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Mathematics Question on Straight lines

The value of kk for which the equation x2−4xy−y2+6x+2y+k=0{{x}^{2}}-4xy-{{y}^{2}}+6x+2y+k=0 represents a pair of straight lines is

A

k=4k=4

B

k=−1k=-1

C

k=−45k=\frac{-4}{5}

D

k=−225k=\frac{-22}{5}

Answer

k=−45k=\frac{-4}{5}

Explanation

Solution

The given equation x2−4xy−y2+6x+2y+k−0{{x}^{2}}-4xy-{{y}^{2}}+6x+2y+k-0 ..(i)
represent the point of straight line, if Δ=0,\Delta =0,
where Δ=abc+2fgh−af2−bg2−ch2=0\Delta =abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 ..(ii)
Comparing E (i) with the following equation
ax2+2hxy+by2+2gx+2fy+c=0a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0
We get, a=1, h=−2, b=−1,  g=3, f=1, c=ka=1,\,h=-2,\,b=-1,\,\,g=3,\,f=1,\,c=k
From E (ii), (1) (−1) (k)+2(1) (3) (−2)−(1) (1)2−(−1)(1)\,(-1)\,(k)+2(1)\,(3)\,(-2)-(1)\,{{(1)}^{2}}-(-1)
(3)2−(k) (−2)2=0{{(3)}^{2}}-(k)\,{{(-2)}^{2}}=0
⇒\Rightarrow −k−12−1+9−4k=0-k-12-1+9-4k=0
⇒\Rightarrow −5k=4  ⇒ k=−4/5-5k=4\,\,\Rightarrow \,k=-4/5