Linesx2+kxy+y2 and x+y−1=0 make the sides of equilateral triangle. (given)
∵x2+kxy+y2=0 represent a pair of straight lines passing through origin. Let m be the slope of one line whose equation is y=mx According to the question, tanθ=|
m1−m2
1+m1m2
|θ= angle between lines ⇒tan60°=|
m−(−1)
1+m(−1)
| [∵ slope of x+y=1 is −1] ⇒√3=
m−1
1−m
⇒(1−m)2=(m+1)2 ⇒2m2−8m+2=0 ⇒m2−4m+1=0 (
y
x
)2−4(
y
x
)+1=0[∵y=mx⇒
y
x
=m] ⇒
y2
x2
−
4y
x
+1=0 ⇒y2−4xy+x2=0 Compare with x2+kxy+y2=0, we get k=−4 ∴k2=(−4)2=16