Coordinate Geometry and Lines Practice Questions — GRE General Test

Question 1Question →

Line $L$ is defined by the equation $2x + y = 6$ in the $xy$ -plane. Which of the following statements about line $L$ must be true? Select all that apply.

Select all that apply

The slope of line

L

is

-2

.

The

x

-intercept of line

L

is

(3, 0)

.

The slope of line

L

is

-\frac{1}{2}

.

A line perpendicular to line

L

has a slope of

-2

.

The

y

-intercept of line

L

is

(0, -6)

.

Show answer & explanation

Answer: The slope of line $L$ is $-2$ .; The $x$ -intercept of line $L$ is $(3, 0)$ .

Answer

The slope of line

L

is

-2

, and the

x

-intercept of line

L

is

(3, 0)

.

The given line equation

2x + y = 6

can be rewritten in slope-intercept form

y = -2x + 6

. This directly shows that the slope is

-2

. Setting

y = 0

gives

2x = 6 \implies x = 3

, so the

x

-intercept is

(3, 0)

. Therefore, both the statement that the slope is

-2

and the statement that the

x

-intercept is

(3, 0)

are correct.

Step-by-Step Solution

1

Convert the equation to slope-intercept form (

y = mx + b

).

y = -2x + 6

This isolates

y

to clearly reveal the slope

m = -2

and the

y

-intercept

(0, 6)

.

2

Find the

x

-intercept by setting

y = 0

.

2x + 0 = 6 \implies x = 3

, giving coordinate

(3, 0)

The

x

-intercept is the point where the line crosses the

x

-axis.

3

Determine perpendicular slope rules.

Perpendicular slope

= -\frac{1}{-2} = \frac{1}{2}

Perpendicular lines have negative reciprocal slopes, not identical slopes.

Key Concept

Linear equations, slope-intercept form, intercepts, and perpendicular slopes in coordinate geometry.

Estimated Time:1m 0s

Question 2Question →

In the $xy$ -plane, line $k$ has a slope of $\frac{2}{5}$ and passes through the point $(5, 8)$ . Line $m$ is perpendicular to line $k$ and has the same $y$ -intercept as line $k$ . What is the $x$ -intercept of line $m$ ?

Show answer & explanation

Answer: 2.4

Answer

2.4

To find the

x

-intercept of line

m

, first determine the equation of line

k

. Using slope-intercept form

y = mx + b

with

m = \frac{2}{5}

and point

(5, 8)

, we get

8 = \frac{2}{5}(5) + b \implies b = 6

. Thus, the

y

-intercept of line

k

(and line

m

) is

(0, 6)

. Next, line

m

is perpendicular to line

k

, so its slope is the negative reciprocal of

\frac{2}{5}

, which is

-\frac{5}{2}

. The equation for line

m

is

y = -\frac{5}{2}x + 6

. Setting

y = 0

to solve for the

x

-intercept yields

0 = -\frac{5}{2}x + 6 \implies \frac{5}{2}x = 6 \implies x = \frac{12}{5} = 2.4

.

Step-by-Step Solution

1

Find the equation and

y

-intercept of line

k

Line

k

has equation

y = \frac{2}{5}x + 6

, with

y

-intercept at

(0, 6)

.

Using point-slope form

y - y_1 = m(x - x_1)

with

m = \frac{2}{5}

and

(x_1, y_1) = (5, 8)

.

2

Calculate the slope of line

m

The slope of line

m

is

-\frac{5}{2}

.

Perpendicular lines have negative reciprocal slopes:

-\frac{1}{2/5} = -\frac{5}{2}

.

3

Construct the equation of line

m

Line

m

has equation

y = -\frac{5}{2}x + 6

.

Line

m

shares the

y

-intercept

(0, 6)

with line

k

.

4

Solve for the

x

-intercept of line

m

x = 2.4

Set

y = 0

in

y = -\frac{5}{2}x + 6

to get

0 = -\frac{5}{2}x + 6 \implies \frac{5}{2}x = 6 \implies x = 2.4

.

Key Concept

Perpendicular Slopes and Line Intercepts

Question 3Question →

In the $xy$ -plane, line $m$ passes through the point $(4, -2)$ and has a $y$ -intercept of $6$ . What is the slope of line $m$ ?

Show answer & explanation

Answer: -2

Answer

The slope of line

m

is

-2

.

The

y

-intercept of

6

indicates that the line passes through

(0, 6)

. Substituting

(0, 6)

and

(4, -2)

into the slope formula yields

m = \frac{6 - (-2)}{0 - 4} = \frac{8}{-4} = -2

.

Step-by-Step Solution

1

Identify the coordinates of two points on line

m

.

The line passes through

(4, -2)

and the

y

-intercept point

(0, 6)

.

The

y

-intercept is the point where

x = 0

.

2

Apply the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

.

Substitute

(x_1, y_1) = (4, -2)

and

(x_2, y_2) = (0, 6)

into the formula:

m = \frac{6 - (-2)}{0 - 4}

.

Slope represents the ratio of vertical change to horizontal change between two points on a line.

3

Simplify the fractional expression.

m = \frac{8}{-4} = -2

.

Dividing positive

8

by negative

4

gives

-2

.

Key Concept

Slope of a line given two points or a point and intercept

Question 4Question →

In the $xy$ -plane, line $k$ passes through the points $(2, 5)$ and $(6, 13)$ . What is the slope of line $k$ ?

2

\frac{1}{2}

-2

-\frac{1}{2}

\frac{9}{4}

Show answer & explanation

Answer: $2$

Answer

The slope of line

k

is

2

.

The slope of a straight line passing through points

(x_1, y_1)

and

(x_2, y_2)

is calculated as

m = \frac{y_2 - y_1}{x_2 - x_1}

. Substituting

(2, 5)

and

(6, 13)

gives

m = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2

.

Step-by-Step Solution

1

Identify the given points on the line.

