What is the an equation of the line that goes through (−1, −3) and is perpendicular to the line #2x + 7y + 5 = 0#?

1 Answer
Dec 3, 2015

#y = 7/2 x + 1/2#

Explanation:

Let's write the equation of the line #2x + 7y + 5 = 0# in standard form first.
To do so, solve the equation for #y#:

#color(white)(xxx) 2x + 7y + 5 = 0#

#<=> 7y = -2x - 5#

#<=> y = -2/7 x - 5/7#

Now, we know that #m_1 = -2/7# is the slope of this line.

To make the line that you need to construct perpendicular to the given line, its slope #m# must fulfill the condition #m = -1/m_1#, so the slope of the line that you need to create is

#m = 7/2#.

Now that you have the slope #m# and your point #(-1, -3)#, you can compute the equation of the line.

The standard form is

#y = mx + n#.

Since you know that the point #(-1, -3)# is on your line, #x = -1# and #y = -3# must fulfill this equation.
Thus, plug #x#, #y# and #m# and solve for #n#:

#-3 = 7/2 * (-1) + n#

#<=> n = 1/2#

The equation of your line is #y = 7/2 x + 1/2#.