**Equation of a Straight Line Worksheet :**

Worksheet given in this section is much useful to the students who would like to practice problems on finding the equation of a line.

Before look at the worksheet, if you would like to know the basic stuff about finding the equation of a line,

**Problem 1 :**

Find the equations of the straight lines parallel to the coordinate axes and passing through the point (3, -4).

**Problem 2 :**

Find the general equation of the straight line whose angle of inclination is 45° and y-intercept is 2/5.

**Problem 3 :**

Find the general equation of the straight line passing through the point (-2, 3) with slope 1/3.

**Problem 4 :**

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

**Problem 5 :**

The vertices of a triangle ABC are A(2, 1), B(-2, 3) and C(4, 5). Find the equation of the median through the vertex A.

**Problem 6 :**

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

**Problem 7 :**

Find the equations of the straight lines each passing through the point (6, -2) and whose sum of the intercepts is 5.

**Problem 8 :**

Find the general equations of the straight lines parallel to x- axis which are at a distance of 5 units from the x-axis.

**Problem 9 :**

Find the slope and y-intercept of the straight line whose equation is 4x - 2y + 1 = 0.

**Problem 10 :**

A straight line has the slope 5. If the line cuts y-axis at -2, find the general equation of the straight line.

**Problem 1 :**

Find the equations of the straight lines parallel to the coordinate axes and passing through the point (3, -4)

**Solution :**

Let L and L' be the straight lines passing through the point (3, -4) and parallel to x-axis and y-axis respectively.

The y-coordinate of every point on the line L is – 4.

Hence, the equation of the line L is

y = - 4

Similarly, the x-coordinate of every point on the straight line L' is 3

Hence, the equation of the line L' is

x = 3

**Problem 2 :**

Find the general equation of the straight line whose angle of inclination is 45° and y-intercept is 2/5.

**Solution :**

From the angle of inclination 45°, we can get the slope.

Slope of the line m = tan45° = 1

Given : y-intercept b = 2/5

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

y = mx + b

Substitute m = 1 and b = 2/5.

y = 1x + 2/5

Multiply each side by 5.

5y = 5x + 2

Subtract 5y from each side.

0 = 5x - 5y + 2

or

5x - 5y + 2 = 0

Hence the general equation of straight line is

5x - 5y + 2 = 0

**Problem 3 :**

Find the general equation of the straight line passing through the point (-2, 3) with slope 1/3.

**Solution :**

Given : Point = (-2, 3) and slope m = 1/3

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

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

Substitute (x_{1} , y_{1}) = (-2 , 3) and m = 1/3.

y - 3 = 1/3 ⋅ (x + 2)

Multiply each side by 3.

3(y - 3) = x + 2

Simplify.

3y - 9 = x + 2

Subtract 3y from each side.

-9 = x - 3y + 2

Add 9 to each side.

0 = x - 3y + 11

So, the general equation of straight line is

x - 3y + 11 = 0

**Problem 4 :**

Find the general equation of the straight 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 - y_{1}) / (y_{2} - y_{1}) = (x - x_{1}) / (x_{2} - x_{1})

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 straight line is

5x + 3y + 2 = 0

**Problem 5 :**

The vertices of a triangle ABC are A(2, 1), B(-2, 3) and C(4, 5). Find the equation of the median through the vertex A.

**Solution :**

Median is a straight line joining a vertex and the midpoint of the opposite side.

Let D be the midpoint of BC.

The median through A is nothing but the line joining two points A (2,1) and D(1, 4).

So, the equation of the median through A is

(y - y_{1}) / (y_{2} - y_{1}) = (x - x_{1}) / (x_{2} - x_{1})

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

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

Simplify.

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

Cross multiply.

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

Simplify.

-y + 1 = 3x - 3

3x + y - 4 = 0

Hence, the equation of the median through A is

3x + y - 4 = 0

**Problem 6 :**

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

Hence the general equation of straight line is

9x + 8y - 6 = 0

**Problem 7 :**

Find the equations of the straight lines each passing through the point (6, -2) and whose sum of the intercepts is 5.

**Solution :**

Let "a" and "b" be the x-intercept and y-intercept of the required straight line respectively.

Given : Sum of the intercepts = 5

So, we have

a + b = 5

Subtract a from each side.

b = 5 - a

Now, equation of the straight line in intercept form is

x/a + y/b = 1

Plugging b = 5 - a, we get

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

Simplify.

[(5-a)x + ay ] / a(5-a) = 1

(5-a)x + ay = a(5-a) -----(1)

The straight line is passing through (6, -2).

So, substitute (x, y) = (6, -2).

(5 - a)6 - 2a = a(5 - a)

30 - 6a - 2a = 5a - a²

a² - 13a + 30 = 0

a² - 13a + 30 = 0

(a - 10)(a - 3) = 0

a = 10 and a = 3

When a = 10**,**

** **(1)-----> (5 - 10)x + 10y = 10(5 - 10)

- 5x + 10y = - 50

5x - 10y - 50 = 0

x - 2y - 10 = 0

When a = 3**,**

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

2x + 3y = 6

Hence, x - 2y - 10 = 0 and 2x + 3y - 6 = 0 are the general equations of the required straight lines.

**Problem 8 :**

Find the general equations of the straight lines parallel to x- axis which are at a distance of 5 units from the x-axis.

**Solution :**

From the given information, we can sketch the two lines as given below.

One line is above the x-axis at a distance of 5 units. And another line is below the x-axis at a distance of 5 units.

Hence, y = 5 and y = -5 are the required straight lines.

**Problem 9 :**

Find the slope and y-intercept of the straight line whose equation is 4x - 2y + 1 = 0.

**Solution :**

Since we want to find the slope and y-intercept, let us write the given equation 4x - 2y + 1 = 0 in slope-intercept form.

4x - 2y + 1 = 0

4x + 1 = 2y

Divide each side by 2.

(4x + 1)/2 = y

2x + 1/2 = y

or

y = 2x + 1/2

The above form is slope intercept form.

If we compare y = 2x + 1/2 and y = mx + b, we get

m = 2 and b = 1/2

Hence, the slope is 2 and y-intercept is 1/2.

**Problem 10 :**

A straight line has the slope 5. If the line cuts y-axis at -2, find the general equation of the straight line.

**Solution :**

Since the line cuts y-axis at "-2", clearly y-intercept is "-2"

So, the slope m = 5 and y-intercept b = -2.

Equation of a straight line in slope-intercept form :

y = mx + b

Substitute m = 5 and b = -2

y = 5x - 2

Subtract y from each side.

5x - y - 2 = 0

Hence, the general equation of the required line is

5x - y - 2 = 0

After having gone through the stuff given above, we hope that the students would have understood, "Equation of a Straight Line Worksheet".

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