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Algebra 1 : Functions and Lines

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Example Questions

Example Question #242 : Slope And Line Equations

What is the slope of the line connected by the points $(\frac{1}{2},-4)$ and $(4,-\frac{3}{4})$ ?

Possible Answers:

$-\frac{19}{18}$

$\frac{13}{14}$

$-\frac{19}{14}$

$\frac{19}{14}$

$\frac{19}{18}$

Correct answer:

$\frac{13}{14}$

Explanation:

Write the slope formula.

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

Substitute the points into the formula.

$m = \frac{-\frac{3}{4}-(-4)}{4-\frac{1}{2}}$

Rewrite the numbers on the top and bottom with a common denominator.

$m = \frac{-\frac{3}{4}-(-4)}{4-\frac{1}{2}} = \frac{-\frac{3}{4}+\frac{16}{4}}{\frac{8}{2}-\frac{1}{2}}$

Simplify the top and bottom.

$m= \frac{\frac{13}{4}}{\frac{7}{2}}$

Rewrite this complex fraction using multiplication.

$\frac{\frac{13}{4}}{\frac{7}{2}} = \frac{13}{4} \div \frac{7}{2} = \frac{13}{4} \times \frac{2}{7}$

Multiply the numerator by numerator and denominator by denominator.

$\frac{13}{4} \times \frac{2}{7} = \frac{26}{28} = \frac{13}{14}$

The slope is $\frac{13}{14}$ .

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Example Question #582 : Functions And Lines

Find the slope of the line: x+6y = 37

Possible Answers:

$-\frac{1}{6}$

$\frac{1}{6}$

$\frac{37}{6}$

$-\frac{37}{6}$

$-\frac{6}{37}$

Correct answer:

$-\frac{1}{6}$

Explanation:

In order to find the slope, we will need to put this equation in slope-intercept form.

Write the slope-intercept form.

y=mx+b

Isolate the y-term by subtracting on both sides.

x+6y -x= 37-x

Simplify both sides.

6y =-x+37

Divide by six on both sides.

$\frac{6y }{6}=\frac{-x+37}{6}$

Simplify both fractions and split the terms on the right side.

The equation in standard form is: $y=-\frac{1}{6}x+\frac{37}{6}$

We can see that the slope is: $-\frac{1}{6}$

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Example Question #583 : Functions And Lines

Find the slope of the given line:

3x+4=-5y+2

Possible Answers:

$\frac{3}{5} = $ slope$$

$\frac{-3}{5} = $ slope$$

undefined

$\frac{5}{3} = $ slope$$

$\frac{-5}{3} = $ slope$$

Correct answer:

$\frac{-3}{5} = $ slope$$

Explanation:

3x+4=-5y+2

First rearrange in the form y = m*x+b where m = slope and b = y-intercept

(3x+4)-2=(-5y+2)-2

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

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

Slope = $\frac{-3}{5}$

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Example Question #1 : Points And Distance Formula

Find the length of the line segment from the origin to the point (3, 4).

Possible Answers:

Correct answer:

Explanation:

Here, we need to use the distance formula between the two points (0, 0) and (3, 4).

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$d=\sqrt{(3-0)^2+(4-0)^2}$

$d=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

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Example Question #1 : How To Find The Length Of A Line With Distance Formula

I have two points, (–8,3) and (6,–1). If I want to connect those two points with a line segment, how long would that line segment need to be?

Possible Answers:

$\frac{5}{2}$

$8\sqrt{2}$

$2\sqrt{53}$

Infinite

Correct answer:

$2\sqrt{53}$

Explanation:

To determine how long the line needs to be to connect those two points, we need to use the distance formula, shown below.

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The two points are (x_1,y_1) and (x_2,y_2) . In our case, the points are (–8, 3) and (6, –1).

$d=\sqrt{(6-(-8))^2+(-1-3)^2}$

$d=\sqrt{14^2+(-4)^2}$

$d=\sqrt{196+14}$

$d=\sqrt{212}$

$d=\sqrt{4(53)}$

$d=2\sqrt{53}$

So in order to connect the two points, the length of the line needs to have $d=2\sqrt{53}$ .

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Example Question #2 : How To Find The Length Of A Line With Distance Formula

What is the distance between the points (7,8) and (10,4) ?

