Lines

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ACT Math › Lines

Questions 1 - 10

Which of the following lines is perpendicular to the line ?

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

Explanation

Perpendicular lines will have slopes that are negative reciprocals of one another. Our first step will be to find the slope of the given line by putting the equation into slope-intercept form.

The slope of this line is . The negative reciprocal will be $+\frac{1}{3}$ , which will be the slope of the perpendicular line.

Now we need to find the answer choice with this slope by converting to slope-intercept form.

$y=\frac{4}{12}x-\frac{6}{12}$

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

This equation has a slope of $+\frac{1}{3}$ , and must be our answer.

Which of the following lines is perpendicular to the line with the given equation:
?

$y = -\frac{1}{3}x + 5$

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

Explanation

First we must recognize that the equation is given in slope-intercept form, where is the slope of the line.

Two lines are perpendicular if and only if the product of their slopes is . In other words, the slope of a line that is perpendicular to a given line is the negative reciprocal of that slope.

Thus, for a line with a given slope of 3, the line perpendicular to that slope must be the negative reciprocal of 3, or $-\frac{1}{3}$ .

To double check that that does indeed give a product of when multiplied by three simply compute the product:

$-\frac{1}{3}\cdot 3= -1$

Suppose the midpoint of a line segment is What are the endpoints of the segment?

Explanation

The midpoint of a line segment is found using the formula $\left ( \frac{x_1+x_2 }{2}, \frac{y_1+y_2}{2}\right )$ .

The midpoint is given as Going through the answer choices, only the points and yield the correct midpoint of .

$\left ( \frac{2+6}{2}, \frac{8+16}{2} \right )=(4,12)$

The midpoint of a line segment is . If one endpoint of the line segment is , what is the other endpoint?

Explanation

The midpoint formula can be used to solve this problem, where the midpoint is the average of the two coordinates.

$\text{midpoint}=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

We are given the midpoint and one endpoint. Plug these values into the formula.

$(4,-1)=(\frac{-4+x}{2},\frac{3+y}{2})$

$4=\frac{-4+x}{2}\ \text{and}\ -1=\frac{3+y}{2}$

Solve for the variables to find the coordinates of the second endpoint.

$8=-4+x\ \text{and}\ -2=3+y$

$12=x\ \text{and}\ -5=y$

The final coordinates of the other endpoint are .

The midpoint of a line segment is . If one endpoint of the line segment is , what is the other endpoint?

Explanation

The midpoint formula can be used to solve this problem, where the midpoint is the average of the two coordinates.

$\text{midpoint}=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

We are given the midpoint and one endpoint. Plug these values into the formula.

$(4,-1)=(\frac{-4+x}{2},\frac{3+y}{2})$

$4=\frac{-4+x}{2}\ \text{and}\ -1=\frac{3+y}{2}$

Solve for the variables to find the coordinates of the second endpoint.

$8=-4+x\ \text{and}\ -2=3+y$

$12=x\ \text{and}\ -5=y$

The final coordinates of the other endpoint are .

What is the equation of a tangent line to

at point ?

Explanation

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

Find the slope of the tangent line to where .

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 is .

Now we plug in , giving us .

Find the slope of the tangent line to where .

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 is .

Now we plug in , giving us .

There is a line defined by the equation below:

There is a second line that passes through the point and is parallel to the line given above. What is the equation of this second line?

$y=-\frac{3}{4}x+2.75$

$y=-\frac{3}{4}x+1.25$

$y=\frac{3}{4}x+2.75$

$y=\frac{3}{4}x+1.25$

$y=\frac{3}{4}x+2.625$

Explanation

Parallel lines have the same slope. Solve for the slope in the first line by converting the equation to slope-intercept form.

3x + 4y = 12

4y = _–_3x + 12

y = –(3/4)x + 3

slope = _–_3/4

We know that the second line will also have a slope of _–_3/4, and we are given the point (1,2). We can set up an equation in slope-intercept form and use these values to solve for the y-intercept.

y = mx + b

2 = _–_3/4(1) + b

2 = _–_3/4 + b

b = 2 + 3/4 = 2.75

Plug the y-intercept back into the equation to get our final answer.

y = –(3/4)x + 2.75

What is the slope of a line which passes through coordinates $\dpi{100} \small (3,7)$ and $\dpi{100} \small (4,12)$ ?