(x_1, y_1) = (2, 5)

and

(x_2, y_2) = (6, 13)

.

These coordinates provide the required values for the slope formula.

2

Apply the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

.

m = \frac{13 - 5}{6 - 2} = \frac{8}{4}

.

The slope is defined as the change in vertical position (

y

) divided by the change in horizontal position (

x

).

3

Simplify the fraction.

m = 2

.

Dividing

8

by

4

yields the simplified slope of the line.

Key Concept

Slope of a Line in Coordinate Geometry

Estimated Time:45s

Question 5Question →

In the $xy$ -plane, line $p$ is defined by the equation $y = -\frac{3}{4}x + 3$ . Which of the following statements about line $p$ must be true? Select all that apply.

Select all that apply

Line

p

passes through the point

(4, 0)

.

Line

p

is perpendicular to any line with a slope of

\frac{4}{3}

.

The

y

-intercept of line

p

is

(0, -3)

.

Line

p

passes through Quadrant III.

The slope of line

p

is

-\frac{4}{3}

.

Show answer & explanation

Answer: Line $p$ passes through the point $(4, 0)$ .; Line $p$ is perpendicular to any line with a slope of $\frac{4}{3}$ .

Answer

The statements asserting that line

p

passes through the point

(4, 0)

and that line

p

is perpendicular to any line with a slope of

\frac{4}{3}

are correct.

Substituting

x = 4

into the line equation gives

y = 0

, showing that

(4, 0)

lies on line

p

. Additionally, the slope of line

p

is

-\frac{3}{4}

, and the negative reciprocal of

-\frac{3}{4}

is

\frac{4}{3}

, making any line with slope

\frac{4}{3}

perpendicular to line

p

.

Step-by-Step Solution

1

Identify the slope and

y

-intercept directly from the equation

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

.

The slope is

m = -\frac{3}{4}

and the

y

-intercept is

(0, 3)

.

The equation is given in standard slope-intercept form

y = mx + b

.

2

Verify point inclusion and perpendicular slope relationship.

Substituting

x = 4

yields

y = -\frac{3}{4}(4) + 3 = 0

, confirming

(4, 0)

is on the line. The negative reciprocal of

-\frac{3}{4}

is

\frac{4}{3}

, confirming the perpendicular line slope.

A point lies on a line if its coordinates satisfy the equation, and perpendicular lines have slopes that multiply to

-1

.

3

Analyze quadrant coverage.

Line

p

connects

(0, 3)

on the positive

y

-axis to

(4, 0)

on the positive

x

-axis, covering Quadrant I. For

x < 0

,

y > 3

(Quadrant II). For

x > 4

,

y < 0

(Quadrant IV). It never enters Quadrant III where both coordinates are negative.

A line with a positive

y

-intercept and negative slope crosses Quadrants I, II, and IV only.

Key Concept

Line properties in coordinate geometry including slope, intercepts, perpendicularity, and quadrant passage.

Question 6Question →

In the $xy$ -plane, line $k$ is defined by the equation $y = 3x - 4$ . Line $L$ is perpendicular to line $k$ and passes through the point $(6, 2)$ . What is the $y$ -intercept of line $L$ ?

-16

-4

4

12

20

Show answer & explanation

Answer: $4$

Answer

The

y

-intercept of line

L

is

4

.

Line

k

has a slope of

3

. Since line

L

is perpendicular to line

k

, the slope of line

L

is

-\frac{1}{3}

. Substituting the point

(6, 2)

into the slope-intercept equation

y = mx + b

gives

2 = -\frac{1}{3}(6) + b

, which simplifies to

2 = -2 + b

, so

b = 4

. Therefore, the

y

-intercept is

4

.

Step-by-Step Solution

1

Find the slope of line

L

The slope of line

L

is

-\frac{1}{3}

.

Line

k

has equation

y = 3x - 4

, so its slope is

3

. Perpendicular lines have slopes that are negative reciprocals.

2

Substitute the point

(6, 2)

and slope

-\frac{1}{3}

into the slope-intercept form

2 = -\frac{1}{3}(6) + b \implies 2 = -2 + b

The equation of a line is

y = mx + b

, where

m

is the slope and

b

is the

y

-intercept.

3

Solve for

b

b = 4

Adding

2

to both sides yields the

y

-intercept.

Key Concept

Perpendicular Line Slopes and Slope-Intercept Form

Question 7Question →

In the $xy$ -plane, line $\ell_1$ is defined by the equation $3x - 4y = 12$ . Line $\ell_2$ is perpendicular to line $\ell_1$ and intersects the positive $y$ -axis at the point $(0, d)$ . If the area of the triangular region bounded by line $\ell_1$ , line $\ell_2$ , and the $y$ -axis is $24$ , what is the value of $d$ ?

5

6

7

9

13

Show answer & explanation

Answer: $7$

Answer

The value of

d

is

7

.

The line

\ell_1

has equation

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

, placing its

y

-intercept at

(0, -3)

. Line

\ell_2

is perpendicular, so its slope is

-\frac{4}{3}

, giving the equation

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

. The vertical base of the triangle along the

y

-axis spans from

(0, -3)

to

(0, d)

, with a length of

d + 3

. The intersection of the two lines occurs at an

x

-coordinate of

\frac{12(d+3)}{25}

, which serves as the height of the triangle. Setting the area

\frac{1}{2} \times (d+3) \times \frac{12(d+3)}{25} = 24

simplifies to

(d+3)^2 = 100

. Because

d > 0

,

d + 3 = 10

, giving

d = 7

.

Step-by-Step Solution

1

Find the slope and

y

-intercept of line

\ell_1

.

Converting

3x - 4y = 12

into slope-intercept form yields

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

. The slope of

\ell_1

is

m_1 = \frac{3}{4}

and its

y

-intercept is

(0, -3)

.

Knowing the slope and

y

-intercept of

\ell_1

is essential to determine the equation of line

\ell_2

and the vertices of the triangular region along the

y

-axis.

