If you wanted to make the graph of y=5x+3 les steep, which equation could you use?
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Answer:
Find another point on the perpendicular line.
Step-by-step explanation:
Given an original line "m", and a point off the line "Q", in order to construct a second line "p", meant to be perpendicular to "m" through the point "Q", fundamentally, the only truly necessary step to construct a perpendicular line through is to find another point on the yet-to-be-found perpendicular line.
Most often, this is accomplished by exploiting the fact that "p" is the set of all points that are equidistant from any pair of points that are symmetric about "p".
Since the symmetry must be about "p", and we don't even know where "p" is, one often finds two points on "m" that are equidistant from "Q".
This can be accomplished by adjusting a compass to a fixed radius (larger than the distance from "Q" to "m"), and making an arc that intersects "m" in two places. Those two places will be equidistant from "Q", and are simultaneously on line "m". Thus, these two points, "A" & "B" are symmetric about "p".
Since "A" & "B" are symmetric about "p", they are equidistant from "p", and are on "m". One could try to find the point of intersection between "p" and "m" through construction, but this is unnecessary. We need only find a second point (besides "Q") that is equidistant from "A" & "B", which will necessarily be a point on "p", to form the line perpendicular to "m".
To do this, fix the compass with any radius, and from "A" make a large arc generally in the direction of "B", and make the same radius arc from "B" in the direction of "A" such that the two arcs intersect at some point that isn't "Q". This point of intersection we can call point "T", and the line QT is line "p", the line perpendicular to the original line, necessarily containing "Q".
