# SLOPE INTERCEPT FORM EQUATION OF A LINE WORKSHEET

Problem 1 :

Find the slope-intercept form equation of the straight line with slope 2 and y-intercept 5.

Problem 2 :

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

Problem 3 :

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

Problem 4 :

Find the point-slope form equation of the straight line passing through the point (-5, -4) with slope 3.

Problem 5 :

Find the point-slope form equation of the straight line passing through the point (1, 2) and parallel to the line whose equation is x + 2y + 3  =  0.

Problem 6 :

Find the point-slope form equation of the straight line passing through the point (-2, 3) and perpendicular to the line whose equation is x - 2y - 6  =  0.

Problem 7 :

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.

Problem 1 :

Find the slope-intercept form equation of the straight line with slope 2 and y-intercept 5.

Solution :

Given : Slope = 2 and y-intercept form = 5.

Equation of the straight line in slope-intercept form :

y  =  mx + b

Substitute m = 2 and b = 5.

y  =  2x + 5

Problem 2 :

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

Solution :

Because 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

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

Problem 3 :

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

Solution :

Because 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 a straight line in slope-intercept form is

y  =  mx + b

Substitute m = 5 and b = -2.

y  =  5x - 2

Problem 4 :

Find the point-slope form equation of the straight line passing through the point (-5, -4) with slope 3.

Solution :

Equation of the straight line in slope-intercept form.

y  =  mx + b

Substitute m = 2/3.

y  =  3x + b -----(1)

Given : The line is passing through the point (-5, -4).

Then,

(1)----> -4  =  3(-5) + b

-4  =  -15 + b

11  =  b

Substitute b = 11 in (1).

(1)-----> y  =  3x + 11

Problem 5 :

Find the point-slope form equation of the straight line passing through the point (1, 2) and parallel to the line whose equation is x + 2y + 3  =  0.

Solution :

Write the equation of the line 'x + 2y + 3 = 0' in slope intercept form.

That is, y = mx + b.

x + 2y + 3  =  0

2y  =  -x - 3

y  =  (-1/2)x - 3/2

So, slope of the given line is -1/2.

Because the required line is parallel to the given line, the slopes are equal.

Then, slope of the required line is -1/2.

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

y  =  (-1/2)x + b -----(1)

Given : The line is passing through the point (1, 2).

Then,

(1)----> 2  =  (-1/2)(1) + b

2  =  -1/2 + b

2 + 1/2  =  b

5/2  =  b

Substitute b = 5/2 in (1).

(1)-----> y  =  (-1/2)x + 5/2

Problem 6 :

Find the point-slope form equation of the straight line passing through the point (-2, 3) and perpendicular to the line whose equation is x - 2y - 6  =  0.

Solution :

Write the equation of the line 'x - 2y - 6 = 0' in slope intercept form.

That is, y = mx + b.

x - 2y - 6  =  0

-2y  =  -x + 6

2y  =  x - 6

y  =  (1/2)x - 3

So, slope of the given line is 1/2.

Because the required line is perpendicular to the given line, product of the slopes is equal to -1.

Let 'm' be the slope of the required line.

Then,

m x (1/2)  =  -1

m/2  =  -1

m  =  -2

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

y  =  -2x + b -----(1)

Given : The line is passing through the point (-2, 3).

Then,

(1)----> 3  =  -2(-2) + b

3  =  4 + b

Subtract 4 from each side.

-1  =  b

Substitute b = -1 in (1).

(1)-----> y  =  -2x - 1

Problem 7 :

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 substitute 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 substitute 95 for x in

y  =  1500x + 100000

Then,

y  =  1500x95 + 100000

y  =  142500  +  100000

y  =  242500

So, the cost of 95 units is \$242500.

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