Homework help please?

Question

Write an equation for the line that is parallel to the given line and that passes through the given point. Write the equation in slope-intercept form.

y = -2/3x + 12 ; (6,-3)

a.y = 3/2x + 1
b.y = -3/2x + 7
c.y = -2/3x + 7
d. y = -2/3x + 1

Answer 1

Slope-intercept form line equation is y = mx + c, where m is slope and c is y-intercept.

comapre the equation with slope-intercept form and find the slope. m = -2/3.

 

Parallel lines has same slope, so slope of the line is -2/3.

THe line equation is y = -2/3x + c

Find the y-intercept value by substituting (6, -3) in the equation.

-3 = -2/3*6 + c

-3 = -4 + c

c = 1

 

The line equation is y = -2/3x + 1

Option D is right choice.

Answer 2

The equation  is y = (-2/3)x + 12 and point is (6,-3)

Above equation write in slope - intercept form y = m₁x + b.

y = (-2/3)x + 12

Compare the equation with slope - intercept form y = m₁x + b.

So, m1 = -2/3 and b = 12

Slopes of parallel lines are equal.

So, a line parallel to it has slope of m₁ = m₂ = -2/3.

You know the slope and a point on the line, so use the point - slope form with (x₁ ,y₁ ) = (6, -3) to write equation of the line.

(y - y₁ ) = m₂(x - x₁ )

(y - (-3)) = -2/3(x - 6)

Apply distributive property: a(b - c) = ab - ac.

y + 3 = (-2/3)x - (-2/3)6

y + 3 = (-2/3)x + 4

Subtract 3 from each side

y + 3 - 3 = (-2/3)x +  4 - 3

y  = (-2/3)x + 1

The equation is y  = (-2/3)x + 1.