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Question
If the equation 2xy+4x−6y+k=0 represents a pair of straight lines, then the value of (k+20) is
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Solution
The general equation of second degree :
ax2+2hxy+by2+2gx+2fy+c=0
Given : 2xy+4x−6y+k=0
Comparing both,
a=0, b=0, h=1, g=2, f=−3, c=k
For two straight lines,
h2−ab≥0 and △=0
△=0+2(−3)(2)(1)−0−0−k(1)2=−12−k
⇒−12−k=0
⇒ k=−12
∴ k+20=−12+20=8
ax2+2hxy+by2+2gx+2fy+c=0
Given : 2xy+4x−6y+k=0
Comparing both,
a=0, b=0, h=1, g=2, f=−3, c=k
For two straight lines,
h2−ab≥0 and △=0
△=0+2(−3)(2)(1)−0−0−k(1)2=−12−k
⇒−12−k=0
⇒ k=−12
∴ k+20=−12+20=8
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