2

Determine the equation of line

\ell_2

.

Since

\ell_2

is perpendicular to

\ell_1

, its slope is the negative reciprocal of

\frac{3}{4}

, which is

m_2 = -\frac{4}{3}

. Given that

\ell_2

intersects the

y

-axis at

(0, d)

, its equation is

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

.

Perpendicular lines have slopes that are negative reciprocals of each other.

3

Find the

x

-coordinate of the intersection point of

\ell_1

and

\ell_2

.

Set the two equations equal:

\frac{3}{4}x - 3 = -\frac{4}{3}x + d \implies \left(\frac{3}{4} + \frac{4}{3}\right)x = d + 3 \implies \frac{25}{12}x = d + 3 \implies x = \frac{12(d+3)}{25}

.

The

x

-coordinate of the intersection point represents the horizontal altitude (height) of the triangle with respect to the vertical base along the

y

-axis.

4

Express the area of the triangular region in terms of

d

and solve for

d

.

The vertical base along the

y

-axis stretches from

(0, -3)

to

(0, d)

, having length

d - (-3) = d + 3

. The height is

h = \frac{12(d+3)}{25}

. Using the area formula

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

, we set

\frac{1}{2}(d+3)\left(\frac{12(d+3)}{25}\right) = 24 \implies \frac{6(d+3)^2}{25} = 24 \implies (d+3)^2 = 100

. Since

d > 0

,

d + 3 = 10

, so

d = 7

.

The problem states that the area of the triangle is

24

.

Key Concept

Perpendicular line slopes and geometric area calculations using coordinate geometry

Estimated Time:2m 30s

Question 8Question →

In the $xy$ -plane, line $m$ is defined by the equation $ax + by = c$ , where $a$ , $b$ , and $c$ are non-zero real numbers such that $ab < 0$ and $ac > 0$ . Line $k$ is perpendicular to line $m$ and intersects line $m$ at its $x$ -intercept. Which of the following statements must be true? Select all such statements.

Select all that apply

Line

m

has a positive slope.

Line

k

passes through Quadrant II.

The

y

-intercept of line

k

is negative.

The point of intersection of line

m

and line

k

lies in Quadrant I.

Line

m

passes through Quadrant II.

Show answer & explanation

Answer: Line $m$ has a positive slope.; Line $k$ passes through Quadrant II.

Answer

The statements asserting that line

m

has a positive slope and that line

k

passes through Quadrant II must be true.

The statement regarding line

m

having a positive slope is correct because

ab < 0

implies

a

and

b

have opposite signs, making

-\frac{a}{b} > 0

. The statement asserting line

k

passes through Quadrant II is correct because line

k

possesses a negative slope

\frac{b}{a} < 0

and a positive

y

-intercept

-\frac{bc}{a^2} > 0

, ensuring it enters Quadrant II.

Step-by-Step Solution

1

Determine the slope and intercepts of line

m

.

Line

m

:

y = -\frac{a}{b}x + \frac{c}{b}

. Slope is

-\frac{a}{b} > 0

because

ab < 0

.

x

-intercept is

\left(\frac{c}{a}, 0\right)

where

\frac{c}{a} > 0

because

ac > 0

.

y

-intercept is

\left(0, \frac{c}{b}\right)

where

\frac{c}{b} < 0

because

b

and

c

have opposite signs.

Converting standard line equations to slope-intercept form exposes the signs of slopes and intercepts based on coefficient products.

2

Determine the slope, equation, and properties of line

k

.

Since line

k

is perpendicular to line

m

, its slope is the negative reciprocal of

-\frac{a}{b}

, which is

\frac{b}{a} < 0

. Line

k

passes through

\left(\frac{c}{a}, 0\right)

, giving equation

y = \frac{b}{a}\left(x - \frac{c}{a}\right) = \frac{b}{a}x - \frac{bc}{a^2}

.

Perpendicular lines have slopes whose product is

-1

.

3

Analyze quadrant coverage for both lines and verify statements.

Line

m

has positive slope and negative

y

-intercept

\implies

passes through Quadrants I, III, IV. Line

k

has negative slope and positive

y

-intercept

-\frac{bc}{a^2} > 0 \implies

passes through Quadrants I, II, IV. Intersection is at

\left(\frac{c}{a}, 0\right)

on the positive

x

-axis.

Systematic sign analysis determines quadrant trajectory and exact axis locations.

Key Concept

Properties of lines, perpendicular slopes, and sign analysis of intercepts in coordinate geometry.

Question 9Question →

In the $xy$ -plane, line $k$ passes through the point $(2, -1)$ and is perpendicular to the line $3x - 2y = 6$ . Line $m$ is parallel to line $k$ . If the distance between line $k$ and line $m$ is $\sqrt{13}$ units and line $m$ has a positive $y$ -intercept, what is the $y$ -intercept of line $m$ ?

\frac{14}{3}

\frac{17}{2}

\frac{5}{2}

4

7

Show answer & explanation

Answer: $\frac{14}{3}$

Answer

\frac{14}{3}

First, find the slope of the given line

3x - 2y = 6

, which is

\frac{3}{2}

. Because line

k

is perpendicular, its slope is

-\frac{2}{3}

. Using the point

(2, -1)

, line

k

has equation

2x + 3y - 1 = 0

. Line

m

is parallel, so it has equation

2x + 3y + C = 0

. The distance between two parallel lines

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

is given by

d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

. Setting

\frac{|C - (-1)|}{\sqrt{2^2 + 3^2}} = \sqrt{13}

yields

|C + 1| = 13

, which gives

C = 12

or

C = -14

. The

y

-intercept of line

m

is

-\frac{C}{3}

. Since the

y

-intercept must be positive,

C

must be negative, so

C = -14

. Therefore, the

y

-intercept is

-\frac{-14}{3} = \frac{14}{3}

.

