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Intermediate Geometry : Lines

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Example Question #1401 : Intermediate Geometry

Find the equation of the line that passes through the points (14, 9) and (-6, 12) .

Possible Answers:

$y=-\frac{15}{26}x-\frac{111}{26}$

$y=\frac{3}{26}x-\frac{113}{13}$

$y=\frac{20}{3}x+\frac{120}{11}$

$y=-\frac{3}{20}x+\frac{111}{10}$

Correct answer:

$y=-\frac{3}{20}x+\frac{111}{10}$

Explanation:

Recall that the slope-intercept form of a line:

y=mx+b ,

where $m=\text{slope}$ and $b=\text{y-intercept}$ .

First, find the slope of the line by using the following formula:

$\text{Slope of Line}=\frac{y_2-y_1}{x_2-x_1}$

$\text{Slope}=\frac{12-9}{-6-14}=-\frac{3}{20}$

Next, find the y-intercept of the line by plugging in of the points into the semi-completed formula.

$y=-\frac{3}{20}x+b$

Plugging in (14, 9) yields the following:

$9=-\frac{3}{20}(14)+b$

Solve for .

$b-\frac{21}{10}=9$

$b=\frac{111}{10}$

The equation of the line is then $y=-\frac{3}{20}x+\frac{111}{10}$ .

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

Find the equation of the line that passes through the points (20, -3) and (-45, 6) .

Possible Answers:

$y=-\frac{9}{65}x-\frac{3}{13}$

$y=-\frac{13}{24}x-\frac{1}{12}$

$y=-\frac{24}{13}x+\frac{4}{13}$

$y=-\frac{65}{4}x-\frac{5}{2}$

Correct answer:

$y=-\frac{9}{65}x-\frac{3}{13}$

Explanation:

Recall that the slope-intercept form of a line:

y=mx+b ,

where $m=\text{slope}$ and $b=\text{y-intercept}$ .

First, find the slope of the line by using the following formula:

$\text{Slope of Line}=\frac{y_2-y_1}{x_2-x_1}$

$\text{Slope}=\frac{6-(-3)}{-45-20}=-\frac{9}{65}$

Next, find the y-intercept of the line by plugging in of the points into the semi-completed formula.

$y=-\frac{9}{65}x+b$

Plugging in (20, -3) yields the following:

$-3=-\frac{9}{65}(20)+b$

Solve for .

$b-\frac{36}{13}=-3$

$b=-\frac{3}{13}$

The equation of the line is then $y=-\frac{9}{65}x-\frac{3}{13}$ .

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

Find the equation of the line that passes through the points (-17, 1) and (-1, 10) .

Possible Answers:

$y=\frac{9}{4}x+\frac{107}{4}$

$y=\frac{16}{9}x+\frac{100}{9}$

$y=\frac{9}{16}x+\frac{169}{16}$

$y=-\frac{9}{16}x+\frac{3}{16}$

Correct answer:

$y=\frac{9}{16}x+\frac{169}{16}$

Explanation:

Start by finding the slope of the line.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\text{Slope}=\frac{10-1}{-1-(-17)}=\frac{9}{16}$

Now, take one of the points and put the equation into point-slope form. Recall the point-slope form of an equation of a line:

y-y_1=m(x-x_1) , where is the slope and (x_1, y_1) is the point.

$y-10=\frac{9}{16}(x+1)$

Now, simplify this equation so that it is in slope-intercept form. Recall the slope-intercept form of an equation of a line:

y=mx+b , where is the slope and is the y-intercept.

$y-10=\frac{9}{16}(x+1)$

$y-10=\frac{9}{16}x+\frac{9}{16}$

$y=\frac{9}{16}x+\frac{169}{16}$

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Example Question #26 : How To Find The Equation Of A Line

Find the equation of a line that passes through the points (4, 12) and (12, 10) .

