Perpendicular lines crossing at a 90 level angle. Because that this to it is in true, one of two people one heat 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 -1/3).

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

The 2nd pair the equations can be rewritten in slope-intercept form as y = -6x + 16 and y = 5x + 20. The slopes space not equal, for this reason we recognize the lines room not paralleland crossing somewhere. Ther slopes space not an adverse reciprocals, for this reason we recognize they are not perpendicular. The systems is (-4/11, 200/11).

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Intersecting lines re-superstructure a suggest in common, dubbed their intersection point.

Perpendicular lines space intersecting lines that space at best angles to 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 that lines with non-zero slopes, to speak y=m1x+b1 andy=m2x+b2, the specific criterion is the m1m2 = -1 (slopes are an adverse reciprocals that each-other).

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

You deserve to substitute the first equation into the 2nd as follows

6x + y = 16

6x + 2x = 16 Substitution.

8x = 16 combine the x-terms.

x = 16/8 = 2 division by 8 on both sides.

y = 2x = 2*2 = 4 Substitution

So the lines space intersecting v the intersection point at (2,4).

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

Now view if you deserve to do this with your 2nd system y = 3x, y – 5x = 20.

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