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Question 71451This question is from textbook Algebra 1 California Edition
: Chapter 7-8 Parallel and Perpendicular Lines
Introducing the Concept: Slope and Line Relationships
Here's my question:
"Write an equation for the line containing the given point and perpendicular to the given line."
27. (0,6); y-3x=4
28. (-2,4); y=2x-3
29. (0,2); 3y-x=0
My son just started homeschooling and I am at a complete loss as to how to provide him with help on this part of the chapter.
Just an FYI: Chapter 7 "Graphs and Linear Equations" 7-8 Exercises Part B #27, #28, #29
Thank you for any help you can provide us!!!!
This question is from textbook Algebra 1 California Edition
Answer by checkley75(3666)   (Show Source):
You can put this solution on YOUR website! Y-3X=4
Y=3X+4 HAS A SLOPE OF 3 SO THE PERPENDICULAR LINE HAS A SLOE OF THE NEGATIVE RECIPROCAL OF 3 WHICH IS -1/3. NOW USIE THE X,Y VALUES TO SOLVE Y=mX+b FOR THE Y INTERCEPT (b).
6=-1/3*0+b
6=b SO THE LINE WITH A SLOPE OF -1/3 PASSING THROUGH (0,6) IS
Y=-1/3X+6
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Y=2X-3 HAS A SLOPE OF 2 THUS THE PERPENDICULAR SLOPE IS -1/2
4=-1/2*-2+b
4+1=b
5=b SO THE LINE WITH A SLOPE OF -1/2 PASSING THROUGH (-2,4) IS
Y=-1/2X+5
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3Y-X=0
3Y=X
Y=X/3 HAS A SLOPE OF 1/3 THUS THE PERPENDICULAR SLOPE IS -3
2=-3*0+b
2=b SO THE LINE WITH A SLOPE OF -3 PASSING THROUGH (0,2) IS
Y=-3X+2
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