Perpendicular Lines - Intermediate Geometry

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Question

Given the equation of a line: $y=\frac{3}{5}x+2$

Which equation given below is perpendicular to the given line?

Answer

When looking at the equation of the given line, we know that the slope is $\frac{3}{5}$ and the y-intercept is . Any line perpendicular to the given line will create a 90 degree angle with the given line.

Now, there are INFINITE lines perpendicular to the given line, all with different y-intercepts. So in other words, the line we are looking for will have no dependence on the y-intercept, as any y-intercept will do. What we do care about is the slope of the line.

The slope of any line perpendicular to a given line has a negative reciprocal of the slope. So for this problem:

$given;slope=\frac{3}{5};\rightarrow;perpendicual;slope=-\frac{5}{3}$

The only line that has this slope is

$y=-\frac{5}{3}x+7$

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Question

Given the equation of a line:

$y=\frac{1}{2}x+5$

Find the equation of a line parallel to the given line.

Answer

Parallel lines will never touch, and therefore they must have the same slope.

Many of the answers are reciprocals or negative slopes, but the slope we are looking for is $\frac{1}{2}$ .

That leaves us with 2 answers. However, one of the answers is the exact same equation for a line as the given equation. Therefore our answer is:

$y=\frac{1}{2}x+1$

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Question

Which of the following lines is perpendicular to the line

Answer

The slope of the given line is 3. The slope of a perpendicular line is the negative inverse of the given line. In this case, that is equal to $-\frac{1}{3}$ . Therefore, the correct answer is:

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

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Question

Which of the following is perpendicular to

Answer

Two lines are perpendicular if and only if their slopes are negative reciprocals. To find the slope, we must put the equation into slope-intercept form, , where equals the slope of the line. We begin by subtracting from each side, giving us . Next, we subtract 32 from each side, giving us . Finally, we divide each side by , giving us . We can now see that the slope is . Therefore, any line perpendicular to must have a slope of . Of the equations above, only has a slope of .

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Question

Which of the following equations is perpendicular to ?

Answer

Convert the given equation to slope-intercept form:

Divide both sides of the equation by :

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

The slope of this function is :

The slope of the perpendicular line will be the negative reciprocal of the original slope. Substitute and solve:

$m_2 = -\frac{1}{m_1}=-\frac{1}{-1}=1$

Only has a slope of .

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Question

Which line is perpendicular to the given line below?

$y=-\frac{5}{2}x-4$

Answer

Two perpendicular lines have slopes that are opposite reciprocals, meaning that the sign changes and the reciprocal of the slope is taken.

The original equation is in slope-intercept form,

where represents the slope.

In this case, the slope of the original is:

$m=-\frac{5}{2}$

After taking the opposite reciprocal, the result is the slope below:

$m=\frac{2}{5}$

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Question

Are the lines of the equations

and

parallel, perpendicular, or neither?

Answer

Any equation of the form , such as , can be graphed by a vertical line; any equation of the form , such as , can be graphed by a horizontal line. A vertical line and a horizontal line are perpendicular to each other.

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Question

Are the lines of the equations

and

parallel, perpendicular, or neither?

Answer

Write each equation in the slope-intercept form by solving for ; the -coefficient is the slope of the line.

Subtract from both sides:

The line of this equation has slope .

Subtract from both sides:

Multiply both sides by $\frac{1}{2}$

$\frac{1}{2} \cdot 2y =\frac{1}{2} \cdot ( -x+ 12)$

$y = \frac{1}{2} \cdot( - x)+ \frac{1}{2} \cdot 12$

$y = - \frac{1}{2} x+ 6$

The line of this equation has slope $-\frac{1}{2}$ .

Two lines are parallel if and only if they have the same slope; this is not the case. They are perpendicular if and only if the product of their slopes is ; this is not the case, since

$-2 \times\left ( -\frac{1}{2} \right ) = 1$ .

The correct response is that the lines are neither parallel nor perpendicular.

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Question

Are the lines of the equations

and

parallel, perpendicular, or neither?

Answer

Write each equation in the slope-intercept form by solving for ; the -coefficient is the slope of the line.

