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

**Answer
1** :

(i) x + 7y = 0

Given:

The equation is x + 7y = 0

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

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

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.

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

**Answer
2** :

(i) 3x + 2y – 12 = 0

Given:

The equation is 3x + 2y – 12 = 0

Equation of line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepts on x axis and 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

Intercept on x – axis is 4

Intercept on y – axis is 6

(ii) 4x – 3y = 6

Given:

The equation is 4x – 3y = 6

Equation of line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepts on x axis and 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

Intercept on x – axis is 3/2

Intercept on y – axis is -2

(iii) 3y + 2 = 0

Given:

The equation is 3y + 2 = 0

Equation of line in intercept form is given by x/a + y/b = 1, where ‘a’ and ‘b’ are intercepts on x axis and 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

Intercept on x – axis is 0

Intercept on y – axis is -2/3

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

**Answer
3** :

(i) x – √3y + 8 = 0

Given:

The equation is x – √3y + 8 = 0

Equation of line in normal form is given by x cos θ + y sinθ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’is perpendicular distance from origin.

So now, x – √3y + 8 = 0

x – √3y = -8

Divide both the sides by √(1^{2} + (√3)^{2})= √(1 + 3) = √4 = 2

x/2 – √3y/2 = -8/2

(-1/2)x + √3/2y = 4

This is in the form of: x cos 120^{o} + y sin120^{o} = 4

∴ 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:

The equation is y – 2 = 0

Equation of line in normal form is given by x cos θ + y sinθ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’is perpendicular distance from origin.

So now, 0 × x + 1 × y = 2

Divide both sides by √(0^{2} + 1^{2}) =√1 = 1

0 (x) + 1 (y) = 2

This is in the form of: x cos 90^{o} + y sin 90^{o} =2

∴ 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 = 2

Angle between perpendicular and positive x – axis = 90°

(iii) x – y = 4

Given:

The equation is x – y + 4 = 0

Equation of line in normal form is given by x cos θ + y sinθ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’is perpendicular distance from origin.

So now, x – y = 4

Divide both the sides by √(1^{2} + 1^{2})= √(1+1) = √2

x/√2 – y/√2 = 4/√2

(1/√2)x + (-1/√2)y = 2√2

This is in the form: x cos 315^{o} + y sin 315^{o} =2√2

∴ 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 = 2√2

Angle between perpendicular and positive x – axis = 315°

**Answer
4** :

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 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.

**Answer
5** :

Given:

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

4x + 3y = 12

4x + 3y – 12 = 0 …. (1)

Now, compare equation (1) with 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)

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

**Answer
6** :

(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 + C_{1} =0 and Ax + By + C_{2} = 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 + C_{1} =0 and Ax + By + C_{2} = 0 is given by

∴ The distance between parallel lines is |p+r|/l√2

**Answer
7** :

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 thegiven line.

The slope of the given line is 3/4

We know that parallel line have same slope.

∴ Slope of other line = m = 3/4

Equation of line having slope m and passing through (x_{1},y_{1}) is given by

y – y_{1} = m (x – x_{1})

∴ Equation of 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

**Answer
8** :

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 mis the slope of the given line.]

Slope of the given line is 1/7

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

Slope of the line perpendicular to the line having a slopeof 1/7 is -1/(1/7) = -7

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

y = -7 (x – 3)

y = -7x + 21

7x + y = 21

∴ The equation is 7x + y = 21

**Answer
9** :

Given:

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

So, y = -√3x + 1 … (1) and

y = -1/√3x + 1/√3 …. (2)

Slope of line (1) is m1 = -√3, while the slope of 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°

**Answer
10** :

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

Name:

Email:

Copyright 2017, All Rights Reserved. A Product Design BY CoreNet Web Technology