**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

Add 15 to each side.

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

Add 1/2 to each side.

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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