Step-by-Step Solution

1

Determine the slope of line

k

.

The line

3x - 2y = 6

has slope

m_1 = \frac{3}{2}

. Because line

k

is perpendicular to it, the slope of line

k

is

m_k = -\frac{2}{3}

.

Perpendicular lines have negative reciprocal slopes.

2

Write the standard form equation of line

k

.

Using point-slope form with

(2, -1)

:

y - (-1) = -\frac{2}{3}(x - 2) \implies 2x + 3y - 1 = 0

.

Standard form

Ax + By + C_1 = 0

is required to apply the distance formula between parallel lines.

3

Set up the equation for line

m

and use the distance formula between parallel lines.

Since line

m

is parallel to line

k

, its equation is

2x + 3y + C = 0

. The distance between line

k

and line

m

is

d = \frac{|C - (-1)|}{\sqrt{2^2 + 3^2}} = \frac{|C + 1|}{\sqrt{13}} = \sqrt{13}

.

The distance between parallel lines

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

is given by

\frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

.

4

Solve for constant

C

and calculate the

y

-intercept of line

m

.

|C + 1| = 13 \implies C + 1 = 13

or

C + 1 = -13

, giving

C = 12

or

C = -14

. The

y

-intercept of

2x + 3y + C = 0

is

-\frac{C}{3}

. For a positive

y

-intercept,

C

must be negative, so

C = -14

. Thus, the

y

-intercept is

-\frac{-14}{3} = \frac{14}{3}

.

The problem states that line

m

has a positive

y

-intercept.

Key Concept

Distance between parallel lines and perpendicular slope relationships in coordinate geometry

Estimated Time:3m 0s

Question 10Question →

In the $xy$ -plane, line $\ell_1$ passes through the point $(3, 7)$ and is perpendicular to line $\ell_2$ , which is defined by the equation $2x - 5y = 10$ . Line $\ell_3$ is parallel to line $\ell_1$ and has a $y$ -intercept that is $6$ units greater than the $y$ -intercept of line $\ell_1$ . If line $\ell_3$ intersects the $x$ -axis at the point $(a, 0)$ , what is the value of $a$ ?

Show answer & explanation

Answer: 8.2

Answer

The value of

a

is 8.2 (or

\frac{41}{5}

).

Converting

2x - 5y = 10

to slope-intercept form yields

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

, so the slope of line

\ell_2

is

\frac{2}{5}

. Line

\ell_1

is perpendicular to

\ell_2

, giving it a slope of

m_1 = -\frac{5}{2}

. Using the point

(3, 7)

, the line equation for

\ell_1

is

y - 7 = -2.5(x - 3)

, which simplifies to

y = -2.5x + 14.5

. Line

\ell_3

is parallel to

\ell_1

, so

m_3 = -2.5

, and its

y

-intercept is

14.5 + 6 = 20.5

. Writing the equation for line

\ell_3

as

y = -2.5x + 20.5

and setting

y = 0

gives

0 = -2.5a + 20.5 \implies 2.5a = 20.5 \implies a = 8.2

.

Step-by-Step Solution

1

Determine the slope of line

\ell_2

The slope of line

\ell_2

is

\frac{2}{5}

.

Convert

2x - 5y = 10

into slope-intercept form

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

.

2

Determine the slope of line

\ell_1

The slope of line

\ell_1

is

-\frac{5}{2}

.

Perpendicular lines have slopes that are negative reciprocals of each other.

3

Find the

y

-intercept of line

\ell_1

The

y

-intercept of

\ell_1

is

\frac{29}{2} = 14.5

.

Apply point-slope form with point

(3, 7)

:

y - 7 = -\frac{5}{2}(x - 3) \implies y = -\frac{5}{2}x + \frac{29}{2}

.

4

Construct the equation for line

\ell_3

The equation of

\ell_3

is

y = -\frac{5}{2}x + \frac{41}{2}

.

Line

\ell_3

has the same slope as

\ell_1

(

-\frac{5}{2}

) and its

y

-intercept is

14.5 + 6 = 20.5 = \frac{41}{2}

.

5

Calculate the

x

-intercept coordinate

a

of line

\ell_3

a = 8.2

.

Substitute

y = 0

into the equation for

\ell_3

:

0 = -\frac{5}{2}a + \frac{41}{2} \implies 5a = 41 \implies a = 8.2

.

Key Concept

Perpendicular and parallel slopes, point-slope equation derivation, line transformations, and intercept determination.

Estimated Time:2m 30s

Question 11Question →

In the $xy$ -plane, line $\ell_1$ passes through the points $(2, k)$ and $(k, 14)$ , where $k$ is a constant. Line $\ell_2$ passes through the point $(k, 14)$ and has a $y$ -intercept at $(0, 22)$ . If line $\ell_1$ is perpendicular to line $\ell_2$ , what is the sum of all possible values of $k$ ?

-22

-14

-6

6

8

Show answer & explanation

Answer: -6

Answer

The sum of all possible values of

k

is -6.

The correct answer is -6 because the slope of line

\ell_1

is

m_1 = \frac{14-k}{k-2}

and the slope of line

\ell_2

is

m_2 = -\frac{8}{k}

. Since the lines are perpendicular, their slopes multiply to

-1

, yielding

\frac{8(14-k)}{k(k-2)} = 1

. Solving the resulting quadratic equation

k^2 + 6k - 112 = 0

yields

k = 8

and

k = -14

. The sum of these values is

8 + (-14) = -6

.

Step-by-Step Solution

1

Calculate the slope of line

\ell_1

in terms of

k

.

m_1 = \frac{14 - k}{k - 2}

for

k \neq 2

.

The slope of a line passing through

(x_1, y_1)

and

(x_2, y_2)

is given by

\frac{y_2 - y_1}{x_2 - x_1}

.

2

Calculate the slope of line

\ell_2

using

(k, 14)

and the

y

-intercept

(0, 22)

.

m_2 = \frac{14 - 22}{k - 0} = -\frac{8}{k}

for

k \neq 0

.

