What is the equation of the line perpendicular to #y=-7/5 # that passes through # (-35,5) #?

2 Answers
Mar 5, 2016

Answer:

#x=-35#

Explanation:

Firstly, let's go over what we already know from the question. We know that the #y#-#"intercept"# is #-7/5# and that the slope, or #m#, is #0#.

Our new equation passes through #(-35,5)#, but the slope will not change since 0 is neither positive nor negative. This means that we need to find the #x-"intercept"#. So, our line will be passing through vertically, and have a undefined slope (we don't have to include #m# in our equation).

In our point, #(-35)# represents our #x-"axis"#, and #(5)# represents our #y-"axis"#. Now, all we have to do is pop the #x-"axis"# #(-35)#into our equation, and we're done!

The line that is perpendicular to #y=−7/5# that passes through #(−35,5)# is #x=-35#.

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Here's a graph of both lines.

Mar 5, 2016

Answer:

solution is, # x+35=0#

Explanation:

#y=-7/5# represents a straight line parallel to x-axis lying at a distance #-7/5# unit from x-axis.
Any straight line perpendicular to this line should be parallel to y-axis and can be represented by the equation #x=c# ,where c = a constant distance of the line from y-axis.
Since the line whose equation to be determined passes through(-35,5) and is parallel to y-axis, it will be at a distance -35 unit from y-axis. Hence its equation should be #x=-35=>x+35=0#