In the xy-plane, line k is defined by the equation 3x - 4y = 12. Line m is parallel to lin… — GRE General Test

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 $m$ ?

Answer: 3.25

Answer

3.25

Rewriting line

k

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

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

Find the

y

-intercept of line

k

Line

k

has a

y

-intercept at

(0, -3)

Setting

x = 0

3x - 4y = 12

gives

-4y = 12 \implies y = -3

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

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

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

Solve for constant

C

C = 13

C = -37

|-12 - C| = 25

yields

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

and

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

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

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