**Problem 1 : **

Find the general equation of a line whose slope is 3 and y-intercept is 4.

**Problem 2 : **

Find the general equation of a line passing through the point (3, 0) with slope -6.

**Problem 3 :**

Find the general equation of a line passing through the point (2, 5) with slope -5.

**Problem 4 :**

Find the general equation of a line passing through the points (-1, 1) and (2, -4).

**Problem 5 :**

If the x-intercept and y-intercept of a line are 2/3 and 3/4 respectively, find the general equation of the line.

**Problem 6 : **

Find the equation of a line which is parallel to x-axis and passing through the point (2, 3).

**Problem 7 : **

Find the general equation of a line which is parallel to the line 3x −7y = 12 and passing through the point (6, 4).

**Problem 8 : **

Find the equation of a line which is perpendicular to the line

y = 4/3 ⋅ x − 7

and passing through the point (7, –1).

**Problem 1 : **

Find the general equation of a line whose slope is 3 and y-intercept is 4.

**Solution : **

Equation of a line in slope-intercept form :

y = mx + b

Substitute 3 for m and 4 for b.

y = 3x + 4.

Subtract y from each side.

0 = 3x - y + 4

or

3x - y + 4 = 0

So, the general equation of the required line is

3x - y + 4 = 0

**Problem 2 : **

Find the general equation of a line passing through the point (3, 0) with slope -6.

**Solution :**

Given : A point and slope

So, the equation of the straight line in point-slope form is

y - y_{1} = m(x - x_{1})

Substitute (x_{1} , y_{1}) = (3 , 0) and m = -6.

y - 0 = -6(x - 3)

Simplify.

y = -6x + 18

Add 6x to each side.

6x + y = 18

Subtract 18 from each side.

6x + y - 18 = 0

So, the general equation of straight line is

6x + y - 18 = 0

**Problem 3 :**

Find the general equation of a line passing through the point (2, 5) with slope -5.

**Solution :**

In this problem, instead of using point-slope form, we can use slope-intercept form also to find the equation of the line.

y = mx + b

Substitute m = -5.

y = -5x + b ----(1)

Substitute (x, y) = (2, 5)

5 = -5(2) + b

Simplify.

5 = -10 + b

Add 10 to each side.

15 = b

Substitute 15 in (1).

(1)----> y = -5x + 15

Add 5x to each side.

5x + y = 15

Subtract 15 from each side.

5x + y - 15 = 0

So, the general equation of straight line is

5x + y - 15 = 0

**Problem 4 :**

Find the general equation of a line passing through the points (-1, 1) and (2, -4).

**Solution :**

Given : Two points on the straight line : (-1, 1) and (2, -4).

So, the equation of the straight line in two-points form is

(y - y₁) / (y₂ - y₁) = (x - x₁) / (x₂ - x₁)

Substitute (x_{1} , y_{1}) = (-1, 1) and (x_{2}, y_{2}) = (2, -4).

(y - 1) / (-4 - 1) = (x + 1) / (2 + 1)

Simplify.

(y - 1) / (-5) = (x + 1) / 3

Cross multiply.

3(y - 1) = -5(x + 1)

3y - 3 = -5x - 5

5x + 3y + 2 = 0

Hence the general equation of the required line is

5x + 3y + 2 = 0

**Problem 5 :**

If the x-intercept and y-intercept of a line are 2/3 and 3/4 respectively, find the general equation of the line.

**Solution :**

Given :

x- intercept "a" = 2/3

y-intercept "b" = 3/4

So, the equation of the straight line in intercept form is

x/a + y/b = 1

Substitute a = 2/3 and b = 3/4.

x / (2/3) + y / (3/4) = 1

Simplify.

3x/2 + 4y/3 = 1

(9x + 8y) / 6 = 1

Multiply each side by 6.

9x + 8y = 6

Subtract 6 from each side from 6.

9x + 8y - 6 = 0

So, the general equation of the required line is

9x + 8y - 6 = 0

**Problem 6 : **

Find the equation of a line which is parallel to x-axis and passing through the point (2, 3).

**Solution : **

Equation of a line parallel to x-axis :

y = k -----(1)

The above line is passing through (2, 3).

So, substitute (x, y) = (2, 3).

3 = k

Substitute k = 3 in (1).

y = 3

So, the equation of the required line is

y = 3

**Problem 7 : **

Find the general equation of a line which is parallel to the line 3x −7y = 12 and passing through the point (6, 4).

**Solution : **

Equation of a line parallel to 3x - 7y = 12 :

3x - 7y + k = 0 ----- (1)

The above line is passing through (6, 4).

So, substitute (x, y) = (6, 4).

3(6) - 7(4) + k = 0

Simplify.

18 - 28 + k = 0

-10 + k = 0

Add 10 to each side.

k = 10

Substitute k = 10 in (1).

(1)-----> 3x - 7y + 10 = 0

So, the general equation of the required line is

3x - 7y + 10 = 0

**Problem 8 : **

Find the equation of a line which is perpendicular to the line

y = 4/3 ⋅ x − 7

and passing through the point (7, –1).

**Solution : **

Write the equation y = 4/3 ⋅ x − 7 in general form.

y = 4/3 ⋅ x − 7

Multiply each side by 3.

3y = 4x - 21

Subtract 3y from each side.

0 = 4x - 3y - 21

or

4x - 3y - 21 = 0

Equation of a line perpendicular to 4x - 3y - 21 = 0.

3x + 4y + k = 0 ----- (1)

The above line is passing through (7, -1).

So, substitute (x, y) = (7, -1).

3(7) + 4(-1) + k = 0

Simplify.

21 - 4 + k = 0

17 + k = 0

Subtract 17 from each side.

k = -17

Substitute k = -17 in (1).

(1)-----> 3x + 4y - 17 = 0

So, the general equation of the required straight line is

3x + 4y - 17 = 0

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