Perpendicular and parallel equations

S

supremacy32

New member
Joined
Aug 29, 2015
Messages
9
I feel like I have these figured out but would like someone to check my work if possible. 2nd guessing Myself on several homework problems.

Give answers in slope-intercept form:
Give the equation of the line that is parallel to 4y=-3x+2 and has a y intercept of 10.

y= (-3/4)x +2/4
y-10=(-3/4)(x-0)
y-10=(-3/4)x
Solution: y=(-3/4)x+10

Give the equation of the line that is perpendicular to 4y=-3x+2 and passes through the point (10,12).

y= (-3/4)x +2/4
m1=-(1/m2)
m=-4/3
y-12=(-4/3)(x-10)
y=(-4/3)x-(40/3)+12
Solution: y=(-4/3)x-(4/3)

Please let me know if this is correct, or if not, where I went wrong and what I can do to fix it. Thanks in advance!
 
Last edited:
supremacy32 said:
I feel like I have these figured out but would like someone to check my work if possible. 2nd guessing Myself on several homework problems.

Give answers in slope-intercept form:
Give the equation of the line that is parallel to 4y=-3x+2 and has a y intercept of 10.

y= (-3/4)x +2/4
y-10=(-3/4)(x-0)
y-10=(-3/4)x
Solution: y=(-3/4)x+10

Give the equation of the line that is perpendicular to 4y=-3x+2 and passes through the point (10,12).

y= (-3/4)x +2/4 m1=-(1/m2) m=-4/3
y-12=(-4/3)(x-10)
y=(-4/3)x-(40/3)+12
Solution: y=(-4/3)x-(4/3)

Please let me know if this is correct, or if not, where I went wrong and what I can do to fix it. Thanks in advance!
Click to expand...
Correct - both. [No .... the second problem has some arithmatic mistakes - corrected below]
 
Last edited by a moderator:
Actually, only the first part about parallel lines is correct. But I did notice one small error in the perpendicular lines part. When dealing with problems of this nature, the best way to check your work is to simply graph your solution. I did that and found a problem. This graph shows the original line and your solution. The original line you were given is plotted in red. The point your line is supposed to go through, (10,12) is the green dot. Your solution is plotted in blue. Notice how your lines are perpendicular, but your blue line goes nowhere near (10,12)? Now how do you suppose you can modify the line such that it is still perpendicular, but passes through (10,12)?

As a hint, recall the definition of perpendicular lines. A line is perpendicular to another line if their slopes are the reciprocals of one another. But what if the lines' slopes were negative reciprocals of one another? Would they still be perpendicular?
 
supremacy32 said:
I feel like I have these figured out but would like someone to check my work if possible. 2nd guessing Myself on several homework problems.

Give answers in slope-intercept form:
Give the equation of the line that is parallel to 4y=-3x+2 and has a y intercept of 10.

y= (-3/4)x +2/4
y-10=(-3/4)(x-0)
y-10=(-3/4)x
Solution: y=(-3/4)x+10

Give the equation of the line that is perpendicular to 4y=-3x+2 and passes through the point (10,12).

y= (-3/4)x +2/4
m1=-(1/m2)
m= + 4/3
y-12=(4/3)(x-10)
y=(4/3)x-(40/3)+12
Solution: y=(4/3)x - (4/3)

Please let me know if this is correct, or if not, where I went wrong and what I can do to fix it. Thanks in advance!
Click to expand...

I missed some algebraic errors -so I'll correct those! (Denis move over... make space)
 
You must log in or register to reply here.
Top