Possible Answers:

$y=\frac{1}{4}x+16$

$y=-\frac{1}{4}x+13$

y=4x+10

y=-4x+8

Correct answer:

$y=-\frac{1}{4}x+13$

Explanation:

Start by finding the slope of the line.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\text{Slope}=\frac{10-12}{12-4}=-\frac{1}{4}$

Now, take one of the points and put the equation into point-slope form. Recall the point-slope form of an equation of a line:

y-y_1=m(x-x_1) , where is the slope and (x_1, y_1) is the point.

$y-12=-\frac{1}{4}(x-4)$

Now, simplify this equation so that it is in slope-intercept form. Recall the slope-intercept form of an equation of a line:

y=mx+b , where is the slope and is the y-intercept.

$y-12=-\frac{1}{4}(x-4)$

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

$y=-\frac{1}{4}x+13$

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Example Question #27 : How To Find The Equation Of A Line

Find the equation of the line that passes through the points $(-\frac{1}{2}, 6)$ and $(\frac{3}{2}, 7)$ .

Possible Answers:

$y=\frac{1}{2}x+\frac{19}{4}$

$y=\frac{1}{2}x+\frac{9}{2}$

$y=\frac{1}{2}x+\frac{25}{4}$

$y=\frac{1}{2}x+\frac{21}{2}$

Correct answer:

$y=\frac{1}{2}x+\frac{25}{4}$

Explanation:

Start by finding the slope of the line.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\text{Slope}=\frac{7-6}{\frac{3}{2}-(-\frac{1}{2})}=\frac{1}{2}$

Now, take one of the points and put the equation into point-slope form. Recall the point-slope form of an equation of a line:

y-y_1=m(x-x_1) , where is the slope and (x_1, y_1) is the point.

$y-7=\frac{1}{2}(x-\frac{3}{2})$

Now, simplify this equation so that it is in slope-intercept form. Recall the slope-intercept form of an equation of a line:

y=mx+b , where is the slope and is the y-intercept.

$y-7=\frac{1}{2}(x-\frac{3}{2})$

$y-7=\frac{1}{2}x-\frac{3}{4}$

$y-7=\frac{1}{2}x+\frac{25}{4}$

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Example Question #28 : How To Find The Equation Of A Line

Find the equation of the line that passes through the points (-21, 6) and (10, -5) .

Possible Answers:

$y=-\frac{11}{31}x-\frac{31}{11}$

$y=-\frac{11}{31}x-\frac{47}{31}$

$y=-\frac{11}{31}x-\frac{41}{11}$

$y=-\frac{11}{31}x-\frac{45}{31}$

Correct answer:

$y=-\frac{11}{31}x-\frac{45}{31}$

Explanation:

Start by finding the slope of the line.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\text{Slope}=\frac{-5-6}{10-(-21)}=-\frac{11}{31}$

Now, take one of the points and put the equation into point-slope form. Recall the point-slope form of an equation of a line:

y-y_1=m(x-x_1) , where is the slope and (x_1, y_1) is the point.

$y+5=-\frac{11}{31}(x-10)$

Now, simplify this equation so that it is in slope-intercept form. Recall the slope-intercept form of an equation of a line:

y=mx+b , where is the slope and is the y-intercept.

$y+5=-\frac{11}{31}(x-10)$

$y+5=-\frac{11}{31}x+\frac{110}{31}$

$y=-\frac{11}{31}x-\frac{45}{31}$

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Example Question #29 : How To Find The Equation Of A Line

Find the equation of the line that passes through the points (16, 16) and (-8, 10) .

Possible Answers:

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

$y=\frac{1}{4}x+10$

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

$y=\frac{1}{4}x-8$

Correct answer:

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

Explanation:

Start by finding the slope of the line.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\text{Slope}=\frac{10-16}{-8-16}=\frac{1}{4}$

Now, take one of the points and put the equation into point-slope form. Recall the point-slope form of an equation of a line:

y-y_1=m(x-x_1) , where is the slope and (x_1, y_1) is the point.

