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Precalculus : Graphing Functions

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12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Find The Distance Between A Point And A Line

Find the distance between point $(\frac{1}{2},\frac{1}{2})$ to the line $x=-\frac{1}{2}$ .

Possible Answers:

$\frac{1}{2}$

$-\frac{1}{4}$

Correct answer:

Explanation:

Distance cannot be a negative number. The function $x=-\frac{1}{2}$ is a vertical line. Subtract the value of the line to the x-value of the given point to find the distance.

$D=\left | (-\frac{1}{2})-\frac{1}{2} \right | = \left | -1\right |=1$

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Example Question #2 : Find The Distance Between A Point And A Line

Find the distance between point (-1,6) to line x=1 .

Possible Answers:

$\frac{1}{2}$

Correct answer:

Explanation:

The line x=1 is vertical covering the first and fourth quadrant on the coordinate plane.

The x-value of (-1,6) is negative one.

Find the perpendicular distance from the point to the line by subtracting the values of the line and the x-value of the point.

Distance cannot be negative.

$\left | 1-(-1)\right |=\left |2 \right |=2$

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Example Question #3 : Find The Distance Between A Point And A Line

How far apart are the line $y = \frac{ 1}{2 } x - 3$ and the point (2, 7 ) ?

Possible Answers:

$\frac{ 18 }{ \sqrt5}$

$\frac{22}{\sqrt5}$

$\frac{17}{\sqrt3}$

$\frac{18}{\sqrt3}$

$\frac{ 3}{\sqrt5}$

Correct answer:

$\frac{ 18 }{ \sqrt5}$

Explanation:

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is (x_0, y_0) and the line is ax + by + c = 0

First, we'll re-write the equation $y = \frac{1}{2} x - 3$ in this form to identify a, b, and c:

$y = \frac{1}{2} x - 3$ subtract half x and add 3 to both sides

$y - \frac{1}{2}x + 3 = 0$ multiply both sides by 2

2y - x + 6 = 0 Now we see that a = -1, b = 2, c = 6 . Plugging these plus (x_0, y_0) = (2, 7 ) into the formula, we get:

$distance = \frac{|-1(2)+2(7)+6 |}{\sqrt{(-1)^2 +( 2)^2 }} = \frac{|-2 + 14+6|}{\sqrt{1+4} } = \frac{18}{\sqrt5}$

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Example Question #2 : Find The Distance Between A Point And A Line

How far apart are the line $y = -\frac{2}{3}x + 8$ and the point (1,3) ?

Possible Answers:

$\frac{35}{\sqrt5}$

$\frac{10.\overline3}{\sqrt{1.\overline4}}$

$\frac{15}{\sqrt{13}}$

$\frac{15}{\sqrt5}$

$\frac{13}{\sqrt{3}}$

Correct answer:

$\frac{13}{\sqrt{3}}$

Explanation:

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is (x_0, y_0) and the line is ax + by + c = 0

First, we'll re-write the equation $y = -\frac{2}{3} x +8$ in this form to identify a, b, and c:

$y = -\frac{2}{3} x+8$ add $\frac{2}{3}x$ to and subtract 8 from both sides

$y + \frac{2}{3}x -8 = 0$ multiply both sides by 3

3y +2x - 24 = 0 Now we see that a = 2, b = 3, c = -24 . Plugging these plus (x_0, y_0) = (1, 3 ) into the formula, we get:

$distance = \frac{|2(1)+3(3)-24 |}{\sqrt{(3)^2 +( 2)^2 }} = \frac{|2 + 9-24|}{\sqrt{9+4} } = \frac{13}{\sqrt{13}}$

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Example Question #1 : Find The Distance Between A Point And A Line

Find the distance between $y = -\frac{4}{5}x - 2$ and (-3,4) .

