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

**Question 1. Reduce the following equations into slope â€“ intercept form and find their slopes and the y â€“ intercepts.**

**(i) x + 7y = 0**

**(ii) 6x + 3y â€“ 5 = 0**

**(iii) y = 0**

**Solution: **

(i) x + 7y = 0Given that,

The equation is x + 7y = 0

Slope â€“ 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 â€“ 5 = 0Given that,

The equation is 6x + 3y â€“ 5 = 0

Slope â€“ 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 = 0Given that,

The equation is y = 0

Slope â€“ 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 Ã— 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 â€“ 12 = 0**

**(ii) 4x â€“ 3y = 6**

**(iii) 3y + 2 = 0**

**Solution: **

(i) 3x + 2y â€“ 12 = 0Given that,

The equation is 3x + 2y â€“ 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 â€“ 3y = 6Given that,

The equation is 4x â€“ 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 â€“ 3y = 6

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

4x/6 â€“ 3y/6 = 6/6

2x/3 â€“ 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 = 0Given 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 â€“ âˆš3y + 8 = 0**

**(ii) y â€“ 2 = 0**

**(iii) x â€“ y = 4**

**Solution: **

(i) x â€“ âˆš3y + 8 = 0Given that,

The equation is x â€“ âˆš3y + 8 = 0

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

Therefore, x â€“ âˆš3y + 8 = 0

x â€“ âˆš3y = -8

Let’s divide both sides by âˆš(1

^{2}+ (âˆš3)^{2}) = âˆš(1 + 3) = âˆš4 = 2, and we will get,x/2 â€“ âˆš3y/2 = -8/2

(-1/2)x + âˆš3/2y = 4

Now the above equation is in the form of x cos 120

^{o}+ y sin 120^{o}= 4

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

Perpendicular distance of line from origin = 4

Angle between perpendicular and positive x â€“ axis = 120Â°

(ii) y â€“ 2 = 0Given that,

The equation is y â€“ 2 = 0

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

Therefore, 0 Ã— x + 1 Ã— y = 2

Let’s divide both sides by âˆš(0

^{2 }+ 1^{2}) = âˆš1 = 1, and we will get,0 (x) + 1 (y) = 2

Now the above equation is in the form of x cos 90

^{o}+ y sin 90^{o}= 2

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

Perpendicular distance of line from origin is 2 and

Angle between perpendicular and positive x â€“ axis is 90Â°

(iii) x â€“ y = 4Given that,

The equation is x â€“ y + 4 = 0

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

Therefore, x â€“ y = 4

Let’s divide both sides by âˆš(1

^{2}+ 1^{2}) = âˆš(1+1) = âˆš2, and we will get,x/âˆš2 â€“ y/âˆš2 = 4/âˆš2

(1/âˆš2)x + (-1/âˆš2)y = 2âˆš2

Now the above equation is in the form of x cos 315

^{o}+ y sin 315^{o}= 2âˆš2

Hence, the above equation is of the form x cos Î¸ + y sin Î¸ = p, where Î¸ = 315Â° and p = 2âˆš2.

Perpendicular distance of line from origin is 2âˆš2 and

Angle between perpendicular and positive x â€“ axis is 315Â°

**Question 4. Find the distance of the point (â€“1, 1) from the line 12(x + 6) = 5(y â€“ 2).**

**Solution: **

Given that,

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

12x + 72 = 5y â€“ 10

12x â€“ 5y + 82 = 0 ——–(i)

Now, after comparing equation (i) with general equation of line Ax + By + C = 0, we get A = 12, B = â€“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| / âˆšA

^{2 }+ B^{2}Given points (x1, y1) are (-1, 1)

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

d = |12 x (-1) + (-5) + 82| / âˆš12

^{2}+ (-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 â€“ 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| / âˆšA

^{2}+ B^{2}4 = |4a + 3 Ã— 0 – 12| / âˆš4

^{2}+ 3^{2}4 = |4a – 12| / 5

|4a â€“ 12| = 4 Ã— 5

Â± (4a â€“ 12) = 20

4a â€“ 12 = 20 or â€“ (4a â€“ 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 â€“ axis are (-2, 0) and (8, 0)

**Question 6. Find the distance between parallel lines**

**(i) 15x + 8y â€“ 34 = 0 and 15x + 8y + 31 = 0**

**(ii) l(x + y) + p = 0 and l (x + y) â€“ r = 0**

**Solution:**

(i) 15x + 8y â€“ 34 = 0 and 15x + 8y + 31 = 0Given that,

The parallel lines are 15x + 8y â€“ 34 = 0 & 15x + 8y + 31 = 0.