$y-16=\frac{1}{4}(x-16)$

Now, simplify this equation so that it is in slope-intercept form. Recall the slope-intercept form of an equation of a line:

y=mx+b , where is the slope and is the y-intercept.

$y-16=\frac{1}{4}(x-16)$

$y-16=\frac{1}{4}x-4$

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

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Example Question #30 : How To Find The Equation Of A Line

Find the equation of the line that passes through the points (-5, 7) and (9, 6) .

Possible Answers:

$y=-\frac{1}{14}x+\frac{46}{7}$

$y=-\frac{1}{14}x+\frac{93}{14}$

$y=-\frac{1}{14}x-\frac{23}{14}$

$y=-\frac{1}{4}x+\frac{15}{14}$

Correct answer:

$y=-\frac{1}{14}x+\frac{93}{14}$

Explanation:

Start by finding the slope of the line.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\text{Slope}=\frac{6-7}{9-(-5)}=-\frac{1}{14}$

Now, take one of the points and put the equation into point-slope form. Recall the point-slope form of an equation of a line:

y-y_1=m(x-x_1) , where is the slope and (x_1, y_1) is the point.

$y-6=-\frac{1}{14}(x-9)$

Now, simplify this equation so that it is in slope-intercept form. Recall the slope-intercept form of an equation of a line:

y=mx+b , where is the slope and is the y-intercept.

$y-6=-\frac{1}{14}(x-9)$

$y-6=-\frac{1}{14}x+\frac{9}{14}$

$y=-\frac{1}{14}x+\frac{93}{14}$

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Example Question #31 : How To Find The Equation Of A Line

Untitled

Which inequality is graphed above?

Possible Answers:

y > x

$y \le x$

y<x

$y \ge x$

Correct answer:

$y \ge x$

Explanation:

The line that serves as the boundary passes through the points (0,0) and (5,5) and can easily be seen to be the line of the equation y = x .

The graph of the inequality includes the line itself, as is demonstrated by the fact that it is solid, so the correct choice is either

$y \le x$

$y \ge x$

From the diagram, we see that, for example, the point (0, 1) (and the rest of the positive -axis) is in the set. Since 1> 0 , this makes the statement $y \ge x$ true, so that is the inequality to choose.

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Example Question #32 : How To Find The Equation Of A Line

Untitled

Refer to the above figures. To the left is the graph of the equation

y = -3x + 7

What inequality is graphed at right?

Possible Answers:

y < -3x + 7

$y \le -3x + 7$

$y \ge -3x + 7$

y > -3x + 7

Correct answer:

$y \ge -3x + 7$

Explanation:

As indicated by the solid line, the graph of the inequality at right includes the line of the equation, so the inequality graphed is either

$y \ge -3x + 7$

$y \le -3x + 7$

To determine which one, we can select a test point and substitute its coordinates in either inequality, testing whether it is true for those values. The easiest test point is (0,0) ; it is not part of the solution region, so we want the inequality that it makes false. Let us select the first inequality:

$y \ge -3x + 7$

$0 \ge -3 (0) + 7$

$0 \ge 7$

(0,0) makes this inequality false, so the graph of the inequality $y \ge -3x + 7$ is the one that does not include the origin. This is the correct choice. (Note that if we had selected the other inequality, we would have seen that (0,0) makes it true; this would have allowed us to draw the same conclusion.)

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Intermediate Geometry : Lines

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #1401 : Intermediate Geometry

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

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

Example Question #26 : How To Find The Equation Of A Line

Example Question #27 : How To Find The Equation Of A Line

Example Question #28 : How To Find The Equation Of A Line

Example Question #29 : How To Find The Equation Of A Line

Example Question #30 : How To Find The Equation Of A Line

Example Question #31 : How To Find The Equation Of A Line

Example Question #32 : How To Find The Equation Of A Line

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