Subtract from both sides:

Multiply both sides by $\frac{1}{2}$ :

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

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

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

The slope is the -coefficient $- \frac{3}{2}$

Add to both sides:

Multiply both sides by $\frac{1}{2}$ :

$\frac{1}{2} \cdot 2y = \frac{1}{2} \cdot (3x+ 17)$

$y= \frac{1}{2} \cdot 3x+ \frac{1}{2} \cdot 17$

$y= \frac{3}{2} x+ \frac{17}{2}$

The slope is the -coefficient $\frac{3}{2}$ .

Two lines are parallel if and only if they have the same slope; this is not the case. They are perpendicular if and only if the product of their slopes is ; this is not the case, since $- \frac{3}{2} \times \frac{3}{2} =- \frac{9}{4}$ . The lines are neither parallel nor perpendicular.

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Question

The slopes of two lines on the coordinate plane are and 4.

True or false: the lines are perpendicular.

Answer

Two lines on the coordinate plane are perpendicular if and only if the product of their slopes is . The product of the slopes of the lines in question is

$- 0.25 \cdot 4 = -1$ ,

so the lines are indeed perpendicular.

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Question

Two lines intersect at the point . One line passes through the point ; the other passes through .

True or false: The lines are perpendicular.

Answer

Two lines are perpendicular if and only if the product of their slopes is . The slope of each line can be found from the coordinates of two points using the slope formula

$m = \frac{y_{2}- y_{1}}{x_{2}-x_{1}}$

To find the slope of the first line, set $x_{1} = 3,y_{1}= 8 , x_{2} = 7, y_{2}= 3$ :

$m = \frac{3-8}{7-3} = \frac{-5}{4} = -\frac{5}{4}$

To find the slope of the second line, set $x_{1} = 3,y_{1}= 8 , x_{2} = -2, y_{2}=1 2$ :

$m = \frac{12-8}{-2-3} = \frac{4}{-5} = -\frac{4}{5}$

The product of the slopes is

$-\frac{5}{4} \left ( -\frac{4}{5} \right )= 1$

As the product is not , the lines are not perpendicular.

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Question

The slopes of two lines are 6 and . True or false: the lines are perpendicular.

Answer

Two lines on the coordinate plane are perpendicular if and only if the product of their slopes is . The product of the slopes of the lines in question is

$- 0.16666 \cdot 6 = -0.99996$

The product is not equal to , so the lines are not perpendicular.

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Question

Which line below is perpendicular to ?

Answer

The definition of a perpendicular line is one that has a negative, reciprocal slope to another.

For this particular problem, we must first manipulate our initial equation into a more easily recognizable and useful form: slope-intercept form or .

$y=-\frac{5}{6}x+6$

According to our formula, our slope for the original line is $-\frac{5}{6}$ . We are looking for an answer that has a perpendicular slope, or an opposite reciprocal. The opposite reciprocal of $-\frac{5}{6}$ is $\frac{6}{5}$ . Flip the original and multiply it by .

Our answer will have a slope of $\frac{6}{5}$ . Search the answer choices for $\frac{6}{5}$ in the position of the equation.

$\dpi{100} y = \frac{6}{5}x + 3$ is our answer.

(As an aside, the negative reciprocal of 4 is $-\frac{1}{4}$ . Place the whole number over one and then flip/negate. This does not apply to the above problem, but should be understood to tackle certain permutations of this problem type where the original slope is an integer.)

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Question

What line is perpendicular to $y=\frac{-1}{2}x-3$ through ?

Answer

$y=\frac{-1}{2}x-3$ is given in the slope-intercept form. So the slope is $\frac{-1}{2}$ and the y-intercept is .

If the lines are perpendicular, then $m_{1}\cdot m_{2}=-1$ so the new slope must be

Next we substitute the new slope and the given point into the slope-intercept form of the equation to calculate the intercept. So the equation to solve becomes so

So the equation of the perpendicular line becomes or in standard form

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Question

Find the equation of the line perpendicular to $y=\frac{-1}{3}x+2$ .

Answer

The definition of a perpendicular line is a line with a negative reciprocal slope and identical intercept.

