Find the equation of the line that is perpendicular to the line
y = x-9 and passes through the point (7,9).

Linear equations are typically organized in slope-intercept form: [tex]y=mx+b[/tex] where m is the slope and b is the y-intercept (the value of y when x is 0)
Perpendicular lines always have slopes that are negative reciprocals (ex. 3 and -1/3, 5/6 and -6/5, etc.)

1) Determine the slope (m)

y=x-9

Rewrite the equation

y=1x-9

Now, we can identify clearly that the slope of the line is 1. The negative reciprocal of 1 is -1, so therefore, the slope of a perpendicular line would be -1. Plug this into [tex]y=mx+b[/tex]:

[tex]y=-1x+b\\y=-x+b[/tex]

2) Determine the y-intercept (b)

[tex]y=-x+b[/tex]

Plug in the given point (7,9) and solve for b

[tex]9=-7+b[/tex]

Add 7 to both sides to isolate b

[tex]9+7=-7+b+7\\16=b[/tex]

Therefore, the y-intercept is 16. Plug this back into [tex]y=-x+b[/tex]:

[tex]y=-x+16[/tex]

I hope this helps!

Find the equation of the line that is perpendicular to the line y = x-9 and passes through the point (7,9).

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Find the equation of the line that is perpendicular to the line
y = x-9 and passes through the point (7,9).