The

y

-intercept

(0, 22)

provides a second point on line

\ell_2

to find its slope.

3

Set up the perpendicularity condition

m_1 \cdot m_2 = -1

.

\left(\frac{14 - k}{k - 2}\right)\left(-\frac{8}{k}\right) = -1 \implies \frac{8(14 - k)}{k(k - 2)} = 1$.

Two non-vertical lines are perpendicular if and only if the product of their slopes equals

-1

.

4

Expand and rearrange the equation into standard quadratic form.

112 - 8k = k^2 - 2k \implies k^2 + 6k - 112 = 0.

Multiplying both sides by

k(k - 2)

clears the denominator to form a quadratic equation.

5

Factor the quadratic equation to find all possible values of

k

.

(k + 14)(k - 8) = 0 \implies k = -14 \text{ or } k = 8.

The quadratic expression factors neatly, giving two valid non-zero values for

k

.

6

Calculate the sum of all possible values of

k

.

(-14) + 8 = -6.

Summing the two solutions gives the required final numerical value.

Key Concept

Perpendicular Slopes and Quadratic Line Equations

Estimated Time:2m 30s

Question 12Question →

In the $xy$ -plane, line $L$ passes through the points $(t, t+2)$ and $(3t-1, 2t+7)$ , where $t$ is a constant. Line $M$ is perpendicular to line $L$ and passes through the point $(4, -1)$ . If the $y$ -intercept of line $M$ is $11$ , what is the value of $t$ ?

-16

-\frac{14}{5}

\frac{8}{5}

-\frac{2}{7}

14

Show answer & explanation

Answer: $-16$

Answer

-16

The slope of line M is computed from its

y

-intercept

(0, 11)

and the point

(4, -1)

as

\frac{11 - (-1)}{0 - 4} = -3

. Because line L is perpendicular to line M, the slope of line L is the negative reciprocal of

-3

, which is

\frac{1}{3}

. Calculating the slope of line L using points

(t, t+2)

and

(3t-1, 2t+7)

gives

\frac{(2t+7)-(t+2)}{(3t-1)-t} = \frac{t+5}{2t-1}

. Setting

\frac{t+5}{2t-1} = \frac{1}{3}

leads to

3t + 15 = 2t - 1

, which simplifies to

t = -16

.

Step-by-Step Solution

1

Determine the slope of line M using its given points.

Line M passes through

(4, -1)

and its

y

-intercept

(0, 11)

. The slope

m_M = \frac{11 - (-1)}{0 - 4} = \frac{12}{-4} = -3

.

The slope of a line through

(x_1, y_1)

and

(x_2, y_2)

is given by

\frac{y_2 - y_1}{x_2 - x_1}

.

2

Determine the required slope of line L.

Since line L is perpendicular to line M, its slope

m_L = -\frac{1}{m_M} = -\frac{1}{-3} = \frac{1}{3}

.

Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other.

3

Express the slope of line L in terms of

t

and solve for

t

.

Using points

(t, t+2)

and

(3t-1, 2t+7)

,

m_L = \frac{(2t+7) - (t+2)}{(3t-1) - t} = \frac{t+5}{2t-1}

. Setting

\frac{t+5}{2t-1} = \frac{1}{3}

yields

3(t+5) = 1(2t-1) \implies 3t + 15 = 2t - 1 \implies t = -16

.

Equating the algebraic slope expression to the numerical slope allows solving for the unknown parameter

t

.

Key Concept

Perpendicular Slopes and Coordinate Line Equations

Estimated Time:2m 0s

Question 13Question →

In the $xy$ -plane, line $L_1$ is given by the equation $ax + 3y = 12$ , where $a$ is a constant. Line $L_2$ is perpendicular to line $L_1$ and passes through the point $(4, -1)$ . If the $x$ -intercept of line $L_2$ is double its $y$ -intercept, and both intercepts of line $L_2$ are non-zero, what is the value of $a$ ?

-6

-\frac{3}{2}

\frac{3}{2}

3

6

Show answer & explanation

Answer: $-6$

Answer

-6

Rearranging

L_1: ax + 3y = 12

into slope-intercept form gives

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

, yielding a slope of

m_1 = -\frac{a}{3}

. The perpendicular line

L_2

has a slope of

m_2 = \frac{3}{a}

. Using the point

(4, -1)

, the equation for

L_2

is

y + 1 = \frac{3}{a}(x - 4)

, which simplifies to

y = \frac{3}{a}x - \frac{12+a}{a}

. Setting

x = 0

gives the

y

-intercept

y_0 = -\frac{12+a}{a}

, and setting

y = 0

gives the

x

-intercept

x_0 = \frac{12+a}{3}

. Equating

x_0 = 2 y_0

yields

\frac{12+a}{3} = -2\frac{12+a}{a}

. Canceling the non-zero term

(12+a)

gives

\frac{1}{3} = -\frac{2}{a}

, which leads to

a = -6

.

Step-by-Step Solution

1

Determine the slope of line

L_1

and line

L_2

Slope of

L_1

is

m_1 = -\frac{a}{3}

, so the slope of perpendicular line

L_2

is

m_2 = \frac{3}{a}

.

Perpendicular lines have slopes that are negative reciprocals of each other.

2

Write the equation of line

L_2

using point-slope form

y - (-1) = \frac{3}{a}(x - 4) \implies y = \frac{3}{a}x - \frac{12}{a} - 1 = \frac{3}{a}x - \left(\frac{12 + a}{a}\right)

.

Line

L_2

passes through

(4, -1)

with slope

m_2 = \frac{3}{a}

.

3

Find the

x

-intercept and

y

-intercept of line

L_2

The

y

-intercept is

y_0 = -\frac{12 + a}{a}

. Setting

y = 0

gives

0 = \frac{3}{a}x - \frac{12 + a}{a} \implies x_0 = \frac{12 + a}{3}

.

