Solveeit Logo
Login

Question

Question: If \(6x^{2} + 11xy - 10y^{2} + x + 31y + k = 0\) represents a pair of straight lines, then \(k =\)...

If 6x2+11xy−10y2+x+31y+k=06x^{2} + 11xy - 10y^{2} + x + 31y + k = 0 represents a pair of straight lines, then k=k =

A

– 15

B

6

C

– 10

D

– 4

Answer

– 15

Explanation

Solution

−6.10k+11.1.314−6(312)2+10(12)2−k(112)2=0- 6.10k + \frac{11.1.31}{4} - 6\left( \frac{31}{2} \right)^{2} + 10\left( \frac{1}{2} \right)^{2} - k\left( \frac{11}{2} \right)^{2} = 0

⇒−k3614=54154⇒k=−15\Rightarrow - k\frac{361}{4} = \frac{5415}{4} \Rightarrow k = - 15.