In the xy-plane, line l passes through the points (-2, -3) and (2, 5). Line k is parallel … — SAT

In the $xy$ -plane, line $l$ passes through the points $(-2, -3)$ and $(2, 5)$ . Line $k$ is parallel to line $l$ and has a $y$ -intercept of $(0, -1)$ . If the point $(a, 9)$ lies on line $k$ , what is the value of $a$ ?

Cevap: 5

Cevap

The value of

a

5

To find the value of

a

, we first determine the slope of line

l

using the two given points,

(-2, -3)

and

(2, 5)

. The slope

m

is given by

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

. Since line

k

is parallel to line

l

, it has the same slope of

2

. The

y

-intercept of line

k

(0, -1)

, so the equation of line

k

y = 2x - 1

. To find the value of

a

, we substitute the point

(a, 9)

into this equation:

9 = 2a - 1

. Solving for

a

gives

10 = 2a

, which simplifies to

a = 5

Adım Adım Çözüm

Calculate the slope of line

l

using the two given points.

The slope of line

l

2

Using the slope formula

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

with the points

(-2, -3)

and

(2, 5)

, we get

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

Find the equation of line

k

The equation of line

k

y = 2x - 1

Line

k

is parallel to line

l

, so it has the same slope,

m = 2

. Its

y

-intercept is

(0, -1)

, which gives the equation

y = 2x - 1

in slope-intercept form.

Solve for

a

by substituting the point

(a, 9)

into the equation of line

k

The value of

a

5

Substituting

x = a

and

y = 9

into

y = 2x - 1

gives

9 = 2a - 1

. Solving for

a

yields

10 = 2a

, or

a = 5

Anahtar Kavram

Linear Functions and Graphs

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