MATH SOLVE

4 months ago

Q:
# What is the equation, in slope-intercept form, of the line that is perpendicular to the line y – 4 = –(x – 6) and passes through the point (−2, −2)?

Accepted Solution

A:

That line is in point-slope form:

[tex]\sf y-y_1=m(x-x_1)[/tex]

Where 'm' is the slope and (x1, y1) is a point on the line.

Perpendicular lines have opposite slopes. To get the opposite, we take the reciprocal and multiply it by -1.

[tex]\sf -1[/tex]

Reciprocal:

[tex]\sf -\dfrac{1}{1}\rightarrow -\dfrac{1}{1}\rightarrow -1[/tex]

Multiply by -1:

[tex]\sf 1[/tex]

We can plug this slope and the point into point-slope form, and then convert it to slope-intercept form.

[tex]\sf y-y_1=m(x-x_1)[/tex]

[tex]\sf y-(-2)=1(x-(-2))[/tex]

Negatives cancel out:

[tex]\sf y+2=1(x+2)[/tex]

Distribute 1 into the parenthesis:

[tex]\sf y+2=x+2[/tex]

Subtract 2 to both sides:

[tex]\boxed{\sf y=x}[/tex]

[tex]\sf y-y_1=m(x-x_1)[/tex]

Where 'm' is the slope and (x1, y1) is a point on the line.

Perpendicular lines have opposite slopes. To get the opposite, we take the reciprocal and multiply it by -1.

[tex]\sf -1[/tex]

Reciprocal:

[tex]\sf -\dfrac{1}{1}\rightarrow -\dfrac{1}{1}\rightarrow -1[/tex]

Multiply by -1:

[tex]\sf 1[/tex]

We can plug this slope and the point into point-slope form, and then convert it to slope-intercept form.

[tex]\sf y-y_1=m(x-x_1)[/tex]

[tex]\sf y-(-2)=1(x-(-2))[/tex]

Negatives cancel out:

[tex]\sf y+2=1(x+2)[/tex]

Distribute 1 into the parenthesis:

[tex]\sf y+2=x+2[/tex]

Subtract 2 to both sides:

[tex]\boxed{\sf y=x}[/tex]