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:

& =\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{aligned}$$ Taking the LCM in the denominator, we get $$\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.