# DIFFERENT FORMS EQUATION OF STRAIGHT LINE

## About "Different forms equation of straight line"

Here we are going to see "Different forms equation of straight line".

Suppose that we have the graph of a straight line and we want to find its equation.

For any straight line, if we want to find the equation, we must have the following information of that straight line.

(i)  Slope and y-intercept

(ii)  One point and slope

(iii) Two points

(iv) Two intercepts (x-intercept and y-intercept)

If we have any one of the five information given above we will be able to find the equation of a straight line using the formulas given below.

Now, let us look at the formulas of different forms equation of straight line.

## Different forms equation of straight line

1. Slope - Intercept form equation of a line

Here,

Slope of the line  =  m

y-intercept  =  b

2.Point - slope form equation of a line

Here,

Slope of the line  =  m

Point  =  (x₁ , y)

3. Two-points form equation of a line

Here, the two points are (x₁ , y) and (x₂ , y)

4.  Intercept form equation of a line

Here,

x- intercept  =  a

y- intercept  =  b

Apart from the above forms of equation of straight line, there are some other ways to get equation of a straight line.

1. If a straight line is passing through a point (0,k) on y-axis and parallel to x-axis, then  the equation of the straight line is   y = k

2. If a straight line is passing through a point (c,0) on x-axis and parallel to y-axis, then  the equation of the straight line is   x = c

3. Equation of x-axis is  y  =  0.

(Because, the value of "y" in all the points  on x-axis is zero)

4. Equation of y-axis is  x  =  0.

(Because, the value of "x" in all the points  on y-axis is zero)

5. General equation of a straight line is

ax + by + c  =  0

## Problems related to different forms equation of straight line

To have better understanding on different forms equation of straight line, let us look st some example problems.

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

Let us look at the next problem on "Different forms equation of straight line"

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

Now, we know the slope m = 1 and y-intercept b = 2/5.

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

y = mx + b

Plugging m = 1 and b = 2/5, we get  y  =  1x + 2/5

y  =  (5x + 2) / 5 -------> 5y = 5x + 2 --------> 0  =  5x - 5y + 2 = 0

Hence the general equation of straight line is 5x - 5y + 2 = 0

Let us look at the next problem on "Different forms equation of straight line"

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 = m(x - x)

Plugging (x₁ , y₁) = (-2 , 3) and m = 1/3, we get

y - 3 = (1/3)(x + 2) ----------> 3y - 9 = x + 2

3y - 9 = x + 2 ------------> 0 = x - 3y + 11

Hence the general equation of straight line is x - 3y + 11 = 0

Let us look at the next problem on "Different forms equation of straight line"

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 are (-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)

Plugging (x₁ , y₁) = (-1, 1) and (x₂, y) = (2, -4), we get

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

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

Let us look at the next problem on "Different forms equation of straight line"

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, equation of the median through A is

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

Plugging (x₁ , y₁) = (2, 1) and (x₂, y) = (1, 4), we get

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

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

- y + 1 = 3x - 3 ----------> 3x + y - 4 = 0

Hence, the equation of the median through A is 3x + y - 4 = 0.

Let us look at the next problem on "Different forms equation of straight line"

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 and y-intercept  "b"  =  3/4

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

x/a  +  y/b  =  1

Plugging  a  =  2/3  and  b  =  3/4, we get

x / (2/3)  +  y / (3/4)  =  1

3x / 2  +  4y / 3  =  1

(9x + 8y)  / 6  =  1

9x + 8y  =  6 -------> 9x + 8y - 6  =  0

Hence the general equation of straight line is 9x + 8y - 6 = 0

Let us look at the next problem on "Different forms equation of a 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.

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 --------> 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

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

(5-a)x + ay  =  a(5-a) ----------(1)

Since the straight line (1) is passing through (6, -2),

we can plug x = 6 and y = -2 in (1)

(1) --------> (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

5x - 10y - 50  =  0 -------> x - 2y - 10  =  0

When a = 3, (1)------->(5 - 3)x + 3y  =  3(5 - 3)

2x + 3y  =  6 --------> 2x + 3y - 6  =  0

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

Let us look at the next problem on "Different forms equation of straight line"

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.

