Find the equation of the line which is perpendicular to the line x/a - y/b = 1 at the point where this line meets y-axis.
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The point where the line (x/a)-(y/b)=1 meets y-axis i.e (0,y)
0-(y/b) = 1
y = -b
hence the required point is P(0,-b)
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The equation of line perpendicular to line
ax+by+c = 0
is ;
bx-ay+k = 0
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The equation of line perpendicular to given line
-(x/b)-(y/a) + k = 0
passing through P(0,-b)
0-(-b/a)+k= 0
k = -b/a
______________________
Hence the required equation is ;
-(x/b)-(y/a) -(b/a) = 0
(x/b)+(y/a)+(b/a) = 0
0-(y/b) = 1
y = -b
hence the required point is P(0,-b)
__________________
The equation of line perpendicular to line
ax+by+c = 0
is ;
bx-ay+k = 0
------------------------------------
The equation of line perpendicular to given line
-(x/b)-(y/a) + k = 0
passing through P(0,-b)
0-(-b/a)+k= 0
k = -b/a
______________________
Hence the required equation is ;
-(x/b)-(y/a) -(b/a) = 0
(x/b)+(y/a)+(b/a) = 0
NeneAmano:
Can you please tell me why have we used
-(x/b)-(y/a)+k=0
because the product of slopes of two perpendicular lines is -1
So, what's k ?
k is a constant
OK
we have to find k by putting the value of given point into the equation
OK
Thanks for helping:)
My pleasure 'ARL' ! (^_^)
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