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Linear equations  (Page 4/6)

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Let L size 12{L} {} be a line with slope m size 12{m} {} , and containing a point ( x 1 , y 1 ) size 12{ \( x rSub { size 8{1} } ,y rSub { size 8{1} } \) } {} . If ( x , y ) size 12{ \( x,y \) } {} is any other point on the line L size 12{L} {} , then by the definition of a slope, we get

m = y − y 1 x − x 1 size 12{m= { {y - y rSub { size 8{1} } } over {x - x rSub { size 8{1} } } } } {}
y − y 1 = m ( x − x 1 ) size 12{y - y rSub { size 8{1} } =m \( x - x rSub { size 8{1} } \) } {}

The last result is referred to as the point-slope form or point-slope formula. If we simplify this formula, we get the equation of the line in the standard form, Ax + By = C size 12{ ital "Ax"+ ital "By"=C} {} .

Using the point-slope formula, find the standard form of an equation of the line that passes through the point (2, 3) and has slope –3/5.

Substituting the point (2, 3) and m = − 3 / 5 size 12{m= - 3/5} {} in the point-slope formula, we get

y − 3 = − 3 / 5 ( x − 2 ) size 12{y - 3= - 3/5 \( x - 2 \) } {}

Multiplying both sides by 5 gives us

5 ( y − 3 ) = − 3 / 5 ( x − 2 ) size 12{5 \( y - 3 \) = - 3/5 \( x - 2 \) } {}
5y − 15 = − 3x + 6 size 12{5y - "15"= - 3x+6} {}
3x + 5y = 21 size 12{3x+5y="21"} {}

Find the standard form of the line that passes through the points (1, -2), and (4, 0).

m = 0 − ( − 2 ) 4 − 1 = 2 3 size 12{m= { {0 - \( - 2 \) } over {4 - 1} } = { {2} over {3} } } {}

The point-slope form is

y − ( − 2 ) = 2 / 3 ( x − 1 ) size 12{y - \( - 2 \) =2/3 \( x - 1 \) } {}

Multiplying both sides by 3 gives us

3 ( y + 2 ) = 2 ( x − 1 ) size 12{3 \( y+2 \) =2 \( x - 1 \) } {}
3y + 6 = 2x − 2 size 12{3y+6=2x - 2} {}
− 2x + 3y = − 8 size 12{ - 2x+3y= - 8} {}
2x − 3y = 8 size 12{2x - 3y=8} {}

We should always be able to convert from one form of an equation to another. That is, if we are given a line in the slope-intercept form, we should be able to express it in the standard form, and vice versa.

Write the equation y = − 2 / 3x + 3 size 12{y= - 2/3x+3} {} in the standard form.

Multiplying both sides of the equation by 3, we get

3y = − 2x + 9 size 12{3y= - 2x+9} {}
2x + 3y = 9 size 12{2x+3y=9} {}

Write the equation 3x − 4y = 10 size 12{3x - 4y="10"} {} in the slope-intercept form.

Solving for y size 12{y} {} , we get

− 4y = − 3x + 10 size 12{ - 4y= - 3x+"10"} {}
y = 3 / 4x − 5 / 2 size 12{y=3/4x - 5/2} {}

Finally, we learn a very quick and easy way to write an equation of a line in the standard form. But first we must learn to find the slope of a line in the standard form by inspection.

By solving for y size 12{y} {} , it can easily be shown that the slope of the line Ax + By = C size 12{ ital "Ax"+ ital "By"=C} {} is − A / B size 12{ - A/B} {} . The reader should verify.

Find the slope of the following lines, by inspection.

  1. 3x − 5y = 10 size 12{3x - 5y="10"} {}
  2. 2x + 7y = 20 size 12{2x+7y="20"} {}
  3. 4x − 3y = 8 size 12{4x - 3y=8} {}
  1. A = 3 size 12{A=3} {} , B = − 5 size 12{B= - 5} {} , therefore, m = − 3 − 5 = 3 5 size 12{m= - { {3} over { - 5} } = { {3} over {5} } } {}
  2. A = 2 size 12{A=2} {} , B = 7 size 12{B=7} {} , therefore, m = − 2 7 size 12{m= - { {2} over {7} } } {}
  3. m = − 4 − 3 = 4 3 size 12{m= - { {4} over { - 3} } = { {4} over {3} } } {}

Now that we know how to find the slope of a line in the standard form by inspection, our job in finding the equation of a line is going to be very easy.

Find an equation of the line that passes through (2, 3) and has slope − 4 / 5 size 12{ - 4/5} {} .

Since the slope of the line is − 4 / 5 size 12{ - 4/5} {} , we know that the left side of the equation is 4x + 5y size 12{4x+5y} {} , and the partial equation is going to be

4x + 5y = c size 12{4x+5y=c} {}

Of course, c size 12{c} {} can easily be found by substituting for x size 12{x} {} and y size 12{y} {} .

4 ( 2 ) + 5 ( 3 ) = c size 12{4 \( 2 \) +5 \( 3 \) =c} {}
23 = c size 12{"23"=c} {}

The desired equation is

4x + 5y = 23 size 12{4x+5y="23"} {} .

If you use this method often enough, you can do these problems very quickly.

Applications

Now that we have learned to determine equations of lines, we get to apply these ideas in real-life equations.

A taxi service charges $0.50 per mile plus a $5 flat fee. What will be the cost of traveling 20 miles? What will be cost of traveling x size 12{x} {} miles?

The cost of traveling 20 miles = y = ( . 50 ) ( 20 ) + 5 = 10 + 5 = 15 size 12{"The cost of travelling 20 miles"=y= \( "." "50" \) \( "20" \) +5="10"+5="15"} {}
The cost of traveling   x    miles = y = ( . 50 ) ( x ) + 5 = . 50 x + 5 size 12{"The cost of travelling "x" mile"=y= \( "." "50" \) \( x \) +5= "." "50"x+5} {}

In this problem, $0.50 per mile is referred to as the variable cost , and the flat charge $5 as the fixed cost . Now if we look at our cost equation y = . 50 x + 5 size 12{y= "." "50"x+5} {} , we can see that the variable cost corresponds to the slope and the fixed cost to the y-intercept.

The variable cost to manufacture a product is $10 and the fixed cost $2500. If x size 12{x} {} represents the number of items manufactured and y size 12{y} {} the total cost, write the cost function.

The fact that the variable cost represents the slope and the fixed cost represents the y-intercept, makes m = 10 size 12{m="10"} {} and y = 2500 size 12{y="2500"} {} .

Therefore, the cost equation is y = 10 x + 2500 size 12{y="10"x+"2500"} {} .
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Source:  OpenStax, Linear equations. OpenStax CNX. Jun 15, 2015 Download for free at https://legacy.cnx.org/content/col11828/1.1
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