Intercepts occur where

x = 0

(for

y

-intercept) and

y = 0

(for

x

-intercept).

4

Set up and solve the given relation

x_0 = 2 y_0

\frac{12 + a}{3} = 2 \left( -\frac{12 + a}{a} \right)

. Since the intercepts are non-zero,

12 + a \neq 0

. Dividing both sides by

(12 + a)

gives

\frac{1}{3} = -\frac{2}{a} \implies a = -6

.

The problem states the

x

-intercept is double the

y

-intercept.

Key Concept

Perpendicular Slopes and Line Intercepts

Estimated Time:2m 0s

Question 14Question →

In the $xy$ -plane, line $m$ has an $x$ -intercept of $-6$ and a $y$ -intercept of $3$ . Line $k$ is perpendicular to line $m$ and passes through the point $(4, 9)$ . What is the $x$ -intercept of line $k$ ?

Show answer & explanation

Answer: 8.5

Answer

8.5

First, determine the slope of line

m

using the given intercepts

(-6,0)

and

(0,3)

, yielding

\frac{3-0}{0-(-6)} = \frac{1}{2}

. Since line

k

is perpendicular to line

m

, its slope is the negative reciprocal of

\frac{1}{2}

, which is

-2

. Using the point-slope equation with point

(4,9)

, the equation of line

k

is

y - 9 = -2(x - 4)

, simplifying to

y = -2x + 17

. Finding the

x

-intercept by setting

y = 0

yields

0 = -2x + 17

, giving

x = 8.5

.

Step-by-Step Solution

1

Calculate the slope of line

m

Slope of line

m

is

\frac{1}{2}

Line

m

passes through the points

(-6,0)

and

(0,3)

.

2

Determine the slope of perpendicular line

k

Slope of line

k

is

-2

Perpendicular lines have negative reciprocal slopes.

3

Derive the equation of line

k

y = -2x + 17

Use point-slope form with given point

(4,9)

and slope

-2

.

4

Solve for the

x

-intercept of line

k

x = 8.5

Set

y = 0

in the linear equation

y = -2x + 17

.

Key Concept

Perpendicular lines, slope calculation from intercepts, and line equations

Question 15Question →

In the $xy$ -plane, line $L_1$ passes through the points $(2, 5)$ and $(6, -3)$ . Line $L_2$ is perpendicular to line $L_1$ and passes through the point $(1, 2)$ . Line $L_3$ is parallel to line $L_2$ and has a $y$ -intercept that is $5$ units greater than the $y$ -intercept of line $L_2$ . What is the $x$ -intercept of line $L_3$ ?

-13

-\frac{5}{2}

\frac{9}{2}

13

15

Show answer & explanation

Answer: $-13$

Answer

The

x

-intercept of line

L_3

is

-13

.

The slope of line

L_1

is

\frac{-3 - 5}{6 - 2} = -2

. The negative reciprocal slope for line

L_2

is

\frac{1}{2}

. Using point-slope form with point

(1,2)

, line

L_2

is

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

, giving a

y

-intercept of

\frac{3}{2}

. Line

L_3

, being parallel to

L_2

, shares slope

\frac{1}{2}

and has

y

-intercept

\frac{3}{2} + 5 = \frac{13}{2}

. Finding the

x

-intercept of

y = \frac{1}{2}x + \frac{13}{2}

by setting

y=0

yields

x = -13

.

Step-by-Step Solution

1

Calculate the slope of line

L_1

The slope

m_1 = \frac{-3 - 5}{6 - 2} = \frac{-8}{4} = -2

.

The slope formula is

m = \frac{y_2 - y_1}{x_2 - x_1}

.

2

Determine the slope and equation of perpendicular line

L_2

The slope

m_2 = -\frac{1}{-2} = \frac{1}{2}

. Since

L_2

passes through

(1, 2)

, its equation is

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

, which simplifies to

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

.

Perpendicular lines have negative reciprocal slopes.

3

Determine the equation of line

L_3

The

y

-intercept of

L_2

is

\frac{3}{2}

. The

y

-intercept of

L_3

is

\frac{3}{2} + 5 = \frac{13}{2}

. Since

L_3

is parallel to

L_2

, its slope is

m_3 = \frac{1}{2}

. Thus, the equation of

L_3

is

y = \frac{1}{2}x + \frac{13}{2}

.

Parallel lines have equal slopes.

4

Find the

x

-intercept of line

L_3

Setting

y = 0

gives

0 = \frac{1}{2}x + \frac{13}{2} \implies \frac{1}{2}x = -\frac{13}{2} \implies x = -13

.

The

x

-intercept is the value of

x

when

y = 0

.

Key Concept

Slope of parallel and perpendicular lines, line equations, and intercept calculations

Question 16Question →

In the $xy$ -plane, line $K$ is defined by the equation $3x - 4y = 12$ . Line $M$ is perpendicular to line $K$ and passes through the point $(6, -1)$ . Which of the following statements regarding line $M$ must be true? Select all that apply.

Select all that apply

Line

M

has a

y

-intercept of

(0, 7)

.

Line

M

passes through the point

(-3, 11)

.

Line

M

has an

x

-intercept of

(7, 0)

.

Line

M

passes through Quadrant III.

The point of intersection between line

K

and line

M

lies in Quadrant IV.

Show answer & explanation

Answer: Line $M$ has a $y$ -intercept of $(0, 7)$ .; Line $M$ passes through the point $(-3, 11)$ .

Answer

The statements asserting that Line

M

has a

y

-intercept of

(0, 7)

and that Line

M

passes through the point

(-3, 11)

are correct.

Line

K

has a slope of

\frac{3}{4}

, making the perpendicular slope of line

M

equal to

-\frac{4}{3}

. Using the point

(6, -1)

, the equation of line

M

is

y = -\frac{4}{3}x + 7

. Evaluating the options: setting

x = 0

gives

y = 7

, confirming the

y

-intercept is

(0, 7)

; substituting

x = -3

yields

y = 11

, confirming

(-3, 11)

lies on line

M

. Both of these statements are true.

