Question

Find the equations of the pair of lines through the origin which are perpendicular to the lines represented $6x^2-5xy+y^2=0$

• A. 3y+x=0
• B. 2y+x=0
• C. x+3y=0
• D. x+2y=0

Answer 1

Answer: (A), (B)

The correct options are
A

3y+x=0

B

2y+x=0

If we compare given equation $6x^2-5xy+y^2=0$ with $a{x}^2+2hxy+b{y}^2=0$

We find the a = 6, b = 1, $h=-\frac{5}{2}$

$h^2-ab=\frac{25}{4}-6=\frac{1}{4}$> 0

GIven equation represents a pair of distinct lines passing through the origin

Now, $6x^2-5xy+y^2=0$

After dividing both sides by $y^2$, we get the quadratic equation in $\frac{y}{x}$.

$(\frac{y}{x}{)}^2-5(\frac{y}{x})+6=0$

$(\frac{Y}{X}{)}^2-3(\frac{Y}{X})-2(\frac{Y}{X})+6=0$

$(\frac{y}{x}-3)(\frac{y}{x}-2)=0$

$\frac{y}{x}-3=0$ or $\frac{y}{x}-2=0$

y = 3x or y = 2x

We need to find the lines which are perpendicular to y = 3x and y = 2x

Since, product of slope of perpendicular lines is -1

Slope of those lines should be $\frac{-1}{3}$ and $\frac{-1}{2}$.

Perpendicular lines with slope $\frac{-1}{3}$ and $\frac{-1}{2}$ passing through origin

y = $\frac{-1}{3}$x and y = $\frac{-1}{2}$x

3y + x = 0 and 2y + x = 0