Solve and check the following systems.

1) y = 2x 2) y = 3x

6x + y = 16 y – 5x = 20

Solve and check the following systems.

1) y = 2x 2) y = 3x

6x + y = 16 y – 5x = 20

Tutors, please sign in to answer this question.

Perpendicular lines intersect at a 90 degree angle. For this to be true, either one line is horizontal (e.g., y = 5) and the other is vertical (e.g., x = 4) or the slopes are negative reciprocals of each other (e.g., 3 and -1/3).

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

The second pair of equations can be rewritten in slope-intercept form as y = -6x + 16 and y = 5x + 20. The slopes are not equal, so we know the lines are not parallel and intersect somewhere. Ther slopes are not negative reciprocals, so we know they are not perpendicular. The solution is (-4/11, 200/11).

Intersecting lines share a point in common, called their intersection point.

Perpendicular lines are intersecting lines that are at right 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 pair of lines with non-zero slopes, say y=m_{1}x+b_{1} and y=m_{2}x+b_{2}, the precise criterion is that m_{1}m_{2} = -1 (slopes are negative reciprocals of each-other).

Here is the solution to the first system {y = 2x, 6x + y = 16}

You can substitute the first equation into the second as follows

6x + y = 16

6x + 2x = 16 Substitution.

8x = 16 Combining the x-terms.

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

y = 2x = 2*2 = 4 Substitution

So the lines are intersecting with the intersection point at (2,4).

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

Now see if you can do this with your second system {y = 3x, y – 5x = 20}.