Consider two lines in a coordinate plane: Line A, represented by the equation 3x + ky = 8 … — ACT

Consider two lines in a coordinate plane: Line $A$ , represented by the equation $3x + ky = 8$ for some constant $k$ , and Line $B$ , which contains the points $(2, -1)$ and $(5, 8)$ . If Line $A$ is perpendicular to Line $B$ , what is the value of $k$ ?

9Cevap
B
-9
C
1
D
-1
E
3

Cevap

The value of

k

9

The correct answer is the value

9

. First, we find the slope of Line

B

using the coordinate points

(2, -1)

and

(5, 8)

, which gives a slope of

3

. Since Line

A

is perpendicular to Line

B

, its slope must be the negative reciprocal of

3

, which is

-\frac{1}{3}

. By rewriting the equation of Line

A

3x + ky = 8

, in slope-intercept form, we identify its slope as

-\frac{3}{k}

. Setting these two slope values equal gives

-\frac{3}{k} = -\frac{1}{3}

, which solves to

k = 9

Adım Adım Çözüm

Calculate the slope of Line

B

using the slope formula

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

with the points

(2, -1)

and

(5, 8)

The slope of Line

B

m_B = \frac{8 - (-1)}{5 - 2} = \frac{9}{3} = 3

Finding the slope of the given line is necessary to determine the perpendicular slope.

Determine the perpendicular slope of Line

A

using the relationship

m_A = -\frac{1}{m_B}

The slope of Line

A

must be

m_A = -\frac{1}{3}

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

Express the slope of Line

A

in terms of

k

by rewriting

3x + ky = 8

in slope-intercept form (

y = mx + b

Subtracting

3x

from both sides gives

ky = -3x + 8

, and dividing by

k

yields

y = -\frac{3}{k}x + \frac{8}{k}

. The slope is

-\frac{3}{k}

This allows us to set up an equation to solve for

k

Equate the slope of Line

A

to the perpendicular slope found in Step 2 and solve for

k

-\frac{3}{k} = -\frac{1}{3} \Rightarrow \frac{3}{k} = \frac{1}{3} \Rightarrow k = 9

Solving this equation gives the value of

k

that makes the lines perpendicular.

Anahtar Kavram

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

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