In the xy-plane, line k passes through the point (2, -1) and is perpendicular to the line … — GRE General Test

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 $m$ ?

$\frac{14}{3}$ Cevap
B
$\frac{17}{2}$
C
$\frac{5}{2}$
D
4
E
7

Cevap

\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

C = -14

. The

y

-intercept of line

m

-\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}

Adım Adım Çözüm

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

m_k = -\frac{2}{3}

Perpendicular lines have negative reciprocal slopes.

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.

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

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}}

Solve for constant

C

and calculate the

y

-intercept of line

m

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

C + 1 = -13

, giving

C = 12

C = -14

. The

y

-intercept of

2x + 3y + C = 0

-\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.

Anahtar Kavram

Distance between parallel lines and perpendicular slope relationships in coordinate geometry

Tahmini Süre:3m 0s

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