What is the equation of the line perpendicular to #y =3x- 7# that contains (6, 8)?

1 Answer
Dec 29, 2016

Answer:

#(y - 8) = -1/3(x - 6)#

or

#y = -1/3x + 10#

Explanation:

Because the line given in the problem is in the slope intercept form we know the slope of this line is #color(red)(3)#

The slope-intercept form of a linear equation is:

#y = color(red)(m)x + color(blue)(b)#

Where #color(red)(m)# is the slope and #color(blue)(b# is the y-intercept value.
This is a weighted average problem.

Two perpendicular lines have a negative inverse slope of each other.

The line perpendicular to a line with slope #color(red)(m)# has a slope of #color(red)(-1/m)#.

Therefore, the line we are looking for has a slope of #color(red)(-1/3)#.

We can now use the point-slope formula to find the equation of the line we are looking for.

The point-slope formula states: #(y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))#

Where #color(blue)(m)# is the slope and #color(red)(((x_1, y_1)))# is a point the line passes through.

We can substitute the slope we calculate and the point we were given to give the equation we are looking for:

#(y - color(red)(8)) = color(blue)(-1/3)(x - color(red)(6))#

If we want to put this in slope-intercept form we can solve for #y#:

#y - color(red)(8) = color(blue)(-1/3)x - (color(blue)(-1/3) xx color(red)(6)))#

#y - color(red)(8) = color(blue)(-1/3)x - (-2)#

#y - color(red)(8) = color(blue)(-1/3)x + 2#

#y - color(red)(8) + 8 = color(blue)(-1/3)x + 2 + 8#

#y - 0 = color(blue)(-1/3)x + 10#

#y = -1/3x + 10#