SOLUTION: Decide whether the pair line is parallel, perpendicular or neither {{{2x+5y=5}}} {{{5x-2y=10}}}

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Question 227363: Decide whether the pair line is parallel, perpendicular or neither
2x%2B5y=5
5x-2y=10


Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
Decide whether the pair line is parallel, perpendicular or neither
2x%2B5y=5
5x-2y=10

Step 1. Two lines are parallel when they have the same slope and two lines are perpendicular when the product of their slopes is -1 or m1*m2=-1 where m1 is the slope of the first line and m2 is the slope of the line perpendicular to the first line.

Step 2. Put above pair of lines in slope-intercept form given as y=mx+b where m is the slope and b is the y-intercept when x=0 or at point (0,b).

Step 3. Let's look at 2x%2B5y=5 and put it in slope-intercept form using the following steps.

Subtract 2x from both sides of the equation

2x%2B5y-2x=5-2x

5y=-2x%2B5

Divide by 5 to both sides of the equation

5y%2F5=-2x%2F5%2B5%2F5

y=-2x%2F5%2B1

Step 4. The equation 2x%2B5y=5 in slope-intercept form is y=-2x%2F5%2B1 with the slope {{m1=-2/5) and y-intercept b=1

Step 5. Let's now look at equation 5x-2y=10 and put it in slope-intercept form.

Add 2y-10 to both sides of the equation

5x-2y%2B2y-10=10%2B2y-10

5x-10=2y

Divide by 2 to both sides of the equation

5x%2F2-10%2F2=2y%2F2

5x%2F2-5=y or y=5x%2F2-5

Step 6. The equation 5x-2y=10 in slope-intercept form is y=5x%2F2-5 with a slope m2=5%2F2 and y-intercept b=-5.

Step 7. The slopes in Step 4 and 6 are m1=-2%2F5 and m2=5%2F2. The product of their slopes m1%2Am2=%28-2%2F5%29%285%2F2%29=-1. The means the lines are perpendicular.

Step 8. ANSWER: The two given lines are perpendicular.

The graphs of these lines are shown below:

graph%28600%2C600%2C+-10%2C10%2C-10%2C10%2C+-2x%2F5%2B1%2C+5x%2F2-5%29

I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

And good luck in your studies!

Respectfully,
Dr J
http://www.FreedomUniversity.TV