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Algebra 1 : Functions and Lines

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10 Diagnostic Tests 557 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #34 : How To Find The Equation Of A Line

Find the equation for the line given these two points:

$\small (10,2)(9,1) , y-intercept = 0$

Possible Answers:

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

$\small y =3x$

$\small \small y=x$

$\small y = \frac{1}{1}x +9$

Correct answer:

$\small \small y=x$

Explanation:

The equation of a line is $\small y=mx+b$ , where "m" is the slope of the line and "b" is the point at which the line intercepts the y-axis.

The first thing we must do is use the formula

${} \frac{y_{1}-y_{2}}{x_{1}-x_{2}}$

to find the slope, or "m."

If we plug in the points of the two coordinates given,

$\small \small \frac{2-1}{10-9} = \frac{1}{1} = 1$ , we find our slope, "m."

Since our y-intercept is given, we now have everything we need for our equation, $\small \small y=x$ .

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Example Question #33 : How To Find The Equation Of A Line

Find the equation of the line containing the points:

$\small (9,10) (3,6) , y-intercept = -3$

Possible Answers:

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

$\small \small y=\frac{4}{6}x -10$

$\small \small \small y=\frac{5}{6}x -5$

$\small \small \small y=\frac{4}{6}x -6$

Correct answer:

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

Explanation:

The equation of a line is $\small y=mx+b$ , where "m" is the slope of the line and "b" is the point at which the line intercepts the y-axis.

The first thing we must do is use the formula

${} \frac{y_{1}-y_{2}}{x_{1}-x_{2}}$ to find the slope, or "m."

If we plug in the points of the two coordinates given,

$\small \small \frac{10-6}{9-3} = \frac{4}{6}=\frac{2}{3}$ ,

we find our slope, "m."

Since our y-intercept is given, we now have everything we need for our equation,

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

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Example Question #34 : How To Find The Equation Of A Line

Find the equation of the lines given the points:

$\small (30,4)(16, -9) , y-intercept = -6$

Possible Answers:

$\small y=\frac{13}{14}x -6$

$\small y=\frac{13}{14}x -100$

$\small y=-\frac{14}{13}x -4$

$\small y=\frac{13}{14}x -9$

$\small y=-\frac{13}{14}x -6$

Correct answer:

$\small y=\frac{13}{14}x -6$

Explanation:

The equation of a line is $\small y=mx+b$ , where "m" is the slope of the line and "b" is the point at which the line intercepts the y-axis.

The first thing we must do is use the formula

$\frac{y_{1}-y_{2}}{x_{1}-x_{2}}{}$

to find the slope, or "m."

If we plug in the points of the two coordinates given,

$\small \small \small \frac{4--9}{30-16} = \frac{13}{14}$ ,

we find our slope, "m."

Since our y-intercept is given, we now have everything we need for our equation,

$\small \small \small \small \small y=\frac{13}{14}x -6$ .

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Example Question #35 : How To Find The Equation Of A Line

$\small \small (3,1)(8,3), y-intercept = 4$

Using the data above, find the equation of the line.

Possible Answers:

$\small y = \frac{4}{5}x+2$

$\small y = \frac{2}{5}x+10$

$\small y = \frac{2}{5}x+4$

$\small y = \frac{2}{5}x+9$

Correct answer:

$\small y = \frac{2}{5}x+4$

Explanation:

The data provided is enough data to for us to find the equation of a line that passes through these points:

$\small \small (3,1)(8,3), y-intercept = 4$ .

An equation representing a line:

$\small y=mx+b$ ,

where m = slope, b = y-intercept.

To find the slope use the following formula,

$\small \frac{y_{2}-y_{1}}{x_{2}-x_{2}}$ ,

so in this problem the slope:

$\small \small \frac{3-1}{8-3} = \frac{2}{5}$ .

The y-intercept is given. So our equation is

$\small \small y = \frac{2}{5}x+4$ .

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Example Question #36 : How To Find The Equation Of A Line

Given this information, find the equation of the line. "This line passes through two points, one of which is $\small (6,6)$ and the other is $\small (-4,-7)$ . The line also passes through the origin.

Possible Answers:

$\small y = \frac{13}{1}x+1$

$\small y = -\frac{13}{10}x+0$

$\small y = \frac{13}{10}x+10$

$\small y = \frac{13}{10}x$

Correct answer:

$\small y = \frac{13}{10}x$

Explanation:

The data provided is enough data to for us to find the equation of a line that passes through these points:

$\small \small \small (6,6)(-4,-7), y-intercept = 0$ .

An equation representing a line:

$\small y=mx+b$ ,

where m = slope, b = y-intercept.

To find the slope use the following formula,

$\small \frac{y_{2}-y_{1}}{x_{2}-x_{2}}$ ,

so in this problem the slope:

$\small \small \small \frac{6--7}{6--4} = \frac{13}{10}$ .

The y-intercept is given. So our equation is

$\small \small \small y = \frac{13}{10}x$ .

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Example Question #371 : Functions And Lines

(10,5)(-14,3), y-intercept = 4

Using the data above, find the equation of a the line that passes through these points and the -intercept.

Possible Answers:

$\small y = \frac{1}{12}x$

$\small y = \frac{4}{12}x+10$

$\small y = \frac{1}{8}x+5$

$\small y = \frac{1}{12}x+4$

Correct answer:

$\small y = \frac{1}{12}x+4$

Explanation:

The data provided is enough data to for us to find the equation of a line that passes through these points:

$\small \small \small (10,5)(-14,3), y-intercept = 4$ .

