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Class 11 NCERT Solutions- Chapter 10 Straight Lines – Exercise 10.3 | Set 1

  • Last Updated : 07 Apr, 2021

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

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(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. Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

(i) x – √3y + 8 = 0

(ii) y – 2 = 0

(iii) x – y = 4

Solution: 

(i) x – √3y + 8 = 0

Given that,



The equation is x – √3y + 8 = 0

As we know that equation of line in normal form is given by x cos θ + y sin θ = p where θ is the angle between perpendicular and positive x axis and p is perpendicular distance from origin respectively.

Therefore, x – √3y + 8 = 0

x – √3y = -8

Let’s divide both sides by √(12 + (√3)2) = √(1 + 3) = √4 = 2, and we will get,

x/2 – √3y/2 = -8/2

(-1/2)x + √3/2y = 4

Now the above equation is in the form of x cos 120o + y sin 120o = 4

Hence, the above equation is of the form x cos θ + y sin θ = p, where θ = 120° and p = 4.

Perpendicular distance of line from origin = 4

Angle between perpendicular and positive x – axis = 120°

(ii) y – 2 = 0

Given that,

The equation is y – 2 = 0

As we know that equation of line in normal form is given by x cos θ + y sin θ = p where θ is the angle between perpendicular and positive x-axis and p is perpendicular distance from origin respectively.

Therefore, 0 × x + 1 × y = 2

Let’s divide both sides by √(02 + 12) = √1 = 1, and we will get,

0 (x) + 1 (y) = 2

Now the above equation is in the form of x cos 90o + y sin 90o = 2

Hence, the above equation is of the form x cos θ + y sin θ = p, where θ = 90° and p = 2.

Perpendicular distance of line from origin is 2 and

Angle between perpendicular and positive x – axis is 90°

(iii) x – y = 4

Given that,

The equation is x – y + 4 = 0

As we know that equation of line in normal form is given by x cos θ + y sin θ = p where θ is the angle between perpendicular and positive x-axis and p is perpendicular distance from origin respectively.

Therefore, x – y = 4

Let’s divide both sides by √(12 + 12) = √(1+1) = √2, and we will get,

x/√2 – y/√2 = 4/√2

(1/√2)x + (-1/√2)y = 2√2



Now the above equation is in the form of x cos 315o + y sin 315o = 2√2

Hence, the above equation is of the form x cos θ + y sin θ = p, where θ = 315° and p = 2√2.

Perpendicular distance of line from origin is 2√2 and

Angle between perpendicular and positive x – axis is 315°

Question 4. 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 5. 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 6. 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 7. 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 8. 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 9. 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 10. 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.




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