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CBSE Class 11-science Answered

Find the value of lambda.
question image
Asked by dineshchem108 | 20 Sep, 2018, 08:57: AM
answered-by-expert Expert Answer
The equations of line are Error converting from MathML to accessible text.x - y + 1 = 0  and x - 2y + 3.

In Error converting from MathML to accessible text.x - y = -1,
If x = o then y = 1
and if y = 0 then x = begin mathsize 16px style fraction numerator negative 1 over denominator straight lambda end fraction end style
 
In x - 2y = -3
 
If x = o then y = begin mathsize 16px style 3 over 2 end style
and if y = 0 then x = -3
 
 
 
Points if intersection with co-ordinate axes are open parentheses size 16px 0 size 16px comma size 16px 1 close parentheses size 16px comma size 16px space open parentheses fraction numerator size 16px minus size 16px 1 over denominator size 16px lambda end fraction size 16px comma size 16px 0 close parentheses size 16px comma size 16px space open parentheses size 16px 0 size 16px comma size 16px 3 over size 16px 2 close parentheses size 16px space size 16px and size 16px space open parentheses size 16px minus size 16px 3 size 16px comma size 16px space size 16px 0 close parentheses size 16px.
General Equation of a Circle is given by,
 
x2 + y2 + 2gx + 2fy + c = 0          ........(a)
 
Substitute points 
 open parentheses size 16px 0 size 16px comma size 16px 1 close parentheses size 16px comma size 16px space open parentheses size 16px 0 size 16px comma size 16px 3 over size 16px 2 close parentheses size 16px space size 16px and size 16px space open parentheses size 16px minus size 16px 3 size 16px comma size 16px space size 16px 0 close parentheses size 16px.
G
in the equation, we get 3 equations as follows.
 
2f + c = -1    ....(i)
6g - c = -9    ....(ii)
3f + c = begin mathsize 16px style fraction numerator negative 9 over denominator 4 end fraction end style...(iii)
Solve them simulteneously, we get
begin mathsize 16px style straight f equals fraction numerator negative 5 over denominator 4 end fraction space space space comma space straight g space equals space 7 over 4 space and space straight c space equals space 3 over 2 end style
Substitute the value in (a), we get
begin mathsize 16px style straight x squared space plus space straight y squared space plus space 7 over 2 straight x space minus space 5 over 2 straight y space plus space 3 over 2 space equals space 0 end style
Substitute the coordinates of point open parentheses fraction numerator size 16px minus size 16px 1 over denominator size 16px lambda end fraction size 16px comma size 16px 0 close parentheses in above equation we get,
begin mathsize 16px style 3 straight lambda squared space minus space 7 straight lambda space plus space 2 space equals space 0 Take space straight lambda space equals fraction numerator negative straight b space plus-or-minus space square root of straight b squared space minus space 4 ac end root over denominator 2 straight a end fraction space comma we space get comma straight lambda space equals space 2 comma space 1 third end style
Answered by Yasmeen Khan | 20 Sep, 2018, 02:04: PM

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