)/(-1 – 3)

= -2/(-4) = 1/2

Let’s 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

Hence, the required equation of the line is 2x + y = 5

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

Solution:

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

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

m1 = (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 (Given)

(b-3) / (a+1) × (3/4) = -1

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

3b – 9 = -4a – 4

4a + 3b = 5 ————(i)

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

3a – 4b = 16 ———-(ii)

after solving equations (i) and (ii), we get

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

Hence, the co-ordinates of the foot of perpendicular is (68/25, -49/25)

Question 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 that,

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

As we know that the equation of line is y = mx + c

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

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

Slope of the given line is m. (Assumption)

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

Hence, the values of m and c are 1/2 and 5/2 respectively.

Question 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 that,

The equations of given lines are

x cos θ – y sin θ = k cos 2θ ———–(i)

x sec θ + y cosec θ = k —————(ii)

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

d = |Ax1 + By1 + C| / √A² + B²

After comparing equation (i) and (ii) we get,

A = cos θ, B = -sin θ and C = -k cos 2θ

Given that p is length of perpendicular from (0, 0) to line (i)

p = |A × 0 + B × 0 + C| / √A² + B²

= |-k cos 2θ| / √cos² θ + sin² θ = k cos 2θ

p = k cos 2θ

Let’s square on both side, and we get,

p² = k² cos² 2θ ————(iii)

Now compare eqn. (ii) with general equation of line i.e. Ax + By + C = 0, and we get,

A = sec θ, B = cosec θ and C = -k

As we know that q is length of perpendicular from (0, 0) to line (ii)

q = |A x 0 + B x 0 + C| / √A² + B² = |C| / √A² + B²

= |-k| / √sec² θ + cosec² θ = k cos θ sin θ

q = k cos θ sin θ

Multiply both sides by 2, 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 ————–(iv)

Now add (iii) and (iv) we get

p² + 4q² = k² cos² 2θ + k² sin² 2θ

p² + 4q² = k² (cos² 2θ + sin² 2θ) (As we know that cos² 2θ + sin² 2θ = 1)

Hence, p² + 4q² = k²

Hence, proved.

Question 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’s assume that AD be the altitude of triangle ABC from vertex A.

Therefore, AD is perpendicular to BC

Given that,

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

Let’s slope of line BC = m1

m1 = (-1 – 2) / (4 – 1)

m1 = -1

Let’s slope of line AD be m2

AD is perpendicular to BC

m1 × m2 = -1

-1 × m2 = -1

m2 = 1

The equation of the line passing through 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

Equation of BC is

y + 1 = -1 × (x – 4)

y + 1 = -x + 4

x + y – 3 = 0 ————-(i)

Perpendicular distance d of a line Ax + By + C = 0 from a point (x1, y1) is given by d = |Ax1 + By1 + C| / √A² + B²

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

Length of AD = |1 × 2 + 1 × 3 – 3| / √1² + 1² = √2 units (A = 1, B = 1 and C = -3)

Hence, the equation and the length of the altitude from vertex A are y – x = 1 and √2 units respectively.

Question 17. If p is the length of 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:

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 ————-(i)

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

d = |Ax1 + By1 + C| / √A² + B²

After comparing eqn. (i) with general equation of line i.e. Ax + By + C = 0 we get,

A = b, B = a and C = -ab

Let’s assume that if p is length of perpendicular from point (x1, y1) = (0, 0) to line (i), we get

p = |A x 0 + B x 0 – ab| / √a² + b² = |-ab| / √a² + b²

Now square on both the sides we get

p² = (-ab)² / a²+ b²

1 / p² = (a² + b²) / a²b²

Hence, 1/p² = 1/a² + 1/b²

Hence, proved.