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ACT Math : Lines

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Example Questions

Example Question #1 : Coordinate Plane

There is a circle on a coordinate plain. Its perimeter passes through the point (1,0)

. At this point meets a tangent line, which also passes through the point (5,7)

. What is the slope of the line perpindicular to this tangent line?

Possible Answers:

$\frac{7}{4}$

$\frac{4}{7}$

$-\frac{4}{7}$

$-\frac{7}{4}$

$\frac{7}{6}$

Correct answer:

$-\frac{4}{7}$

Explanation:

In this kind of problem, it's important to keep track of information given about your line of interest. In this case, the coordinates given set up the stage for us to be able to get to our line of focus - the line perpendicular to the tangent line. In order to determine the perpendicular line's slope, the tangent line's slope must be calculated. Keeping in mind that:

$slope =\frac{rise}{run}=\frac{\Delta y}{\Delta x} =\frac{y_2-y_1}{x_2-x_1}$

where y2,x2 and y1,x1 are assigned arbitrarily as long as the order of assignment is maintained.

$\frac{y_2-y_1}{x_2-x_1}=\frac{7-0}{5-1}=\frac{7}{4}$ which is the slope of the tangent line.

To calculate the perpendicular line, we have to remember that the product of the tangent slope and the perpendicular slope will equal -1.

$\frac{7}{4} x \frac{?}{?} = -1$ , the perpendicular slope can then be calculated as $\frac{-4}{7}$

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Example Question #1 : Coordinate Plane

Find the slope of the tangent line to y=2x^2-16x+21 where x=2 .

Possible Answers:

Correct answer:

Explanation:

To find the slope of the tangent line, we must take the derivative.

By using the Power Rule we will be able to find the derivative:

$f(x)=a^n \rightarrow f'(x)=na^{n-1}$

Therefore derivative of 2x^2-16x+21 is 4x-16 .

Now we plug in x=2 , giving us 8-16=-8 .

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Example Question #1 : How To Find The Slope Of A Tangent Line

A line runs tangent to a circle at the point (4,2) . The line runs through the origin. Find the slope of the tangent line.

Possible Answers:

0.5

0.33

Cannot be determined

Correct answer:

0.5

Explanation:

The only two bits of information that are given for the tangent line is that it runs through the points (4,2) and (0,0) . With these two points, the line's slope can be easily calculated through the equation: $m=\frac{rise}{run}=\frac{\Delta y}{\Delta x}= \frac{y_{1}-y_{2}}{x_{1}-x_{2}}$

where is slope, is the -coordinate of the points, and is the -coordinates of the points.

Slope can be calculated through substituting in for the given values:

$m=\frac{2-0}{4-0}$

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

m= 0.5

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Example Question #1 : Tangent Lines

Circle A is centered about the origin and has a radius of 5. What is the equation of the line that is tangent to Circle A at the point (–3,4)?

Possible Answers:

3x – 4y = –1

3x – 4y = –25

3x + 4y = 7

–3x + 4y = 1

Correct answer:

3x – 4y = –25

Explanation:

The line must be perpendicular to the radius at the point (–3,4). The slope of the radius is given by Actmath_7_113_q7

The radius has endpoints (–3,4) and the center of the circle (0,0), so its slope is –4/3.

The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is ¾.

The equation of the line is y – 4 = (3/4)(x – (–3))

Rearranging gives us: 3x – 4y = -25

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Example Question #1 : Algebra

Give the equation, in slope-intercept form, of the line tangent to the circle of the equation

$x^{2}+ y^{2} = 225$

at the point (12, -9) .

Possible Answers:

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

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

$y = \frac{4}{3} x- 25$

None of the other responses gives the correct answer.

$y = \frac{3} {4}x-18$

Correct answer:

$y = \frac{4}{3} x- 25$

Explanation:

The graph of the equation $x^{2}+ y^{2} = 225$ is a circle with center (0,0) .

A tangent to this circle at a given point is perpendicular to the radius to that point. The radius with endpoints (0,0) and (12, -9) will have slope

$\frac{ y_{2} - y_{1} }{x_{2} - x_{1} } =\frac{-9 -0}{12-0} =\frac{-9 }{12 }=- \frac{3}{4}$ ,

so the tangent line has the opposite of the reciprocal of this, or $\frac{4}{3}$ , as its slope.

The tangent line therefore has equation

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

$y - (-9)= \frac{4}{3} (x-12 )$

$y +9= \frac{4}{3} x- 16$

$y = \frac{4}{3} x- 25$

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Example Question #2 : Algebra

Give the equation, in slope-intercept form, of the line tangent to the circle of the equation

$x^{2}-6x + y^{2} - 91 = 0$

at the point (-3, 8) .

