Perpendicular lines crossing at a 90 degree angle. Because that this to be true, one of two people one line is horizontal (e.g., y = 5) and also the various other is vertical (e.g., x = 4) or the slopes are an adverse reciprocals that each other (e.g., 3 and also -1/3).

You are watching: Difference between perpendicular and intersecting lines

y = 2x and y = 3x intersect at the origin. (In both cases, if x = 0, climate y = 0.) The slopes space not an adverse reciprocals, so these lines space not perpendicular.

The 2nd pair that equations can be rewritten in slope-intercept form as y = -6x + 16 and also y = 5x + 20. The slopes space not equal, so we understand the lines room not paralleland intersect somewhere. Ther slopes are not an unfavorable reciprocals, so we know they room not perpendicular. The solution is (-4/11, 200/11).

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Intersecting lines re-superstructure a allude in common, referred to as their intersection point.

Perpendicular lines room intersecting lines that are at ideal angles come each-other.

For lines in the x-y plane, every vertical (x = constant) line is perpendicular to every horizontal (y = constant) line. For any type of pair of lines with non-zero slopes, say y=m1x+b1 andy=m2x+b2, the precise criterion is the m1m2 = -1 (slopes are an adverse reciprocals that each-other).

Here is the systems to the an initial system y = 2x, 6x + y = 16

You can substitute the very first equation into the 2nd as follows

6x + y = 16

6x + 2x = 16 Substitution.

8x = 16 combining the x-terms.

x = 16/8 = 2 department by 8 top top both sides.

y = 2x = 2*2 = 4 Substitution

So the lines room intersecting v the intersection suggest at (2,4).

Also since isolating y in the second equation provides y = -6x + 16 and -6*2 = -12≠ -1, they space not perpendicular.

Now watch if you can do this through your second system y = 3x, y – 5x = 20.

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