Step-by-Step Solution

1

Determine the slope of line

K

and the perpendicular slope of line

M

.

Line

K

in slope-intercept form is

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

, so its slope is

m_K = \frac{3}{4}

. The perpendicular slope for line

M

is the negative reciprocal:

m_M = -\frac{4}{3}

.

Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other.

2

Find the equation of line

M

using point-slope form with point

(6, -1)

.

y - (-1) = -\frac{4}{3}(x - 6) \implies y + 1 = -\frac{4}{3}x + 8 \implies y = -\frac{4}{3}x + 7

.

Knowing the slope and a point on the line allows determination of the line's exact linear equation.

3

Evaluate the statements using the equation of line

M

.

1.

y

-intercept: set

x = 0 \implies y = 7

, so

(0,7)

is correct.
2. Point

(-3, 11)

:

y = -\frac{4}{3}(-3) + 7 = 4 + 7 = 11

, so

(-3, 11)

is on the line.
3.

x

-intercept: set

y = 0 \implies -\frac{4}{3}x + 7 = 0 \implies x = \frac{21}{4} = 5.25

, so

(7,0)

is incorrect.
4. Quadrants: A line with negative slope and positive

y

-intercept covers Quadrants I, II, and IV only, so passing through Quadrant III is false.
5. Intersection with line

K

: set

\frac{3}{4}x - 3 = -\frac{4}{3}x + 7 \implies \frac{25}{12}x = 10 \implies x = 4.8

,

y = 0.6

, which is in Quadrant I, not Quadrant IV.

Direct algebraic verification confirms which geometric properties hold true for line

M

.

Key Concept

Perpendicular Slopes and Linear Properties in Coordinate Geometry

Question 17Question →

In the $xy$ -plane, line $k$ passes through the points $(-3, 5)$ and $(1, -3)$ . Line $m$ is perpendicular to line $k$ at line $k$ 's $x$ -intercept. What is the $y$ -intercept of line $m$ ?

\frac{1}{4}

-\frac{1}{4}

10

-1

-4

Show answer & explanation

Answer: $\frac{1}{4}$

Answer

The

y

-intercept of line

m

is

\frac{1}{4}

.

The line

k

has a slope of

-2

and an

x

-intercept of

\left(-\frac{1}{2}, 0\right)

. A line perpendicular to line

k

must have a slope of

\frac{1}{2}

. Substituting the point

\left(-\frac{1}{2}, 0\right)

into the line equation yields

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

, so the

y

-intercept is

\frac{1}{4}

.

Step-by-Step Solution

1

Calculate the slope of line

k

Slope

m_k = \frac{-3 - 5}{1 - (-3)} = \frac{-8}{4} = -2

.

The slope formula is

m = \frac{y_2 - y_1}{x_2 - x_1}

.

2

Find the equation of line

k

and determine its

x

-intercept

Line

k

equation:

y - (-3) = -2(x - 1) \implies y = -2x - 1

. Setting

y = 0

gives

0 = -2x - 1 \implies x = -\frac{1}{2}

. The

x

-intercept is

\left(-\frac{1}{2}, 0\right)

.

The

x

-intercept is the point where the line crosses the

x

-axis (

y = 0

).

3

Find the slope of line

m

Slope

m_m = -\frac{1}{m_k} = -\frac{1}{-2} = \frac{1}{2}

.

Perpendicular lines have slopes that are negative reciprocals of each other.

4

Determine the equation of line

m

and its

y

-intercept

Using point-slope form with

\left(-\frac{1}{2}, 0\right)

and slope

\frac{1}{2}

:

y - 0 = \frac{1}{2}\left(x - \left(-\frac{1}{2}\right)\right) \implies y = \frac{1}{2}x + \frac{1}{4}

. Setting

x = 0

gives

y = \frac{1}{4}

.

The

y

-intercept is the constant term

b

when written in slope-intercept form

y = mx + b

.

Key Concept

Perpendicular line slopes and intercept calculations

Estimated Time:2m 0s

Question 18Question →

In the $xy$ -plane, line $k$ is defined by the equation $3x - 4y = 12$ . Line $m$ is parallel to line $k$ , and the perpendicular distance between line $k$ and line $m$ is $5$ units. If the $y$ -intercept of line $m$ is greater than the $y$ -intercept of line $k$ , what is the $y$ -intercept of line $m$ ?

Show answer & explanation

Answer: 3.25

Answer

3.25

Rewriting line

k

as

3x - 4y - 12 = 0

shows its

y

-intercept is

-3

. Line

m

is parallel, so its equation is

3x - 4y + C = 0

. Using the formula for perpendicular distance between parallel lines

d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

, we have

\frac{|-12 - C|}{\sqrt{3^2 + (-4)^2}} = 5

. This simplifies to

|-12 - C| = 25

, giving

C = 13

or

C = -37

. Setting

x = 0

for line

m

gives

y = -\frac{C}{-4} = \frac{C}{4}

. For

C = 13

, the

y

-intercept is

\frac{13}{4} = 3.25

. Since

3.25 > -3

, this meets all criteria.

Step-by-Step Solution

1

Find the

y

-intercept of line

k

Line

k

has a

y

-intercept at

(0, -3)

.

Setting

x = 0

in

3x - 4y = 12

gives

-4y = 12 \implies y = -3

.

2

Formulate the general equation for line

m

Line

m

has the equation

3x - 4y + C = 0

.

Parallel lines share the same linear coefficients

A = 3

and

B = -4

.

3

Set up the distance formula between parallel lines

\frac{|-12 - C|}{5} = 5

The distance between

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

is

d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

.

