Question

The point of intersection of lines represented by the equation $3x^2+8xy-3y^2+29x-3y+18=0$ is

• A. $\left(-\frac{3}{2},-\frac{5}{2}\right)$
• B. $\left(\frac{2}{3},\frac{5}{2}\right)$
• C. $\left(3,-5\right)$
• D. None of these

Answer 1

The correct option is A

$\left(-\frac{3}{2},-\frac{5}{2}\right)$

Explanation for the correct option:

Comparing above equation with general second degree equation.

Given:

$3x^2+8xy-3y^2+29x-3y+18=0$

$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

We get, $a=3,h=4,b=-3,g=\frac{29}{2},f=-\frac{3}{2},c=18$

We know that, Point of intersection is:$\left(\frac{\left(\mathrm{bg}-\mathrm{fh}\right)}{\left(h^2-\mathrm{ab}\right)},\frac{\left(\mathrm{af}-\mathrm{gh}\right)}{\left(h^2-\mathrm{ab}\right)}\right)$

$=\left(\frac{\left(-3\right)\times {\displaystyle \frac{29}{2}}-{\displaystyle \frac{-3}{2}}\times 4}{4^2-3\times \left(-3\right)},\frac{3\times {\displaystyle \frac{-3}{2}}-{\displaystyle \frac{29}{2}}\times 4}{4^2-3\times \left(-3\right)}\right)$

$=\left(\frac{{\displaystyle \frac{-75}{2}}}{25},\frac{-{\displaystyle \frac{125}{2}}}{25}\right)$

$$\begin{gathered} =\left(-\frac{75}{50},-\frac{125}{50}\right) \\ =\left(-\frac{3}{2},-\frac{5}{2}\right) \end{gathered}$$

Hence, option $A$ is the correct answer.