Find the lines whose combined equation is $6 x^{2} + 5 x y - 4 y^{2} + 7 x + 13 y - 3 = 0$ using the concept of parallel lines through the origin.

Open in App

Solution

$\begin{matrix} 6 x^{2} + 5 x y - 4 y^{2} + 7 x + 13 y - 3 = 0 \Rightarrow 6 x^{2} + (5 y + 7) x - (4 y^{2} - 13 y + 3) = 0 \Rightarrow x = \frac{- (5 y + 7) \pm \sqrt{{(5 y + 7)}^{2} + 4 \times 6 \times (4 y^{2} - 13 y + 3)}}{2 \times 6} O n s o l v i n g, w e w i l l g e t x = \frac{- (5 y + 7) \pm (11 y - 11)}{12} x = \frac{- (5 y + 7) + 11 y - 11}{12} o r x = \frac{- (5 y + 7) - 11 y + 11}{12} x = \frac{6 y - 18}{12} o r x = \frac{- 16 y + 4}{12} \Rightarrow 12 x = 6 y - 18 o r 12 x = - 16 y + 4 \Rightarrow 2 x - y + 3 = 03 x + 4 y - 1 = 0 \end{matrix}$

Suggest Corrections

Similar questions

If $θ$ is an angle between the lines given by the equation $6 x^{2} + 5 x y - 4 y^{2} + 7 x + 13 y - 3 = 0$ , then equation of the line passing through the point of intersection of these lines and making an angle $θ$ with the positive x - axis is

Q. The equation

x^{2} - 5 x y + 4 y^{2} = 0

represents two lines passing through the origin which are

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

a x^{2} + 2 h x y + b y^{2} = 0

represents a pair of straight lines through origin & angle between them is given by

tan θ = \frac{2 \sqrt{h^{2} - a b}}{a + b}

. If the lines are perpendicular then

a + b = 0

and the equation of bisectors is given by

\frac{x^{2} - y^{2}}{a - b} = \frac{x y}{h}

The general equation of second degree given by

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

represent a pair of straight lines if

△ = 0

∣ ∣ ∣ ∣ \begin{matrix} a & h & g h & b & f g & f & c \end{matrix} ∣ ∣ ∣ ∣ = 0

a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0

On the basis of above information answer the following question

If the angle between the lines represented by

6 x^{2} + 5 x y - 4 y^{2} + 7 x + 13 y - 3 = 0

{tan}^{- 1} (m)

a^{2} + b^{2} - a b - a - b + 1 \leq 0

, then

5 a + 6 b

is equal to

Join BYJU'S Learning Program

Find the lines whose combined equation is 6x2+5xy−4y2+7x+13y−3=0 using the concept of parallel lines through the origin.

6x2+5xy−4y2+7x+13y−3=0⇒6x2+(5y+7)x−(4y2−13y+3)=0⇒x=−(5y+7)±√(5y+7)2+4×6×(4y2−13y+3)2×6Onsolving,wewillgetx=−(5y+7)±(11y−11)12x=−(5y+7)+11y−1112orx=−(5y+7)−11y+1112x=6y−1812orx=−16y+412⇒12x=6y−18or12x=−16y+4⇒2x−y+3=03x+4y−1=0

Find the lines whose combined equation is $6 x^{2} + 5 x y - 4 y^{2} + 7 x + 13 y - 3 = 0$ using the concept of parallel lines through the origin.