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x-3y=4
slope=1/3
the equation of the line
y-y1 =m (x-x1)
y+3=1/3( x-5)
3y+9=x-5
3y- x+14= 0 ...........is the required equation
hope it helps
slope=1/3
the equation of the line
y-y1 =m (x-x1)
y+3=1/3( x-5)
3y+9=x-5
3y- x+14= 0 ...........is the required equation
hope it helps
Answered by
1
given eqn is
3y=x-4
y =[1/3]x -[4/3]
we know that Y =mx+b
hence slope = m°= 1/3
since both lines are parallel, hence
m=m°=1/3=slope of another line
to find ewuation of line with slope m and passing through point (5,-3)
(Y-Y°)=m(X-X°)
given Y°= -3 and X° = 5„ m = 1/3
then
(Y+3)= 1/3(X-5)
3y+9= X-5
X-3y-14=0
3y=x-4
y =[1/3]x -[4/3]
we know that Y =mx+b
hence slope = m°= 1/3
since both lines are parallel, hence
m=m°=1/3=slope of another line
to find ewuation of line with slope m and passing through point (5,-3)
(Y-Y°)=m(X-X°)
given Y°= -3 and X° = 5„ m = 1/3
then
(Y+3)= 1/3(X-5)
3y+9= X-5
X-3y-14=0
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