Possible Answers:

$\frac{18}{5}$

$\frac{18}{\sqrt{41}}$

$\frac{11}{\sqrt{42}}$

$\frac{42}{5}$

$\frac{42}{\sqrt{41}}$

Correct answer:

$\frac{18}{\sqrt{41}}$

Explanation:

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is (x_0, y_0) and the line is ax + by + c = 0

First, we'll re-write the equation $y = -\frac{4}{5} x -2$ in this form to identify , , and :

$y = -\frac{4}{5} x-2$ add $\frac{4}{5}x$ and to both sides

$y + \frac{4}{5}x +2 = 0$ multiply both sides by

5 y + 4 x + 10 = 0 Now we see that a = 4, b = 5, c = 10 . Plugging these plus (x_0, y_0) = (-3,4 ) into the formula, we get:

$distance = \frac{|4(-3)+5(4)+10 |}{\sqrt{(4)^2 +(5)^2 }} = \frac{|-12+20+10|}{\sqrt{16+25} } = \frac{18}{\sqrt{41}}$

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Example Question #3 : Find The Distance Between A Point And A Line

Find the distance between y = 4x+1 and (2, -3)

Possible Answers:

$\frac{10}{\sqrt{17}}$

$\frac{12}{\sqrt5}$

$\frac{13}{\sqrt{17}}$

$\frac{12}{\sqrt{17}}$

$\frac{13}{\sqrt{13}}$

Correct answer:

$\frac{12}{\sqrt{17}}$

Explanation:

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is (x_0, y_0) and the line is ax + by + c = 0

First, we'll re-write the equation y =4x+1 in this form to identify , , and :

y =4x+1 subtract and from both sides

y - 4x - 1 = 0

Now we see that a = -4, b = 1, c = -1 . Plugging these plus (x_0, y_0) = (2,-3) into the formula, we get:

$distance = \frac{|-4(2)+1(-3)-1 |}{\sqrt{(1)^2 +(-4)^2 }} = \frac{|-8 -3-1|}{\sqrt{1+16} } = \frac{12}{\sqrt{17}}$

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Example Question #3 : Find The Distance Between A Point And A Line

Find the distance between $y = \frac{7}{2} x - 14$ and (-6,2)

Possible Answers:

$\frac{2}{\sqrt{40}}$

$\frac{2}{\sqrt{53}}$

$\frac{45}{\sqrt{53}}$

$\frac{7}{\sqrt{53}}$

$\frac{45}{\sqrt{40}}$

Correct answer:

$\frac{45}{\sqrt{53}}$

Explanation:

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is (x_0, y_0) and the line is ax + by + c = 0

First, we'll re-write the equation $y = \frac{7}{2} x -14$ in this form to identify , , and :

$y = \frac{7}{2} x-14$ subtract $\frac{7}{2}x$ from and add to both sides

$y -\frac{7}{2}x +14 = 0$ multiply both sides by

2y -7x +28 = 0 Now we see that a = -7, b = 2, c =28 . Plugging these plus (x_0, y_0) = (-6,2 ) into the formula, we get:

$distance = \frac{|-7(-6)+2(2)+28 |}{\sqrt{(-7)^2 +( 2)^2 }} = \frac{|13 + 4+28|}{\sqrt{49+4} } = \frac{45}{\sqrt{53}}$

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Example Question #11 : Coordinate Geometry

Find the distance between $y = \frac{12}{5} x - 3$ and the point (3, -4)

Possible Answers:

$\frac{48}{5}$

$\frac{48}{13}$

$\frac{41}{5}$

$\frac{71}{13}$

$\frac{41}{13}$

Correct answer:

$\frac{41}{13}$

Explanation:

To find the distance, use the formula $distance = \frac{|ax_{0}+by_{0}+c |}{\sqrt{a^2 + b^2 }}$ where the point is (x_0, y_0) and the line is ax + by + c = 0

First, we'll re-write the equation $y = \frac{12}{5} x-3$ in this form to identify , , and :

