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Question 88068: write an equation of a line that is parallel to y=-6x+4 and passes through(2,10).
write an equation of line which is perpendicular to y=(-2/3)x-5 and passes through point (6,3).
write the equation of three lines so that the y-intercept of each is -3.
Answer by checkley75(3666)   (Show Source):
You can put this solution on YOUR website! Y=-6X+4 NOW WE USE THE SLOPE(-6) BECAUSE THE LINE WE NEED IS PARALLEL TO THIS LINE THUS THEY HAVE THE SAME SLOPE AND THE X/Y COORDINATES (2,10) TO FIND THE Y INTERCEPT OF THIS LINE THUS:
2=-6*2+b
2=-12+b
b=2+12
b=14 THE Y INTERCEPT.
THIS GIVES US THE LINE EQUATION OF:
Y=-6X+14 ANSWER.
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Y=-2X/3-5 THIS LINE HAS A SLOPE OF (-2/3) THUS THE LINE PERPENDICULAR TO THIS LINE MUST HAVE A SLOPE OF (3/2) THE NEGATIVE RECIPRICAL.
NOW USING THE POINTS(6,3) WE USE THE LINE FORMULA Y=mX+b TOP FIND THE Y INTERCEPT(b) THUS:
3=3/2*6+b
3=9+b
b=3-9
b=-6 SO THE LINE EQUATION OF THIS PERPENDICULAR LINE THROUGH (6,3) IS:
Y=2X/3-6 ANSWER.
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Y=2X-3
Y=-5X/7-3
Y=18X/35-3
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