Question

Equation $12x^2-10xy+2y^2+11x-5y+2=0$ represent a pair of straight lines. Which of the following is/are true about these lines?

• A. Slope of line 1 is -2
• B. Slope of line 1 is 2
• C. Slope of line 2 is 3
• D. Slope of line 2 is -3

Answer 1

Answer: (B), (C)

The correct options are
B

Slope of line 1 is 2

C

Slope of line 2 is 3

Given equation is $12x^2-10xy+2y^2+11x-5y+2=0$

Quadratic equation in x

$12x^2+(11-10y)x+2y^2-5y+2=0$

$x=\frac{-(11-10y)\pm \sqrt{(11-10y{)}^2-4\times 12(2y^2-5y+2)}}{2\times 12}$

$x=\frac{-(11-10y)\pm \sqrt{121+100y^2-220y-96y^2+240y-96}}{2\times 12}$

= $\frac{10y-11\pm \sqrt{4y^2+20y+25}}{24}$

= $\frac{10y-11\pm \sqrt{(2y+5{)}^2}}{24}$

24x = 10y - 11 $\pm$ (2y + 5)

24x = 10y - 11 + 2y + 5 and 24x = 10y - 11 - (2y + 5)

24x = 12y - 6 and 24x = 8y - 16

24x - 12y + 6 = 0

4x - 2y + 1 = 0 and 3x - y + 2 = 0

Slope of line 1 =$\frac{-4}{-2}$= 2

Slope of line 2 =$\frac{-3}{-1}$= 3

Slope of lines are 2 and 3

Option B and C are correct.