Find the value of k, if the lines joining the origin to the point of intersection 2xsquare -2xy +3y square
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Step-by-step explanation:
Given joint equation of the line is kx
2
−5xy−6y
2
=0
It is in the form of ax
2
+2hxy+by
2
=0
a=k;h=−
2
5
;b=−6
The auxiliary equation of the given line is of the form bm
2
+2hm+a=0
=>(−6)m
2
−5m+k=0
=>6m
2
+5m−k=0....(i)
One of the line is 4x+3y=0. Its slope is −
3
4
. Now, −
3
4
must be one of the roots of the auxiliary equation.
Substituting m=−
3
4
in equation (i)
6(
3
−4
)
2
+5(
3
−4
)−k=0
=>6(
9
16
)+5(
3
−4
)−k=0
=>32−20−3k=0
=>−3k=−12
=>k=4
∴
k
1
=
4
1
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