$y = \frac{12}{5} x-3$ subtract $\frac{12}{5}x$ from and add to both sides

$y -\frac{12}{5}x +3 = 0$ multiply both sides by

5y -12x +15 = 0 Now we see that a = -12, b = 5, c = 15 . Plugging these plus (x_0, y_0) = (3,-4) into the formula, we get:

$distance = \frac{|-12(3)+5(-4)+15 |}{\sqrt{(5)^2 +( 12)^2 }} = \frac{|-36 + -20+15|}{\sqrt{25+144} } = \frac{41}{\sqrt{169}} = \frac{41}{13}$

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Example Question #1 : Find The Distance Between Two Parallel Lines

Find the distance between the two lines.

$\\ y_1=2.65x-13 \\ y_2=2.65x+4$

Possible Answers:

d=2.65

d=-17

d=9

d=6

Correct answer:

d=6

Explanation:

Since the slope of the two lines are equivalent, we know that the lines are parallel. Therefore, they are separated by a constant distance. We can then find the distance between the two lines by using the formula for the distance from a point to a nonvertical line:

$d=\frac{|Ax_1 +By_1+C|}{\sqrt{A^{2}+B^{2}}}$

First, we need to take one of the line and convert it to standard form.

y_1=2.65x-13

2.65x_1-y_1-13=0 where $A=2.65,B=-1, \textup{and }C=-13$

Now we can substitute A, B, and C into our distance equation along with a point, $\left ( x_1,y_1\right )$ , from the other line. We can pick any point we want, as long as it is on line y_2 . Just plug in a number for x, and solve for y. I will use the y-intercept, where x = 0, because it is easy to calculate:

y=2.65(0))+4

y=0+4

y=4

Now we have a point, (0,4) , that is on the line y_2 . So let's plug our values for $x_1,y_1,A,B,\textup{and }C}$ :

$d=\frac{|(2.65)\cdot (0) +(-1)\cdot (4)+(-13)|}{\sqrt{(2.65)^{2}+(-1)^{2}}}=\frac{|0-4-13|}{\sqrt{(7.0225+1)}}=\frac{|-17|}{\sqrt{8.0225}}$

$=\frac{17}{2.83}=6.002\approx6$

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Example Question #1 : Find The Distance Between Two Parallel Lines

Find the distance between y = 2x -3 and y = 2x+1

Possible Answers:

$\frac{6}{\sqrt5}$

$\frac{4}{\sqrt5}$

Correct answer:

$\frac{4}{\sqrt5}$

Explanation:

To find the distance, choose any point on one of the lines. Plugging in 2 into the first equation can generate our first point:

y = 2(2) - 3 = 4 -3 = 1 this gives us the point (2, 1 )

We can find the distance between this point and the other line by putting the second line into the form ax+by+c=0 :

y = 2x+1 subtract the whole right side from both sides

- 2x + y - 1 = 0 now we see that a = -2, b = 1, c = -1

We can plug the coefficients and the point into the formula

$distance = \frac{|ax + by+c|}{\sqrt{a^2 + b^2 }}$ where (x,y) represents the point.

$\frac{|-2(2) + 1(1)-1| }{ \sqrt{1^2 + 2^2 }} = \frac{|-4+1-1|}{\sqrt{1+4}} = \frac{4}{\sqrt{5}}$

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Precalculus : Graphing Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Find The Distance Between A Point And A Line

Example Question #2 : Find The Distance Between A Point And A Line

Example Question #3 : Find The Distance Between A Point And A Line

Example Question #2 : Find The Distance Between A Point And A Line

Example Question #1 : Find The Distance Between A Point And A Line

Example Question #3 : Find The Distance Between A Point And A Line

Example Question #3 : Find The Distance Between A Point And A Line

Example Question #11 : Coordinate Geometry

Example Question #1 : Find The Distance Between Two Parallel Lines

Example Question #1 : Find The Distance Between Two Parallel Lines

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