Question

Find the joint equation of pair of lines passing through (2,-1) and parallel to 2x^2+3xy-9y^2=0

Answer 1

$2x^2+3xy-9y^2=0,$

Desired equation of pair of lines passing through (2, -1) is :

$ax^2+2hxy+by^2+2gx+2fy+c=0,\ \ \ equation\ 3\\\\a=2,\ b=-9,\ h=\frac{3}{2},\\\\Point\ of\ intersection\ of\ L1\ and\ L2\ is:$

$(\frac{hf-bg}{ab-h^2}, \frac{gh-af}{ab-h^2})\\\\ ab-h^2=-18-9/4=-81/4\\\\\frac{3f/2+9g}{-81/4}=2,\ \ => 3f+18g=-81,\ \ --equation\ 1\\\\\frac{3g/2-2f}{-81/4}=-1,\ \ 3g-4f=81/2,\ \ --equation\ 2$

solving equation 1 and 2, we get:
   $81 g = -324+243/2\ \ => g=-4+3/2=-\frac{5}{2}\\\\f=-12\\\\Substitute\ x=2,\ y=-1\ in\ equation\ 3\ and\ solve\ for\ c.\\\\c=-7\\\\2x^2+3xy-9y^2-5x-24y-7=0$

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Alternately,

$Let\ Slopes\ of\ lines:\ m_1\ and\ m_2.\\\\m_1m_2=\frac{2}{-9}=-2/9, \ \ \ m_1+m_2=-\frac{-3}{-9}=\frac{1}{3}\\\\(m_1-m_2)^2=\frac{1}{3^2}-4\frac{-2}{9}=1\\\\ m_1-m_2=+1\ or\ -1 \\\\m_1=\frac{2}{3},\ m_2=-\frac{1}{3},\ \ \ OR,\ \ m_1=-\frac{1}{3},\ m_2=\frac{2}{3}$

The lines are L1 =  y - m₁ x - c₁ = 0  and  y - m₂ x   - c₂ = 0 

These two lines pass through the point (2, -1)  
    c₁ = -1 -2m₁   and  c₂ = -1 - 2 m₂ 

multiply these equations and compare with the standard form of equation of pair of straight lines, we get

    2 g = m₁ c₂ + m₂ c₁   ,      2 f = -  (c₁ + c₂ )   and       c = c₁ c₂ 

then you the answer