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Algebra 1 : Equations of Lines

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Example Questions

Example Question #111 : Equations Of Lines

Give, in slope-intercept form, the equation of a line through the points (-2, 5) and (-3, 8) .

Possible Answers:

y = -3x -1

y = -3x -7

y = -3x +5

$y = -\frac{1}{3}x -\frac{1}{3}$

$y = -\frac{1}{3}x +8$

Correct answer:

y = -3x -1

Explanation:

First, use the slope formula to find the slope, setting $x_{1} = -2, y_{1} = 5,x_{2} = -3, y_{2} = 8$ .

$m = \frac{y _{2} - y_{1}}{x _{2} - x_{1}} = \frac{8 -5 }{-3-(-2)} = \frac{3 }{-1} = -3$

We can write the equation in slope-intercept form as

y = mx + b .

Replace m = -3 :

y = -3x + b

We can find by substituting for x,y using either point - we will choose (-2, 5) :

5= -3 (-2) + b

5= 6 + b

5-6 = 6 -6 + b

b = -1

The equation is y = -3x -1 .

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Example Question #112 : Equations Of Lines

Give, in slope-intercept form, the equation of a line through the points $\left (3.5, 5.5 \right )$ and (5.5, -2.5) .

Possible Answers:

y = -0.25x + 5.5

y = -4x + 2.5

y = -4x + 3.5

y = -4x + 19.5

y = -0.25x + 13.5

Correct answer:

y = -4x + 19.5

Explanation:

First, use the slope formula to find the slope, setting $x_{1} = 3.5, y_{1} = 5.5,x_{2} = 5.5, y_{2} = -2.5$ .

$m = \frac{y _{2} - y_{1}}{x _{2} - x_{1}} = \frac{-2.5 -5.5}{5.5-3.5} = \frac{-8 }{2} = -4$

We can write the equation in slope-intercept form as

y = mx + b .

Replace m = -4 :

y = -4x + b

We can find by substituting for x,y using either point - we will choose (3.5, 5.5) :

5.5= -4 (3.5) + b

5.5= -14 +b

5.5+14 = -14 +14+b

b = 19.5

The equation is y = -4x + 19.5 .

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Example Question #21 : How To Find The Equation Of A Line

Find the equation of the line that is parallel to y=3x+2 and contains the point (0,1).

Possible Answers:

$y=\frac{1}{3}x-1$

$y=\frac{-1}{3}x+1$

y=-3x-1

y=3x+1

Correct answer:

y=3x+1

Explanation:

To find the equation of a line, we need to know the slope and a point that passes through the line. We can then use the equation y-y_1=m(x-x_1) where m is the slope of the line, and $(x_{1}, y_{1})$ a point on the line. For parallel lines, the slopes are the same. The slope of y=3x+2 is 3, so the slope of the parallel line will be 3 as well. We know that the parallel line needs to contain the point (0,1), so we have all of the information we need. We can now use the equation $y-y_1=m(x-x_1)\Rightarrow y-1=3(x-0)\Rightarrow y=3x+1.$

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Example Question #21 : Slope And Line Equations

Write an equation in the form y=mx+b for the line that fits the following points:

(4,3), (6,6), (10,12)

Possible Answers:

$y=-\frac{3}{2}x-6$

$y=\frac{2}{5}x-3$

y=3x-3

$y=\frac{5}{2}x-2$

$y=\frac{3}{2}x-3$

Correct answer:

$y=\frac{3}{2}x-3$

Explanation:

The equation of a line is written in the following format:

y=mx+b

1) The first step, then, is to find the slope, .

is equal to the change in divided by the change in .

So,

$\frac{6-3}{6-4}=\frac{3}{2}$

$y=\frac{3}{2}x+b$

2) Next step is to find . We can find values for and from any one of the given points, plug them in, and solve for .

Let's use (4,3)

So,

$3=\frac{3}{2}(4)+b$

3=6+b

b=-3

Then we just fill in our value for , and we have as a function of .

$y=\frac{3}{2}x-3$

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Example Question #22 : Slope And Line Equations

Which of these lines has a slope of 4 and a y-intercept of 6?

Possible Answers:

None of the other answers

y=2x+3

y=4x+6

6y+4x=10

y=6x+4

Correct answer:

y=4x+6

Explanation:

When an equation is in the y=mx+b form, its slope is and its y-intercept is . Thus, we need an equation with an of 4 and a of 6, which would be

y=4x+6

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Example Question #24 : How To Find The Equation Of A Line

Find the equation of the line with point $\left ( -5,3 \right )$ and slope of .