An equation representing a line:

$\small y=mx+b$ ,

where m = slope, b = y-intercept.

To find the slope we use the following formula,

$\small \frac{y_{2}-y_{1}}{x_{2}-x_{2}}$ ,

so in this problem the slope:

$\small \small \small \frac{5-3}{10--14} = \frac{2}{24} = \frac{1}{12}$ .

The y-intercept is given. So our equation is

$\small \small \small \small y = \frac{1}{12}x+4$ .

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Example Question #38 : How To Find The Equation Of A Line

$\small (-2,5)(2,5), y-intercept = 5$

Using the data above, find an equation for the line that passes through these points and the -intercept.

Possible Answers:

$\small y =13$

$\small y = 3x$

$\small y = 5$

$\small y = 5x+9$

Correct answer:

$\small y = 5$

Explanation:

The data provided is enough data to for us to find the equation of a line that passes through these points:

$\small \small \small \small \small (-2,5)(2,5), y-intercept = 5$ .

An equation representing a line:
$\small y=mx+b$ ,

where m = slope, b = y-intercept.

To find the slope use the following formula,

$\small \frac{y_{2}-y_{1}}{x_{2}-x_{2}}$ ,

so in this problem the slope:

$\small \small \small \small \frac{5-5}{2--2} = \frac{0}{4}$ .

The y-intercept is given. From this slope that is found, we see that the numerator is zero, this means that there is no slope, thus the line must be a horizontal line.

Our formula then is:

$\small y = 5$ .

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Example Question #41 : How To Find The Equation Of A Line

Given the points (3,2) and (6,4) .

Find the slope-intercept form of the line that contains these points.

Possible Answers:

$y=\frac{2}{3}x-4$

$y=\frac{5}{2}x-\frac{9}{2}$

$y=\frac{3}{2}x-9$

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

$y=\frac{3}{4}x+\frac{9}{2}$

Correct answer:

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

Explanation:

Use the given points and plug them into slope formula:

$m=\frac{\left ( y_{2}-y_{1} \right )}{(x_{2}-x_{1})}$

Remember points are written in the following format:

(x,y)

Substitute.

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

$m=\frac{2}{3}$

Now, that we have the slope of the line we can insert values into the point-slope formula:

$(y-y_{1})=m(x-x_{1})$

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

Distribute the fraction through the quantity on the left side of the equation.

$(y-2)= \left (\frac{2}{3}*\frac{x}{1} \right ) - \left ( \frac{3}{1} *\frac{2}{3}\right )$

$(y-2)= \left (\frac{2}{3}x\right ) - \left ( \frac{2}{1}\right )$

$y-2= \left (\frac{2}{3}x\right ) - 2$

Add to both sides.

$y= \left (\frac{2}{3}x\right )$

The slope-intercept form is written as:

y=mx+b

Where is the slope and is the y-intercept.

In our equation our slope is $\frac{2}{3}$ and our y-intercept is .

The equation of the line that contains these points is:

$y=\frac{2}{3}x+0$

Simplify.

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

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Example Question #41 : Slope And Line Equations

Write the slope-intercept form of the equation of the line described.

Passes through the point (3,1) , perpendicular to $y=-\frac{2}5{}x-3$ .

Possible Answers:

$y=-\frac{2}{5}x+\frac{11}{5}$

$y=\frac{5}{2}x-\frac{13}{2}$

$y=-\frac{5}{2}x+\frac{17}{2}$

$y=-\frac{2}5{}x+\frac{13}{5}$

$y=-\frac{1}{2}x+\frac{11}{5}$

Correct answer:

$y=\frac{5}{2}x-\frac{13}{2}$

Explanation:

The slope-intercept equation of a line is in the form y=mx+b .

A line that is perpendicular has a slope that is the opposite reciprocal of the given line.

Slope of perpendicular line:

$m=\frac{5}{2}$

Using the point slope formula,

$y-y_{1}=m(x-x_{1})$

where $x_{1}=3\: \:and\: \:y_{1}=1$

we get the following equation.

$y-1=\frac{5}{2}(x-3)$

$y=\frac{5}{2}x-\frac{15}{2}+1$

$y=\frac{5}{2}x-\frac{13}{2}$

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Example Question #131 : Equations Of Lines

Find the equation of the line with a slope of 2 that passes through the point (4,6).

Possible Answers:

y=2x+4

y=2x-2

y=2x+2

y=2x-14

Correct answer:

y=2x-2

Explanation:

To solve this problem, we need to remember point-slope formula:

$y-y_{1}=m(x-x_{1})$

Then we plug in m=2 and (x1,y1)=(4,6) and solve:

y-6=2(x-4)

y-6=2x-8

y=2x-2

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Algebra 1 : Functions and Lines

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #34 : How To Find The Equation Of A Line

Example Question #33 : How To Find The Equation Of A Line

Example Question #34 : How To Find The Equation Of A Line

Example Question #35 : How To Find The Equation Of A Line

Example Question #36 : How To Find The Equation Of A Line

Example Question #371 : Functions And Lines

Example Question #38 : How To Find The Equation Of A Line

Example Question #41 : How To Find The Equation Of A Line

Example Question #41 : Slope And Line Equations

Example Question #131 : Equations Of Lines

All Algebra 1 Resources

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