Perpendicular Lines

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PSAT Math › Perpendicular Lines

Questions 1 - 10

What line is perpendicular to and passes through ?

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

$y = \frac{-3}{5}x + 2$

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

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

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

Explanation

Convert the given equation to slope-intercept form.

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

The slope of this line is $-\frac{3}{5}$ . The slope of the line perpendicular to this one will have a slope equal to the negative reciprocal.

The perpendicular slope is $\frac{5}{3}$ .

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

$2 = \frac{5}{3}(3) + b$

So the equation of the perpendicular line is $y = \frac{5}{3}x - 3$ .

What line is perpendicular to and passes through ?

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

$y = \frac{-3}{5}x + 2$

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

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

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

Explanation

Convert the given equation to slope-intercept form.

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

The slope of this line is $-\frac{3}{5}$ . The slope of the line perpendicular to this one will have a slope equal to the negative reciprocal.

The perpendicular slope is $\frac{5}{3}$ .

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

$2 = \frac{5}{3}(3) + b$

So the equation of the perpendicular line is $y = \frac{5}{3}x - 3$ .

Find the equation of the line that is perpendicular to $\dpi{100} \small \frac{x}{2}+3=y$ and passes through (5, 6).

$\dpi{100} \small -2x+16=y$

$\dpi{100} \small \frac{1}{2}x+16=y$

$\dpi{100} \small -\frac{1}{2}x+16=y$

$\dpi{100} \small 2x-4=y$

$\dpi{100} \small -2x-4=y$

Explanation

We know that the slope of the original line is $\dpi{100} \small \frac{1}{2}$

Thus the slope of the perpendicular line is the negative reciprocal of $\dpi{100} \small \frac{1}{2}$ , or –2.

Then we plug the slope and point (5, 6) into the form $\dpi{100} \small m\left ( x-x_{1} \right )=y-y_{1}$ , which yields $\dpi{100} \small -2\left ( x-5 \right )=y-6$

When we simplify this, we arrive at $\dpi{100} \small -2x+16=y$

Line includes the points and . Line includes the points and . Which of the following statements is true of these lines?

The lines are distinct but neither parallel nor perpendicular.

The lines are parallel.

The lines are perpendicular.

The lines are identical.

More information is needed to answer this question.

Explanation

We calculate the slopes of the lines using the slope formula.

The slope of line is

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{-3 - (-6)}{4-8} = \frac{3}{-4} = - \frac{3}{4}$

The slope of line is

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{3- 9}{-20- (-12)} = \frac{-6}{-8} = \frac{3}{4}$

Parallel lines and identical lines must have the same slope, so these can be eliminated as choices. The slopes of perpendicular lines must have product . The slopes have product

$- \frac{3}{4} \times \frac{3}{4} = - \frac{9}{16} \neq -1$

so they are not perpendicular.

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

Find the equation of the line that is perpendicular to $\dpi{100} \small \frac{x}{2}+3=y$ and passes through (5, 6).

$\dpi{100} \small -2x+16=y$

$\dpi{100} \small \frac{1}{2}x+16=y$

$\dpi{100} \small -\frac{1}{2}x+16=y$

$\dpi{100} \small 2x-4=y$

$\dpi{100} \small -2x-4=y$

Explanation

We know that the slope of the original line is $\dpi{100} \small \frac{1}{2}$

Thus the slope of the perpendicular line is the negative reciprocal of $\dpi{100} \small \frac{1}{2}$ , or –2.

Then we plug the slope and point (5, 6) into the form $\dpi{100} \small m\left ( x-x_{1} \right )=y-y_{1}$ , which yields $\dpi{100} \small -2\left ( x-5 \right )=y-6$

When we simplify this, we arrive at $\dpi{100} \small -2x+16=y$

Line includes the points and . Line includes the points and . Which of the following statements is true of these lines?

The lines are distinct but neither parallel nor perpendicular.

The lines are parallel.

The lines are perpendicular.

The lines are identical.

More information is needed to answer this question.

Explanation

We calculate the slopes of the lines using the slope formula.

The slope of line is

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{-3 - (-6)}{4-8} = \frac{3}{-4} = - \frac{3}{4}$

The slope of line is

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{3- 9}{-20- (-12)} = \frac{-6}{-8} = \frac{3}{4}$

Parallel lines and identical lines must have the same slope, so these can be eliminated as choices. The slopes of perpendicular lines must have product . The slopes have product

$- \frac{3}{4} \times \frac{3}{4} = - \frac{9}{16} \neq -1$

so they are not perpendicular.

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

If a line has an equation of $2y=3x+3$ , what is the slope of a line that is perpendicular to the line?

$-\frac{2}{3}$

$-\frac{3}{2}$

$\frac{3}{2}$

$3$

$-2$

Explanation

Putting the first equation in slope-intercept form yields $y=\frac{3}{2}x+\frac{3}{2}$ .

A perpendicular line has a slope that is the negative inverse. In this case, $-\frac{2}{3}$ .

If a line has an equation of $2y=3x+3$ , what is the slope of a line that is perpendicular to the line?

$-\frac{2}{3}$

$-\frac{3}{2}$

$\frac{3}{2}$

$3$

$-2$

Explanation

Putting the first equation in slope-intercept form yields $y=\frac{3}{2}x+\frac{3}{2}$ .

A perpendicular line has a slope that is the negative inverse. In this case, $-\frac{2}{3}$ .

The equation of line p is y= 1/4x +6. If line k contains the point (3,5) and is perpendicular to line p, find the equation of line k.

y = 4x - 17

y = 1/4x + 17

y = -4x + 17

y = 3x + 5

Explanation

Using the slope intercept formula, we can see the slope of line p is ¼. Since line k is perpendicular to line p it must have a slope that is the negative reciprocal. (-4/1) If we set up the formula y=mx+b, using the given point and a slope of (-4), we can solve for our b or y-intercept. In this case it would be 17.

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The equation of line p is y= 1/4x +6. If line k contains the point (3,5) and is perpendicular to line p, find the equation of line k.

y = 4x - 17

y = 1/4x + 17

y = -4x + 17

y = 3x + 5