Possible Answers:

y=-4x+23

y=4x+23

y=4x-13

y=4x-23

Correct answer:

y=4x+23

Explanation:

Use the point-slope formula to find the equation:

$Y-Y_{1}=M(X-X_{1})$

y-3=4(x+5)

y-3 = 4x+20

y-3+3=4x+20+3

y=4x+23

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Example Question #25 : How To Find The Equation Of A Line

Which of the following equations in standard form describes a line passing through the point $\left ( -2,5 \right )$ with a slope of $-3\ ?$

Possible Answers:

3x + y = 11

x + 3y = 6

-3x + y = 11

3x + y = - 1

-3x - 4y = - 1

Correct answer:

3x + y = - 1

Explanation:

Start by using point-slope form:

y-5 = -3 (x-(-2))

y-5 = -3(x+2)

Multiply the right side by the distributive property:

y-5 = -3x -6

Then, convert to standard form:

3x + y = - 1

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Example Question #25 : How To Find The Equation Of A Line

Rewrite in slope-intercept form: 5x+20y=60 .

Possible Answers:

y=-4x+3

$y=\frac{1}{4}x+3$

$y=-\frac{1}{4}x+3$

y=4x+3

y=-4x-3

Correct answer:

$y=-\frac{1}{4}x+3$

Explanation:

Slope-intercept form is y=mx+b , where is the slope and is the y-intercept. To rewrite the original equation in slope-intercept form, you must isolate the variable:

5x+20y=60

-5x+20y=60-5x

20y=-5x+60

Now, divide each side by 20 so that stands alone and simplify, and you are left with the slope-intercept form of the equation:

$y=-\frac{5}{20}x+\frac{60}{20}$

$y=-\frac{1}{4}x+3$

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Example Question #2 : Solving Linear Functions

Find the equation for the line goes through the two points below.

(1,-2)

(-3, 6)

Possible Answers:

y = -2x

y = 2x -4

y = 4x - 2

$y = \frac{1}{2}x$

$y = \frac{1}{2}x + 4$

Correct answer:

y = -2x

Explanation:

Let $P_{1} = (1, \; -2) \; \; and \; \; P_{2} = (-3, \; 6)$ .

First, calculate the slope between the two points.

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{6 - (-2)}{-3 - 1} = \frac{8}{-4} = -2$

Next, use the slope-intercept form to calculate the intercept. We are able to plug in our value for the slope, as well the the values for P_1 .

y = mx + b

-2 = -2(1) + b

b=0

Using slope-intercept form, where we know m=-2 and b=0 , we can see that the equation for this line is y = -2x .

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Example Question #111 : Equations Of Lines

What is the equation of the line connecting the points (-3,0) and (2,5) ?

Possible Answers:

y=-x-3

Equation is undefined or does not exist.

y=x-3

y=x+3

y=3x+1

Correct answer:

y=x+3

Explanation:

To find the equation of this line, we need to know its slope and y-intercept. Let's find the slope first using our general slope formula.

$m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}$

The points are and . In this case, our points are (–3,0) and (2,5). Therefore, we can calculate the slope as the following:

$m=\frac{5-0}{2-(-3)}=\frac{5}{5}=1$

Our slope is 1, so plug that into the equation of the line:

$y=mx+b \,\,\, \rightarrow \,\,\,y=(1)x+b \,\,\, \rightarrow \,\,\, y=x+b$

We still need to find b, the y-intercept. To find this, we pick one of our points (either (–3,0) or (2,5)) and plug it into our equation. We'll use (–3,0).

y=x+b

0=-3+b

Solve for b.

b=3

The equation is therefore written as y=x+3 .

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Algebra 1 : Equations of Lines

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #111 : Equations Of Lines

Example Question #112 : Equations Of Lines

Example Question #21 : How To Find The Equation Of A Line

Example Question #21 : Slope And Line Equations

Example Question #22 : Slope And Line Equations

Example Question #24 : How To Find The Equation Of A Line

Example Question #25 : How To Find The Equation Of A Line

Example Question #25 : How To Find The Equation Of A Line

Example Question #2 : Solving Linear Functions

Example Question #111 : Equations Of Lines

All Algebra 1 Resources

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