Question

The point of intersection of the lines represented by equation \[2{{\left( x+2 \right)}^{2}}+3\left( x+2 \right)\left( y-2 \right)-2{{\left( y-2 \right)}^{2}}=0\] is:
(a) \[\left( 2,2 \right)\]
(b) \[\left( -2,-2 \right)\]
(c) \[\left( -2,2 \right)\]
(d) \[\left( 2,-2 \right)\]

Answer 1

Hint: For any given pair of straight lines represented by the equation \[a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0\]. The formula for point of intersection for these pair of straight lines is given as: \[\left( {{x}_{1}},{{y}_{1}} \right)=\left( \sqrt{\dfrac{{{f}^{2}}-bc}{{{h}^{2}}-ab}},\sqrt{\dfrac{{{g}^{2}}-ac}{{{h}^{2}}-ab}} \right)\].

Complete step by step solution:
The given equation of pair of lines is,
\[2{{\left( x+2 \right)}^{2}}+3\left( x+2 \right)\left( y-2 \right)-2{{\left( y-2 \right)}^{2}}=0\]
By expanding the whole square and solving it further, we will have:
\[\Rightarrow 2\left( {{x}^{2}}+4+4x \right)+3\left( xy-2x+2y-4 \right)-2\left( {{y}^{2}}+4-4y \right)=0\]
Opening the bracket, we get
\[\Rightarrow 2{{x}^{2}}+8+8x+3xy-6x+6y-12-2{{y}^{2}}-8-8y=0\]
Combining the like terms, we have
\[\Rightarrow 2{{x}^{2}}-2{{y}^{2}}+2x+3xy+14y-12=0\].
So, the above equation represents a pair of straight lines now.
Now we know, for any given pair of straight lines represented by the equation \[a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0\]:
The formula for point of intersection for these pair of straight lines is given as:
\[\left( {{x}_{1}},{{y}_{1}} \right)=\left( \sqrt{\dfrac{{{f}^{2}}-bc}{{{h}^{2}}-ab}},\sqrt{\dfrac{{{g}^{2}}-ac}{{{h}^{2}}-ab}} \right)\]
From the obtained equation, \[2{{x}^{2}}-2{{y}^{2}}+2x+3xy+14y-12=0\], comparing it with the general form of equation \[a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0\], we get
\[a=2,b=-2,h=\dfrac{3}{2},g=1,f=7,c=-12\].
Substituting the above determined values in the point of intersection of pair of straight lines formula, we have:
\[\begin{align}
  & =\left( \sqrt{\dfrac{{{(7)}^{2}}-(-2)(-12)}{{{\left( \dfrac{3}{2} \right)}^{2}}-(2)(-2)}},\sqrt{\dfrac{{{(1)}^{2}}-(2)(-12)}{{{\left( \dfrac{3}{2} \right)}^{2}}-(2)(-2)}} \right) \\
 & \Rightarrow =\left( \sqrt{\dfrac{49-24}{\dfrac{9}{4}+4}},\sqrt{\dfrac{1+24}{\dfrac{9}{4}+4}} \right) \\
\end{align}\]
Taking the LCM in the denominator, we get
\[\Rightarrow =\left( \sqrt{\dfrac{25}{\dfrac{9+16}{4}}},\sqrt{\dfrac{25}{\dfrac{9+16}{4}}} \right)\]
\[\Rightarrow =\left( \sqrt{\dfrac{25(4)}{25}},\sqrt{\dfrac{25(4)}{25}} \right)\]
Cancelling the like terms and clearing the square roots in the above equation to finally have:
\[=\left( 2,2 \right)\]
Therefore, the point of intersection of \[2{{\left( x+2 \right)}^{2}}+3\left( x+2 \right)\left( y-2 \right)-2{{\left( y-2 \right)}^{2}}=0\] is equal to \[=\left( 2,2 \right)\].
Hence, option (a) is the correct answer.

Note: Expand the equation carefully and compare the coefficients of ‘g’ and ‘f’ attentively or else you may end up with a different answer. Alternatively, you can find out the two lines that are represented by these pair of straight lines and compute the intersection point, but that would be a lengthy process.