4

Solve for constant

C

C = 13

or

C = -37

|-12 - C| = 25

yields

-12 - C = 25 \implies C = -37

and

-12 - C = -25 \implies C = 13

.

5

Determine the required

y

-intercept

y = 3.25

For

C = 13

, the

y

-intercept is

\frac{13}{4} = 3.25

, which is greater than

-3

.

Key Concept

Perpendicular distance between parallel lines and line intercept calculation

Estimated Time:2m 30s

Question 19Question →

In the $xy$ -plane, line $L$ passes through the point $(4, 2)$ and has a negative slope $m$ . Line $L$ intersects the positive $x$ -axis at point $P$ and the positive $y$ -axis at point $Q$ . If the area of triangle $OPQ$ , where $O$ is the origin, is equal to $16$ , what is the slope of line $L$ ?

-2

-1

-\frac{1}{2}

-\frac{1}{4}

\frac{1}{2}

Show answer & explanation

Answer: $-\frac{1}{2}$

Answer

The slope of line

L

is

-\frac{1}{2}

.

The line equation in point-slope form is

y - 2 = m(x - 4)

. Setting

y = 0

yields the

x

-intercept

P\left(4 - \frac{2}{m}, 0\right)

, and setting

x = 0

yields the

y

-intercept

Q(0, 2 - 4m)

. The area of right triangle

OPQ

is

\frac{1}{2}\left(4 - \frac{2}{m}\right)(2 - 4m) = 16

. Expanding this equation leads to

4m^2 + 4m + 1 = 0

, which factors as

(2m + 1)^2 = 0

, giving

m = -\frac{1}{2}

.

Step-by-Step Solution

1

Write the point-slope form of line

L

using point

(4, 2)

and slope

m

.

The equation of line

L

is

y - 2 = m(x - 4)

.

Point-slope form allows expressing intercepts directly in terms of the unknown slope

m

.

2

Find the coordinates of the

x

-intercept

P

and

y

-intercept

Q

.

Setting

y = 0

gives

x_P = 4 - \frac{2}{m}

. Setting

x = 0

gives

y_Q = 2 - 4m

.

Intersections with the axes define the base and height of right triangle

OPQ

.

3

Set up the area equation for triangle

OPQ

.

\text{Area} = \frac{1}{2} \cdot x_P \cdot y_Q = \frac{1}{2} \left(4 - \frac{2}{m}\right)(2 - 4m) = 16.

Triangle

OPQ

is a right triangle with base along the

x

-axis and height along the

y

-axis.

4

Solve the algebraic equation for slope

m

.

Expanding gives

\frac{1}{2}\left(8 - 16m - \frac{4}{m} + 8\right) = 16 \implies 16 - 16m - \frac{4}{m} = 32 \implies -16m - 16 - \frac{4}{m} = 0

. Dividing by

-4

and multiplying by

m

yields

4m^2 + 4m + 1 = 0 \implies (2m + 1)^2 = 0 \implies m = -\frac{1}{2}

.

Factoring the perfect square quadratic equation yields the single real value for

m

.

Key Concept

Coordinate Geometry: Intercepts and Area of Triangles

Alternative Method

Use the intercept form of a line

\frac{x}{a} + \frac{y}{b} = 1

. The area condition gives

ab = 32

. Substituting

(4, 2)

gives

\frac{4}{a} + \frac{2}{b} = 1

. Substituting

b = \frac{32}{a}

leads to

\frac{4}{a} + \frac{2a}{32} = 1 \implies \frac{4}{a} + \frac{a}{16} = 1 \implies a^2 - 16a + 64 = 0 \implies (a-8)^2 = 0 \implies a = 8

. Then

b = 4

, and the slope is

m = -\frac{b}{a} = -\frac{4}{8} = -\frac{1}{2}

.

Estimated Time:2m 0s

Question 20Question →

In the $xy$ -plane, line $k$ has an $x$ -intercept of $(8, 0)$ and a $y$ -intercept of $(0, 6)$ . Line $p$ is perpendicular to line $k$ and intersects line $k$ at its $y$ -intercept. What is the $x$ -intercept of line $p$ ?

-8

-\frac{9}{2}

\frac{9}{2}

8

-\frac{32}{3}

Show answer & explanation

Answer: $-\frac{9}{2}$

Answer

-\frac{9}{2}

The line

k

passes through

(8, 0)

and

(0, 6)

, giving a slope of

m_k = \frac{6 - 0}{0 - 8} = -\frac{3}{4}

. A line perpendicular to

k

must have a slope equal to the negative reciprocal of

-\frac{3}{4}

, which is

\frac{4}{3}

. Because line

p

intersects line

k

at

(0, 6)

, its

y

-intercept is also

6

, making its equation

y = \frac{4}{3}x + 6

. Setting

y = 0

gives

0 = \frac{4}{3}x + 6

, which solves to

x = -\frac{9}{2}

.

Step-by-Step Solution

1

Calculate the slope of line

k

Slope of line

k

is

m_k = \frac{6 - 0}{0 - 8} = -\frac{6}{8} = -\frac{3}{4}

The slope formula is

m = \frac{y_2 - y_1}{x_2 - x_1}

using given intercepts

(8, 0)

and

(0, 6)

.

2

Determine the slope of line

p

Slope of line

p

is

m_p = -\frac{1}{m_k} = \frac{4}{3}

Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other.

3

Write the equation of line

p

Equation of line

p

is

y = \frac{4}{3}x + 6

Line

p

intersects line

k

at its

y

-intercept

(0, 6)

, so line

p

has a

y

-intercept of

6

.

4

Find the

x

-intercept of line

p

x = -\frac{9}{2}

(or

-4.5

)

Set

y = 0

in the line equation:

0 = \frac{4}{3}x + 6 \implies \frac{4}{3}x = -6 \implies x = -6 \cdot \frac{3}{4} = -\frac{9}{2}

.

Key Concept

Perpendicular Slopes and Line Intercepts

Estimated Time:1m 30s

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