How do you write an equation of a line that contains the given point (-5,5) and is perpendicular to the given line y=-5x+9?

To find the equation of the line perpendicular to the given line and going through the given point we will use the point-slope formula. We have been given a point so what we are missing is the slope.

The given line is in slope-intercept form.

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.

Therefore we know the slope of the given line, #color(red)(m)# is #color(red)(-5)# - it is the coefficient of the #x# term.

A perpendicular line will have a slope which is the negative inverse of this line, or #color(red)(m = - 1/-5 = 1/5#.

We can now use the point-slope formula to write the equation for the perpendicular line.

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.

Substituting the point we were given and the slope we calculated gives:

#(y - color(red)(5)) = color(blue)(1/5)(x - color(red)(-5))#

#(y - color(red)(5)) = color(blue)(1/5)(x + color(red)(5))#

or, we can solve for #y# to convert to the more familiar slope-intercept form:

#y - color(red)(5) = color(blue)(1/5)x + (color(blue)(1/5) xx color(red)(5))#

#y - color(red)(5) = color(blue)(1/5)x + 1#

#y - color(red)(5) + 5 = color(blue)(1/5)x + 1 + 5#

#y - 0 = color(blue)(1/5)x + 6#

#y = color(blue)(1/5)x + 6#