Urgent Question!

Question

What is an equation of the line that passes through the point (-7,3) and is perpendicular to the line x-y=9?

Answer 1

Equation of a line can be written using any point (x₁,y₁) on the line and the slope (m) of the line in

Point-slope form:

y-y₁ = m (x-x₁)

You have the point (-7,3). Plug this in for x₁ and y₁ to get: y-3=m(x-(-7) and simplify to: y-3 =m(x+7)

Now you need the slope, m.

Perpendicular lines have opposite reciprocal slopes

Opposite reciprocals look like this: a⁄b and -b/a

We can use the equation of the perpendicular line x-y=9 to find the slope we need.

Change x-y =9 to slope -intercept form (y=mx+b):

x - y = 9

-1x -1x

-y = -1x +9

-1 -1

y = 1x - 9 Slope =1 , Opposite reciprocal slope for the perpendicular line is -1

(written as a reciprocal it's -1/1, which is just -1)

Return to the point slope equation from above y-3 =m(x+7)

and put in the slope -1 for m to get: y-3 =-1(x+7)

(You can simplify if need, but it's not necessary according the question you asked)

Answer 2

If x - y = 9, then y = x - 9. This line has slope 1.

Any line perpendicular to the line with equation y = x - 9 has slope

-1 / (slope of the line with equation x - y = 9) = -1 / 1 = -1

Equation of line with slope -1 and containing the point (-7,3) is y - 3 = -1(x - (-7)).

Simplify to get y = -x - 4