How do you write an equation of a line passing through (-2, -2), perpendicular to #y=x+1#?

1 Answer
Apr 14, 2017

Answer:

See the entire solution process below:

Explanation:

Because the equation given in the problem is in slope-intercept form we can easily determine the slope. 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.

#y = color(red)(1)x + color(blue)(1)#

Therefore the slope of the line from the problem is: #color(red)(m = 1)#

Let's call the slope of a perpendicular line #m_p#.

The slope of a perpendicular line is: #m_p = -1/m#

Substituting #1# for #m# gives the slope of the perpendicular line in this problem as:

#m_p = -1/1 = -1#

Because we are given a point we can use the point-slope formula to find an equation for the 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 slope we calculated and the point from the problem gives:

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

Solution 1) #(y + color(red)(2)) = color(blue)(-1)(x + color(red)(2))#

We can also substitute the values from the problem and the slope we calculated into the slope-intercept formula and solve for #b# to find the equation in slope-intercept form:

#-2 = (color(red)(-1) xx -2) + color(blue)(b)#

#-2 = 2 + color(blue)(b)#

#color(red)(-2) - 2 = color(red)(-2) - 2 + color(blue)(b)#

#-4 = 0 + color(blue)(b)#

#-4 = color(blue)(b)#

#y = color(red)(-1)x - color(blue)(4)#

Solution 2) #y = color(red)(-)x - color(blue)(4)#