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PSAT Math › Lines

Questions 1 - 10

$\textup{What is the slope of a line perpendicular to the line with the equation}$

$3y+2x=6\textup{ ?}$

$\frac{3}{2}$

$\frac{-3}{2}$

$\frac{-2}{3}$

$\frac{2}{3}$

$\textup{Connot be determined with this information.}$

Explanation

$\textup{Get into }y=mx+b\textup{ format: }3y=-2x+6:,:y=-\frac{2}{3}x+2$

$\textup{Perpendicular slope is negative reciprocal of }-\frac{2}{3}: \rightarrow \rightarrow \frac{3}{2}$

One line has four collinear points in order from left to right A, B, C, D. If AB = 10’, CD was twice as long as AB, and AC = 25’, how long is AD?

45'

40'

50'

35'

30'

Explanation

AB = 10 ’

BC = AC – AB = 25’ – 10’ = 15’

CD = 2 * AB = 2 * 10’ = 20 ’

AD = AB + BC + CD = 10’ + 15’ + 20’ = 45’

One line has four collinear points in order from left to right A, B, C, D. If AB = 10’, CD was twice as long as AB, and AC = 25’, how long is AD?

45'

40'

50'

35'

30'

Explanation

AB = 10 ’

BC = AC – AB = 25’ – 10’ = 15’

CD = 2 * AB = 2 * 10’ = 20 ’

AD = AB + BC + CD = 10’ + 15’ + 20’ = 45’

A line segment has endpoints (0,4) and (5,6). What are the coordinates of the midpoint?

(0,4)

(0,6)

(2.5,-5)

(2.5,5)

(3,9)

Explanation

A line segment has endpoints (0,4) and (5,6). To find the midpoint, use the midpoint formula:

X: (x1+x2)/2 = (0+5)/2 = 2.5

Y: (y1+y2)/2 = (4+6)/2 = 5

The coordinates of the midpoint are (2.5,5).

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

$\textup{What is the slope of a line perpendicular to the line with the equation}$

$3y+2x=6\textup{ ?}$

$\frac{3}{2}$

$\frac{-3}{2}$

$\frac{-2}{3}$

$\frac{2}{3}$

$\textup{Connot be determined with this information.}$

Explanation

$\textup{Get into }y=mx+b\textup{ format: }3y=-2x+6:,:y=-\frac{2}{3}x+2$

$\textup{Perpendicular slope is negative reciprocal of }-\frac{2}{3}: \rightarrow \rightarrow \frac{3}{2}$

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

A line segment has endpoints (0,4) and (5,6). What are the coordinates of the midpoint?

(0,4)

(0,6)

(2.5,-5)

(2.5,5)

(3,9)

Explanation

A line segment has endpoints (0,4) and (5,6). To find the midpoint, use the midpoint formula:

X: (x1+x2)/2 = (0+5)/2 = 2.5

Y: (y1+y2)/2 = (4+6)/2 = 5

The coordinates of the midpoint are (2.5,5).

A line segment has endpoints (0,4) and (5,6). What are the coordinates of the midpoint?

(0,4)

(0,6)

(2.5,-5)

(2.5,5)

(3,9)

Explanation

A line segment has endpoints (0,4) and (5,6). To find the midpoint, use the midpoint formula:

X: (x1+x2)/2 = (0+5)/2 = 2.5

Y: (y1+y2)/2 = (4+6)/2 = 5

The coordinates of the midpoint are (2.5,5).

$\textup{What is the slope of a line perpendicular to the line with the equation}$

$3y+2x=6\textup{ ?}$

$\frac{3}{2}$

$\frac{-3}{2}$

$\frac{-2}{3}$

$\frac{2}{3}$

$\textup{Connot be determined with this information.}$

Explanation

$\textup{Get into }y=mx+b\textup{ format: }3y=-2x+6:,:y=-\frac{2}{3}x+2$

$\textup{Perpendicular slope is negative reciprocal of }-\frac{2}{3}: \rightarrow \rightarrow \frac{3}{2}$

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