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High School Math : Algebra I

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Example Questions

Example Question #21 : Lines

What is the length of a line with endpoints at (6,13) and (9,17) ?

Possible Answers:

$\frac{3}{4}$

$\frac{4}{3}$

625

Correct answer:

Explanation:

The formula for the length of a line is very similiar to the pythagorean theorem:

(x_2-x_1)^2+(y_2-y_1)^2=l^2

Plug in our given numbers to solve:

(x_2-x_1)^2+(y_2-y_1)^2=l^2

(17-13)^2+(9-6)^2=l^2

4^2+3^2=l^2

16+9=l^2

25=l^2

$\sqrt{25}=\sqrt{l^2}$

5=l

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Example Question #22 : Lines

Find the distance between points (2,-4) and (14,-2) .

Possible Answers:

148

$\sqrt{148}$

144

$\sqrt{144}$

Correct answer:

$\sqrt{148}$

Explanation:

Use the distance formula:

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

Substitute the given points into the formula:

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

$d=\sqrt{148}$

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

What is the length of a line with endpoints of (2,-4) and (7,9) ?

Possible Answers:

3.7

13.9

17.1

14 7

Correct answer:

13.9

Explanation:

The distance formula is just a reworking of the Pythagorean theorem: $l=\sqrt{\Delta x^2+\Delta y^2}$

Expand that.

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

Plug in our given values.

$l=\sqrt{(7-2)^2+(9--4)^2}$

$l=\sqrt{(5)^2+(13)^2}$

$l=\sqrt{25+169}$

$l=\sqrt{194}$

l=13.9

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Example Question #1366 : High School Math

What is the distance between (1,5) and (6,17) ?

Possible Answers:

Correct answer:

Explanation:

Let $P_{1}=(1,5)$ and $P_{2} = (6,17)$ .

The distance formula is given by $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Substitute in the given points:

$d = \sqrt{(6-1)^{2}+(17-5)^{2}}=\sqrt{(5)^{2}+(12)^{2}}=\sqrt{25+144}=\sqrt{169}=13$

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Example Question #23 : Algebra I

A line segment has endpoints at (2,3) and (-7,3) . What is the distance of this segment?

Possible Answers:

6.36

4.5

Correct answer:

Explanation:

To find the distance, we use the distance formula: $d^2=\Delta x^2+\Delta y^2$ .

Expand that:

d^2=(x_2-x_1)^2+(y_2-y_1)^2

Plug in our given values.

d^2=(-7-2)^2+(3-3)^2

d^2=(-9)^2+(0)^2

d^2=81+0

$\sqrt{d^2}=\sqrt{81}$

d=9

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Example Question #23 : Lines

If a line has a length of , and the endpoints are (0,0) and (4,y) , what is the value of ?

Possible Answers:

1.5

Correct answer:

Explanation:

The formula for the length of a line, l, is the distance formula, which is very similar to the Pythagorean Theorem.

(x_2-x_1)^2+(y_2-y_1)^2=l^2

Note that the problem has already given us a value for the length of the line. That means l=5 . Plug in all of the given values and solve for the missing term.

(4-0)^2+(y-0)^2=5^2

4^2+y^2=5^2

16+y^2=25

Subtract $\small 16$ from both sides.

16+y^2-16=25-16

y^2=9

$\sqrt{y^2}=\sqrt{9}$

y=3

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Example Question #24 : Lines

What is the distance between points (-2,4) and (12,3) ?

Possible Answers:

$\sqrt{196}$

144

197

$\sqrt{197}$

Correct answer:

$\sqrt{197}$

Explanation:

Use the distance formula:

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

Plug in the given points:

$d=\sqrt{(12-(-2)))^2+(3-4)^2}$

$d=\sqrt{197}$

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Example Question #1 : How To Find The Equation Of A Perpendicular Line

Write an equation in slope-intercept form for the line that passes through $\left ( 3,1 \right )$ and that is perpendicular to a line which passes through the two points $\left ( 3,1 \right )$ and $\left ( -2,5 \right )$ .

Possible Answers:

$y=\frac{5}{4}X-\frac{11}{4}$

$y=-\frac{4}{5}X+11$

$y=\frac{5}{4}X+\frac{11}{4}$

y=5X-4

Correct answer:

$y=\frac{5}{4}X-\frac{11}{4}$

Explanation:

Find the slope of the line through the two points. It is $-\frac{4}{5}$ .

Since the slope of a perpendicular line is the negative reciprocal of the original line, the new line's slope is $\frac{5}{4}$ . Plug the slope and one of the points into the point-slope formula $y-y_{1}=m(x-x_{1})$ . Isolate for .

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Example Question #2 : Perpendicular Lines

Find the equation of a line perpendicular to y=-3x+10

Possible Answers:

y=3X+2

$y=\frac{1}{3}X+8$

y=-3X-5

y=-3X+10

Correct answer:

$y=\frac{1}{3}X+8$

Explanation:

Since a perpendicular line has a slope that is the negative reciprocal of the original line, the new slope is $\frac{1}{3}$ . There is only one answer with the correct slope.

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Example Question #1 : How To Find The Equation Of A Perpendicular Line

Find the equation (in slope-intercept form) of a line perpendicular to 3X+2Y=11 .

Possible Answers:

$y=\frac{2}{3}X+\frac{11}{2}$

$y=2X+\frac{11}{2}$

$y=3X+\frac{11}{2}$

$y=-\frac{3}{2}X+11$

Correct answer:

$y=\frac{2}{3}X+\frac{11}{2}$

Explanation:

First, find the slope of the original line, which is $-\frac{3}{2}$ . You can do this by isolating for so that the equation is in slope-intercept form. Once you find the slope, just replace the in the original equation withe the negative reciprocal (perpendicular lines have a negative reciprocal slope for each other). Thus, your answer is

$y=\frac{2}{3}X+\frac{11}{2}$

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High School Math : Algebra I

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #21 : Lines

Example Question #22 : Lines

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

Example Question #1366 : High School Math

Example Question #23 : Algebra I

Example Question #23 : Lines

Example Question #24 : Lines

Example Question #1 : How To Find The Equation Of A Perpendicular Line

Example Question #2 : Perpendicular Lines

Example Question #1 : How To Find The Equation Of A Perpendicular Line

All High School Math Resources

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