The line that contains the point Q(1,-2) and is parallel to the line whose equation is y-4=2/3(x-3)
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Answer:
given,
line y-4=2/3(x-3)
=>2x-3y+6=0
parallel line to the above equation is
2x-3y+k=0 is contains point Q(1,-2)
=>2(1)-3(-2)+k=0
=>2+6+k=0
=>k=-8
there fore
the required line equations is 2x-3y-8=0
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