Write an equation of the line that passes through (9,6) and is perpendicular to the line whose equation is y = -1/3x +7
y-y1=m(x-x1)
y-6=3(x-9)
y-6=3x-27
y=3x-21
that WAS my work. those 4 lines
my choices are:
A)y=-1/3x+9
B)y=-3x+33
C)y=3x-21
D)y=1/3x+3
You got it right. Thanks for showing you work!
Nice job
To find the equation of a line that is perpendicular to a given line, we need to determine the slope of the given line and then find its negative reciprocal.
The slope of the given line is -1/3. The negative reciprocal of -1/3 is 3.
Since the line we want to find passes through the point (9,6), we can use the point-slope form of a line to write its equation.
The point-slope form is: y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope. Substituting the values we know, we get:
y - 6 = 3(x - 9)
Now, we can simplify this equation:
y - 6 = 3x - 27
y = 3x - 27 + 6
y = 3x - 21
So, the equation of the line that passes through (9,6) and is perpendicular to the line y = -1/3x + 7 is y = 3x - 21.