If the straight lines represented by 9x^2 – 12xy + ky^2 = 0 be perpendicular to each other then find the value of k.
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ANSWER
5x
2
+12xy−6y
2
+4x−2y+3=0 ------ ( 1 )
x+ky−1=0
x+ky=1 ----- ( 2 )
Hamoginizing ( 1 ) with ( 2 )
⇒ 5x
2
+12xy−6y
2
+4x(x+ky)−2y(x+ky)+3(x+ky)
2
=0
⇒ 5x
2
+12xy−6y
2
+4x
2
+4kxy−2xy−2ky
2
+3(x
2
+kxy+k
2
y
2
)=0
⇒ 9x
2
+10xy−6y
2
+4kxy−2ky
2
+3x
2
+3kxy+3k
2
y
2
=0
⇒ 12x
2
+(10+4k+6k)xy+(3k
2
−2k−6)y
2
=0
⇒ 12x
2
+(10+10k)xy+(3k
2
−2k−6)y
2
=0
Pair of lines are equally inclined to the axes.
∴ Coefficient of xy=0
⇒ 10k+10=0
⇒ 10k=−10
⇒ k=−1
∴ ∣k∣=∣−1∣=1
Step-by-step explanation:
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