Let us look at the next problem on "Different forms equation of straight line"

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

(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

Let us look at the next problem on "Different forms equation of straight line"

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"

Now, we know that slope m = 5 and y-intercept b = -2.

Equation of straight in slope-intercept form is

y  =  mx + b

Plugging m  =  5  and  b  =  -2,  we get    y  = 5x - 2

y  = 5x - 2 ---------> 5x - y - 2  =  0

Hence, the general equation of the required line is 5x - y - 2 = 0

On this web page "Different forms equation of straight line", next we are going to see, how to use slope-intercept form equation of a line can be used as linear cost function.

## Using slope intercept form as linear cost function

Slope intercept form equation of a line is one of the forms of "Different forms of equation of straight line".

This slope intercept form of " Different forms equation of straight line" can also be used as "Linear cost function".

"Linear cost function" is the function where the cost curve of a particular product will be a straight line. It is a confusable topic for some students who study the topic "Different forms equation of straight line" in algebra in both school and college level math.

Mostly this function used to find the total cost of "n" units of the products produced.

For any product, if the cost curve is linear, the linear cost function of the product will be in the form of

y = Ax + B

Here, "y" stands for total cost

"x" stands for number of units.

"A" stands for cost of one unit of the product

"B" stands for fixed cost.

Linear cost function is called as bi parametric function. Here the two parameters are "A" and "B".

Once the two parameters "A" and "B" are known, the complete function can be known.

In "Different forms equation of straight line", next we are going to see, how word problems can be solved using linear cost function.

## How to solve word problems on linear cost function ?

One of the major advantages of "Different forms equation of straight line" is solving word problems.

The following steps are involved in solving word problems on linear cost function.

Step 1 :

First we have to go through the question carefully and understand the information given in the question.

After having gone through the question, we have to conclude whether the information given in the question fits linear-cost function.

If the information fits the linear-cost function, we have to follow step 2

Step 2 :

Target : We have to know what has to be found.

In linear-cost function, mostly the target would be to find either the value of "y" (total cost) or "x" (number of units).

Step 3 :

In step 3,  we have to calculate the two constants "A"  and "B" from the information given in the questions. It has been shown clearly in the example problem given below.

Step 4 :

Once the values of "A" and "B"  in   y = Ax + B are found, the linear-cost function would be completely known.

Step 5 :

After step 4, based on the target of the question, we have to find either the value of "y" or "x" for the given input.

For example, if the value of "x" (number of units) is given, we can find the value of "y" (total cost).

If the value of "y" (total cost) is given, we can find the value of "x" (number of units).

In "Different forms equation of straight line", next we are going to see an example word problem on linear cost function.

## Example Problem

To understand the applications of "Different forms equation of straight line", let us go through the following problem on linear cost function.

Problem :

A manufacturer produces 80 units of a particular product at a cost of \$ 220000 and 125 units at a cost of \$ 287500. Assuming the cost curve to be linear, find the cost of 95 units.

Solution :

Step 1 :

When we go through the question, it is very clear that the cost curve is linear.

And the function which best fits the given information will be a linear-cost function.

That is,                              y = Ax + B

Here "y" --------> Total cost

"x" --------> Number of units

Step 2 :

Target : We have to find the value of "y" for "x = 95"

Step 3 :

From the question, we have

x = 80        and       y = 220000

x = 75        and       y = 287500

Step 4 :

When we plug the above values of "x" and "y" in y = Ax + B, we get

220000 = 80A + B

287500 = 75A + B

Step 5 :

When we solve the above two linear equations for A and B, we get                      A  =  1500   and    B  =  100000

Step 6 :

From A = 1500 and B = 100000, the linear-cost function for the given information is
y  = 1500x  +  100000

Step 7 :

To estimate the value of "y" for "x = 95", we have to plug "x = 95" in "y = 1500x + 100000"

y = 1500x95 + 100000

y = 142500  +  100000

y = 242500

Hence, the cost of 95 units is \$ 242500

We understand the applications of "Different forms equation of straight line" through the word problem explained on linear cost function.

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