Therefore we need a line with slope 3 and intercept 2.

This means the only fitting line is .

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Question

Which one of these equations is perpendicular to:

Answer

To find the perpendicular line to

we need to find the negative reciprocal of the slope of the above equation.

So the slope of the above equation is since changes by when is incremented.

The negative reciprocal is:

$reciprocal=\frac{-1}{slope}$

$reciprocal=\frac{1}{-1}$

So we are looking for an equation with a .

Only satisfies this condition.

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Question

Suppose a line is represented by a function . Find the equation of a perpendicular line that intersects the point .

Answer

Determine the slope of the function . The slope is:

The slope of a perpendicular line is the negative reciprocal of the original slope. Determine the value of the slope perpendicular to the original function.

$m2 = - \frac{1}{m1} = -\frac{1}{2}$

Plug in the given point and the slope to the slope-intercept form to find the y-intercept.

$2=\left(-\frac{1}{2}\right)(1)+b$

$b=2\frac{1}{2}$

Substitute the slope of the perpendicular line and the new y-intercept back in the slope-intercept equation, .

The correct answer is: $f(x)= -\frac{1}{2}x+2\frac{1}{2}$

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Question

Suppose a perpendicular line passes through and point . Find the equation of the perpendicular line.

Answer

Find the slope from the given equation . The slope is: .

The slope of the perpendicular line is the negative reciprocal of the original slope.

$m2=-\frac{1}{m1}=-\frac{1}{2}$

Plug in the perpendicular slope and the given point to the slope-intercept equation.

$5=\left(-\frac{1}{2}\right)(-1)+b$

$5=\frac{1}{2}+b$

$b=4\frac{1}{2}$

Plug in the perpendicular slope and the y-intercept into the slope-intercept equation to get the equation of the perpendicular line.

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

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Question

Find a line perpendicular , but passing through the point .

Answer

Since we need a line perpendicular to we know our slope must be $m=-\frac{1}{6}$ . This is because perpendicular lines have slopes that are negative reciprocals of each other. In order for our new line to pass through the point we must use the point slope formula. Be sure to use the perpendicular slope.

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

$y-2=\frac{-1}{6}(x-7)$

$y=\frac{-1}{6}x+\frac{7}{6}+2$

$y=\frac{-1}{6}x+\frac{7}{6}+\frac{12}{6}$

$\mathbf{y=-\frac{1}{6}x+\frac{19}{6}}$

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Question

A line is perpendicular to the line of the equation

and passes through the point .

Give the equation of the line.

Answer

A line perpendicular to another line will have as its slope the opposite of the reciprocal of the slope of the latter. Therefore, it is necessary to find the slope of the line of the equation

Rewrite the equation in slope-intercept form . , the coefficient of , will be the slope of the line.

Add to both sides:

Multiply both sides by $\frac{1}{7}$ , distributing on the right:

$\frac{1}{7} \cdot 7y = \frac{1}{7} \cdot( -5x+32)$

$y = -\frac{5}{7}x+\frac{32}{7}$

The slope of this line is $-\frac{5}{7}$ . The slope of the first line will be the opposite of the reciprocal of this, or $\frac{7}{5}$ . The slope-intercept form of the equation of this line will be

$y = \frac{7}{5}x+b$ .

To find , set and and solve:

$8 = \frac{7}{5} (-2)+b$

$8 = - \frac{14}{5} +b$

$\frac{40}{5} = - \frac{14}{5} +b$

Add $\frac{14}{5}$ to both sides:

$\frac{40}{5} + \frac{14}{5} = - \frac{14}{5} +b + \frac{14}{5}$

$\frac{54}{5} = b$

The equation, in slope-intercept form, is $y = \frac{7}{5}x+ \frac{54}{5}$ .

To rewrite in standard form with integer coefficients:

Multiply both sides by 5:

$5 y =5 \left (\frac{7}{5}x+ \frac{54}{5} \right )$

$5 y =5 \left (\frac{7}{5}x \right ) + 5 \left ( \frac{54}{5} \right )$

Add to both sides:

or

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