Coordinate Plane - ACT Math

Card 0 of 2502

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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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

$m=2x_{1}+5x_{1}^4$

.

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 the slope by plugging 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 the slope by plugging our $x_{1}$ into the derivative equation and solving.

Thus, the slope is

$m=4x_{1}^3$

.

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 tagent 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 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

There is a circle on a coordinate plain. Its perimeter passes through the point . At this point meets a tangent line, which also passes through the point . What is the slope of the line perpindicular to this tangent line?

Answer

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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Question

Find the slope of the tangent line to where .

Answer

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 is .

Now we plug in , giving us .

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Question

A line runs tangent to a circle at the point . The line runs through the origin. Find the slope of the tangent line.

Answer

The only two bits of information that are given for the tangent line is that it runs through the points and . 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:

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

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

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Question

Which of the following lines is parallel to:

Answer

First write the equation in slope intercept form. Add to both sides to get . Now divide both sides by to get . The slope of this line is , so any line that also has a slope of would be parallel to it. The correct answer is $x - \frac{1}{3}y = 7$ .

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Question

Which pair of linear equations represent parallel lines?

Answer

Parallel lines will always have equal slopes. The slope can be found quickly by observing the equation in slope-intercept form and seeing which number falls in the " $m$ " spot in the linear equation $(y=mx+b)$ ,

We are looking for an answer choice in which both equations have the same $m$ value. Both lines in the correct answer have a slope of 2, therefore they are parallel.

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