y+6=0 is parrallel to which axis
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Step-by-step explanation:
Given (k−3)x−(4−k
2
)y+k
2
−7k+6=0
a) parallel to x axis
∴y=p p is constant
∴ no x term
so k-3=0
⇒k=3
b) parallel to y axis ∴x=p⇒ no y terms
(k−3)x−(4−k
2
)y+k
2
−7k+6=0
⇒−(4−k
2
)y=0
⇒k
2
=4 ⇒k=±2
c) passing through origin (0,0)
∴x=0 y = 0
(k−3)x−(4−k
2
)y+k
2
−7k+6=0
⇒(k−3)×0−(4−k
2
)×0+k
2
−7k+6=0
⇒k
2
−7k+6=0
⇒k
2
−6k−k+6=0
⇒k(k−6)−1(k−6)=0
(k−6)(k−1)=0
∴k=6,1
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