Line l is perpendicular to x+3y-3=0 and it passes through 1,7
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Answer: x- 3y = 3
⇒(-x/3) + 1 = y [∵y=mx+c]
∴m of given line=(-1/3)
M of required line= -1/(-1/3)= 3
By slope point form,
y-y₁= m [ x- x₁]
y-7= (3)[x-1]
y-7=3x-3
∴ Equation of the line 3x-y+4=0
P.S.: m denotes slope of x+3y-3=0 while m denotes slope of the required line.
Hope that helps!
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