Lines - ACT Math

Card 0 of 1305

Question

What is the slope of the line:

$\frac{14}{3}x=\frac{1}{6}y-7$

Answer

First put the question in slope intercept form (y = mx + b):

–(1/6)y = –(14/3)x – 7 =>

y = 6(14/3)x – 7

y = 28x – 7.

The slope is 28.

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Question

If 2x – 4y = 10, what is the slope of the line?

Answer

First put the equation into slope-intercept form, solving for y: 2x – 4y = 10 → –4y = –2x + 10 → y = 1/2*x – 5/2. So the slope is 1/2.

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Question

What is the slope of the line with equation 4_x_ – 16_y_ = 24?

Answer

The equation of a line is:

y = mx + b, where m is the slope

4_x_ – 16_y_ = 24

–16_y_ = –4_x_ + 24

y = (–4_x_)/(–16) + 24/(–16)

y = (1/4)x – 1.5

Slope = 1/4

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Question

What is the midpoint of MN between the points M(2, 6) and N (8, 4)?

Answer

The midpoint formula is equal to . Add the x-values together and divide them by 2, and do the same for the y-values.

x: (2 + 8) / 2 = 10 / 2 = 5

y: (6 + 4) / 2 = 10 / 2 = 5

The midpoint of MN is (5,5).

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Question

Which of the following is the equation of a line parallel to the line .

Answer

Parallel lines have equivalent slopes, so the correct answer is .

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Question

What line is perpendicular to x + 3_y_ = 6 and travels through point (1,5)?

Answer

Convert the equation to slope intercept form to get y = –1/3_x_ + 2. The old slope is –1/3 and the new slope is 3. Perpendicular slopes must be opposite reciprocals of each other: _m_1 * _m_2 = –1

With the new slope, use the slope intercept form and the point to calculate the intercept: y = mx + b or 5 = 3(1) + b, so b = 2

So y = 3_x_ + 2

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Question

What line is perpendicular to 2x + y = 3 at (1,1)?

Answer

Find the slope of the given line. The perpendicular slope will be the opposite reciprocal of the original slope. Use the slope-intercept form (y = mx + b) and substitute in the given point and the new slope to find the intercept, b. Convert back to standard form of an equation: ax + by = c.

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Question

What is the slope of the line perpendicular to the line given by the equation

6x – 9y +14 = 0

Answer

First rearrange the equation so that it is in slope-intercept form, resulting in y=2/3 x + 14/9. The slope of this line is 2/3, so the slope of the line perpendicular will have the opposite reciprocal as a slope, which is -3/2.

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Question

What is the slope of the line perpendicular to the line represented by the equation y = -2x+3?

Answer

Perpendicular lines have slopes that are the opposite of the reciprocal of each other. In this case, the slope of the first line is -2. The reciprocal of -2 is -1/2, so the opposite of the reciprocal is therefore 1/2.

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Question

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)?

Answer

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

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: 3_x_ – 4_y_ = -25

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Question

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 .

Answer

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 .

A tangent to this circle at a given point is perpendicular to the radius to that point. The radius with endpoints and 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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Question

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 .

Answer

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

A tangent to this circle at a given point is perpendicular to the radius to that point. The radius with endpoints and 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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Question

What is the equation of a tangent line to

at point ?

Answer

To find an equation tangent to

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

So,

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

$m=2x_{1}+4$

.

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,

Once we rearrange, the equation is

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Question

What is the equation of a tangent line to

at point

?

Answer

To find an equation tangent to

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

So,

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

$m=3x_{1}^2$

.

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

Once we rearrange, the equation is

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Question

Find the tangent line equation to

at point

?

Answer

To find an equation tangent to

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

So,

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

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

Thus, the slope is

$m=2x_{1}$

.

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

We plug into

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

Therefore our equation is

Once we rearrange, the equation is

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Question

What is the tangent line equation of

at point

?

Answer

To find an equation tangent to

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

So,

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

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

.

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

Once we rearrange, the equation is

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Question

Find the equation of a tangent line to

for the point

?

Answer

To find an equation tangent to

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

So,

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

$m=2x_{1}$

.

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

Once we rearrange, the equation is

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Question

What is the equation of a tangent line to

at the point

?

Answer

To find an equation tangent to

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

So,

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

$m=25x_{1}^4$

.

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

Once we rearrange, the equation is

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Question

Find the equation of a tangent line to

at the point

?

Answer

To find an equation tangent to

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

So,

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

$m=2x_{1}$

.

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

Once we rearrange, the equation is

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Question

What is the equation of a tangent line to

at the point

?

Answer

To find an equation tangent to

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

So,

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

$m=2x_{1}+1$

.

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

Once we rearrange, the equation is

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