A line 5x+3y+15=0 meets y-axis at point P. Find the co-ordinates of point P. Find the equation of a line through P and perpendicular to x-3y+4=0.
Answers
Answered by
22
1)to find coordinates of P
Put x=0 in the given equation
3y+15=0
3y=-15
y=-15/3
y= -5
P=(0,-5)
2)
Slope of a line x-3y+4=0 is
m1=-a/b= 1/3
Slope of a line perpendicular to x-3y+4=0 is
m2= -3
Therefore
Required equation is passing through p(0,-5) and whose slope = -3
y=mx+b here m slope, b is y intercept
y=-3x-5
y+3x+5=0
3x+y+5=0
Put x=0 in the given equation
3y+15=0
3y=-15
y=-15/3
y= -5
P=(0,-5)
2)
Slope of a line x-3y+4=0 is
m1=-a/b= 1/3
Slope of a line perpendicular to x-3y+4=0 is
m2= -3
Therefore
Required equation is passing through p(0,-5) and whose slope = -3
y=mx+b here m slope, b is y intercept
y=-3x-5
y+3x+5=0
3x+y+5=0
NeneAmaano:
Its 3x+4y+9=0
sry, u'r wrong
a/c to the problem my solution is correct
plz check
My book says the answer is 3X+4y+9=0
Point P is correct but only equation is wrong
check the problem again , is there any mistakes in the equations
plz check any typing errors in the problem u posted
thanks
Thank u , sir
Answered by
21
Hope this helps......
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i did the same thing
yeah sir ur solution is authentic but u have told me to give gud explanation so i give and got result
thanka
it is not abt solution, she said my answer is wrong
kk
:)
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