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

2 Answers
Mar 5, 2016

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.

enter image source here

Here's a graph of both lines.

Mar 5, 2016

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