By using the formula, the distance d between parallel lines Ax + By + C

_{1}= 0 and Ax + By + C_{2}= 0 is given by,d = |C

_{1}– C_{2}| / âˆšA^{2}+ B^{2}From given equation we get, A = 15, B = 8, C

_{1}= -34, C_{2}= 31Now apply the formula and calculate distance between parallel lines,

d = |-34 – 31| / âˆš15

^{2}+ 8^{2}= |-65| / âˆš225 + 64= 65 / 17

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

(ii) l(x + y) + p = 0 and l (x + y) â€“ r = 0Given that,

The parallel lines are l (x + y) + p = 0 and l (x + y) â€“ r = 0.

lx + ly + p = 0 and lx + ly â€“ r = 0

By using the formula,

By using the formula, the distance d between parallel lines Ax + By + C

_{1}= 0 and Ax + By + C_{2}= 0 is given by,d = |C

_{1}– C_{2}| / âˆšA^{2}+ B^{2}From given equation we get, A = l, B = l, C

_{1}= p, C_{2}= -rNow apply the formula and calculate distance between parallel lines,

d = |p – (-r)| / âˆšl

^{2}+ l^{2}= |p+ r| / âˆš2 l = |p+r|/lâˆš2

Hence, the distance between parallel lines is |p+r|/lâˆš2

**Question 7. Find equation of the line parallel to the line 3x âˆ’ 4y + 2 = 0 and passing through the point (â€“2, 3).**

**Solution: **

Given that,

The line is 3x â€“ 4y + 2 = 0

Therefore, y = 3x/4 + 2/4 = 3x/4 + Â½

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 (x

_{1}, y_{1}) is given by, y â€“ y_{1}= m (x â€“ x_{1})Now put the value of slope 3/4 and points (-2, 3) in above formula, and we get,

y â€“ 3 = Â¾ (x â€“ (-2))

4y â€“ 3 Ã— 4 = 3x + 3 Ã— 2

3x â€“ 4y = 18

Hence, the equation is 3x â€“ 4y = 18

**Question 8. Find equation of the line perpendicular to the line x â€“ 7y + 5 = 0 and having x intercept 3.**

**Solution: **

Given that,

The equation of line is x â€“ 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 â€“ d)

y = -7 (x â€“ 3)

y = -7x + 21

7x + y = 21

Hence, the equation is 7x + y = 21

**Question 9. Find angles between the lines âˆš3x + y = 1 and x + âˆš3y = 1.**

**Solution: **

Given that,

The lines are âˆš3x + y = 1 and x + âˆš3y = 1

Therefore, y = -âˆš3x + 1 ———(i) &

y = -1/âˆš3x + 1/âˆš3 ——–(ii)

Slope of line (i) is -âˆš3, while the slope of line (ii) is -1/âˆš3

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

As we know that,

tanÎ¸ = |m1 – m2| / |1 + m

_{1}m_{2}|Put the values of m

_{1}and m_{2}formula, and we will get,= |-âˆš3 – (-1/âˆš3)| / |1 + (-âˆš3)(-1/âˆš3)| = 1 / âˆš3

Î¸ = 30Â°

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

**Question 10. The line through the points (h, 3) and (4, 1) intersects the line 7x âˆ’ 9y âˆ’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 m

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

Let’s slope of line 7x â€“ 9y â€“ 19 = 0 be m

_{2}7x â€“ 9y â€“ 19 = 0

Therefore, y = 7/9x â€“ 19/9

m

_{2}= 7/9Given that, the given lines are perpendicular

m

_{1}Ã— m_{2}= -1-2/(4-h) Ã— 7/9 = -1

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

-14 = -1 Ã— (36 â€“ 9h)

36 â€“ 9h = 14

9h = 36 â€“ 14

h = 22/9

Hence, the value of h is 22/9.