In the standard (x, y) coordinate plane, a line L_1 is perpendicular to the line that cont… — ACT

In the standard $(x, y)$ coordinate plane, a line $L_1$ is perpendicular to the line that contains the points $(3, 5)$ and $(-1, 8)$ . If $L_1$ is also parallel to the line defined by the equation $ax + 6y = 15$ , what is the value of the constant $a$ ?

Cevap: -8

Cevap

The value of the constant

a

-8

The slope of the line containing

(3, 5)

and

(-1, 8)

-\frac{3}{4}

. The slope of a line perpendicular to it is the negative reciprocal, which is

\frac{4}{3}

. Because line

L_1

is parallel to the line

ax + 6y = 15

, they must have equal slopes. The slope of

ax + 6y = 15

-\frac{a}{6}

. Setting the two slopes equal gives

-\frac{a}{6} = \frac{4}{3}

, which yields

a = -8

Adım Adım Çözüm

Calculate the slope of the line containing the points

(3, 5)

and

(-1, 8)

The slope is

-\frac{3}{4}

Using the slope formula

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

for the points

(3, 5)

and

(-1, 8)

, we get

m = \frac{8 - 5}{-1 - 3} = \frac{3}{-4} = -\frac{3}{4}

Determine the slope of line

L_1

using the perpendicular relationship.

The slope of

L_1

\frac{4}{3}

Since line

L_1

is perpendicular to the line with slope

-\frac{3}{4}

, its slope must be the negative reciprocal, which is

-\frac{1}{-3/4} = \frac{4}{3}

Express the slope of the line

ax + 6y = 15

in terms of

a

The slope is

-\frac{a}{6}

Rewriting the equation

ax + 6y = 15

in slope-intercept form (

y = mx + b

) gives

6y = -ax + 15

, which simplifies to

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

. The slope is the coefficient of

x

, which is

-\frac{a}{6}

Set the slope of

L_1

equal to the slope of the parallel line to solve for

a

a = -8

Because line

L_1

is parallel to the line

ax + 6y = 15

, their slopes are equal:

-\frac{a}{6} = \frac{4}{3}

. Multiplying both sides by

-6

gives

a = -8

Anahtar Kavram

Parallel lines have equal slopes, and perpendicular lines have slopes that are negative reciprocals of each other.

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