Possible Answers:

$\sqrt{20}$

$\sqrt{7}$

Correct answer:

Explanation:

To solve problems like this, we simply need to use the distance formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ . Plugging in the and values from our points yields $d = \sqrt{(10-7)^{2}+(4-8)^{2}}$ , or $d = \sqrt{(3)^{2}+(-4)^{2}}$ . Solving this radical gives us a value of $d = \sqrt{25}$ , or 5.

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Example Question #1 : How To Find The Length Of A Line With Distance Formula

Find the length of the line segment with endpoints at (2,9) and (7,6) .

Possible Answers:

None of the other answers are correct.

Correct answer:

None of the other answers are correct.

Explanation:

Use the distance formula, with $x_{1} =2,x_{2} =7,y_{1} =9,y_{2} =6$ :

$d = \sqrt{ \left (x_{1} -x_{2} \right ) ^{2} + \left ( y_{1} -y_{2 \right ) ^{2} }}$

$= \sqrt{ \left (2 -7\right ) ^{2} + \left (9 -6 \right ) ^{2} }$

$= \sqrt{ \left (-5 \right ) ^{2} + 3 ^{2} } = \sqrt{34} \approx 5.8$

Therefore, none of the integer answer choices are correct.

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Example Question #5 : How To Find The Length Of A Line With Distance Formula

Find the distance between the two points (-4,2) and (2,1) .

Possible Answers:

$\sqrt{37}$

$\sqrt{5}$

Correct answer:

$\sqrt{37}$

Explanation:

The distance between two points can be found with the equation $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ . Substituting in values you get $\sqrt{(2+4)^2+(1-2)^2}$ . This means that the answer is $\sqrt{37}$ .

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Example Question #6 : How To Find The Length Of A Line With Distance Formula

Find the distance between the midpoints of line A with the points (-3,4) and (3,4) and line. B with the points (-2,-2) and (2,2) .

Possible Answers:

$\frac{3}{2}$

$\frac{1}{2}$

Correct answer:

Explanation:

Use the midpoint formula:

$\frac{x_{2}+x_{1}}{2}, \frac{y_{2}+y_{1}}{2}$

Remember points are written in the following format:

(x,y)

Substitute for line A

$\frac{\left ( -3+3 \right )}{2} , \frac{\left ( 4+0 \right )}{2}$

$\frac{\left ( 0\right )}{2} , \frac{\left ( 4\right )}{2}$

The midpoint of line A is (0,4) .

Substitute for line B.

$\frac{\left ( -2+2 \right )}{2} , \frac{\left ( -2 +2 \right )}{2}$

$\frac{\left ( 0 \right )}{2} , \frac{\left ( 0 \right )}{2}$

The midpoint of line B is (0,0) .

Now we can find the distance between these two points using the distance formula:

$d= \sqrt{\left ( x_{2} - x_{1} \right )^{2} + \left ( y_{2}-y_{1} \right )^{2}}$

Substitute the using the known values for lines A and B.

$d= \sqrt{\left ( 0-0 \right )^{2} + \left ( 0-4 \right )^{2}}$

Simplify.

$d= \sqrt{\left ( 0\right )^{2} + \left ( -4 \right )^{2}}$

$d= \sqrt{\left ( -4 )^{2}}$

$d= \sqrt{16}$

The distance between the two midpoints of lines A and B is .

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Example Question #7 : How To Find The Length Of A Line With Distance Formula

Find the distance between the following points:

$(3,1), \: (-2,7)$

Possible Answers:

$\sqrt{65}$

$\left(\frac{1}{2},4\right)$

$\sqrt{29}$

$\sqrt{61}$

Correct answer:

$\sqrt{61}$

Explanation:

Use the equation to calculated the distance between two points:

$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

where

$(3,1)=(x_1,y_1), \: (-2,7)=(x_2,y_2)$

we can find the distance.

$d=\sqrt{(-2-3)^{2}+(7-1)^{2}}$

$=\sqrt{25+36}$

$=\sqrt{61}$

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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 #242 : Slope And Line Equations

Example Question #582 : Functions And Lines

Example Question #583 : Functions And Lines

Example Question #1 : Points And Distance Formula

Example Question #1 : How To Find The Length Of A Line With Distance Formula

Example Question #2 : How To Find The Length Of A Line With Distance Formula

Example Question #1 : How To Find The Length Of A Line With Distance Formula

Example Question #5 : How To Find The Length Of A Line With Distance Formula

Example Question #6 : How To Find The Length Of A Line With Distance Formula

Example Question #7 : How To Find The Length Of A Line With Distance Formula

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