Possible Answers:

$y = \frac{4} {3}x+12$

$y =- \frac{3}{4} x+\frac{23}{4}$

None of the other responses gives the correct answer.

$y = \frac{3}{4} x+\frac{41}{4}$

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

Correct answer:

$y = \frac{3}{4} x+\frac{41}{4}$

Explanation:

Rewrite the equation of the circle in standard form to find its center:

$x^{2}-6x + y^{2} - 91 = 0$

$x^{2}-6x + y^{2} = 91$

Complete the square:

$x^{2}-6x+9 + y^{2} = 91+9$

$\left (x-3 \right )^{2} + y^{2} = 100$

The center is (3,0) .

A tangent to this circle at a given point is perpendicular to the radius to that point. The radius with endpoints (3,0) and (-3, 8) will have slope

$\frac{ y_{2} - y_{1} }{x_{2} - x_{1} } = \frac{8 -0}{-3-3} = \frac{8}{-6} = - \frac{4}{3}$ ,

so the tangent line has the opposite of the reciprocal of this, or $\frac{3}{4}$ , as its slope.

The tangent line therefore has equation

$y - 8= \frac{3}{4} (x- (-3))$

$y - 8= \frac{3}{4} (x+3)$

$y - 8= \frac{3}{4} x+\frac{9}{4}$

$y = \frac{3}{4} x+\frac{41}{4}$

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Example Question #3 : How To Find The Equation Of A Tangent Line

What is the equation of a tangent line to

y=x^2+4x+2

at point (1,7) ?

Possible Answers:

y=3x+1

y=x+6

y=6x+1

y=4x+1

y=x^2+4x+2

Correct answer:

y=6x+1

Explanation:

To find an equation tangent to

y=x^2+4x+2

we need to find the first derviative of this equation with respect to to get the slope of the tangent line.

So,

y'=2x+4

due to power rule $y=x^n \rightarrow y'=nx^{n-1}$ .

First we need to find our slope by plugging our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=2x_{1}+4$

m=6 .

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug the point into

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

Therefore our equation becomes,

y-7=6(x-1)

Once we rearrange, the equation is

y=6x+1

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Example Question #2 : Algebra

What is the tangent line equation of

y=x^4+4x

at point

(1,5) ?

Possible Answers:

y=4x^3+4

y=8x-3

y=8x+5

y=4

y=8x+8

Correct answer:

y=8x-3

Explanation:

To find an equation tangent to

y=x^4+4x

we need to find the first derviative of this equation with respect to to get the slope of the tangent line.

So,

y'=4x^3+4

due to power rule $y=ax^n \rightarrow y'=nax^{n-1}$ .

First we need to find the slope by plugging our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=4x_{1}^3+4$

m=8 .

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug it into

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

Therefore our equation is

y-5=8(x-1)

Once we rearrange, the equation is

y=8x-3

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Example Question #4 : Lines

Find the equation of a tangent line to

y=x^2

for the point

(0,0) ?

Possible Answers:

y=2

y=2x

y=0

y=1

y=x^2

Correct answer:

y=0

Explanation:

To find an equation tangent to

y=x^2

we need to find the first derviative of this equation with respect to to get the slope of the tangent line.

So,

y'=2x

due to power rule $y=x^n \rightarrow y'=nx^{n-1}$ .

First we need to find the slope by plugging in our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=2x_{1}$

m=0 .

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug our point into

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

Therefore our equation is

y-0=0(x-0)

Once we rearrange, the equation is

y=0

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Example Question #3 : How To Find The Equation Of A Tangent Line

What is the equation of a tangent line to

y=5x^5+2

at the point

(1,7) ?

Possible Answers:

y=27x

y=25x-18

y=18x-25

y=25x^4

y=25x^4+2

Correct answer:

y=25x-18

Explanation:

To find an equation tangent to

y=5x^5+2

we need to find the first derviative of this equation with respect to to get the slope of the tangent line.

So,

y'=25x^4

due to power rule $y=ax^n \rightarrow y'=nax^{n-1}$ .

First we need to find our slope by plugging in our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=25x_{1}^4$

m=25 .

To find the equation of a tangent line of a given point $(x_{1},y_{1})$ we plug the point into

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

Therefore our equation is

y-7=25(x-1)

Once we rearrange, the equation is

y=25x-18

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ACT Math : Lines

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1 : Coordinate Plane

Example Question #1 : Coordinate Plane

Example Question #1 : How To Find The Slope Of A Tangent Line

Example Question #1 : Tangent Lines

Example Question #1 : Algebra

Example Question #2 : Algebra

Example Question #3 : How To Find The Equation Of A Tangent Line

Example Question #2 : Algebra

Example Question #4 : Lines

Example Question #3 : How To Find The Equation Of A Tangent Line

All ACT Math Resources

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