Question

If the equation $hxy + gx + fy + c = 0$ represents a pair of straight lines , then
1. $fh = cg$
2. $fg = ch$
3. ${h^2} = gf$
4. $fgh = c$

Answer 1

A straight line is a line which is not curved or bent. So, a line that extends to both sides till infinity and has no curves is called a straight line. The equations of two or more lines can be expressed together by an equation of degree higher than one. As we see that a linear equation in x and y represents a straight line, the product of two linear equations represents two straight lines, that is a pair of straight lines.

Complete step-by-step solution:
We know that the equation $a{x^2} + 2hxy + b{y^2} = 0$represents a pair of straight lines passing through origin and hence it can be written as product of two linear factors, $a{x^2} + 2hxy + b{y^2} = (lx + my)(px + qy)$ where $lp = a$ , $mq = b$ and $lq + mp = 2h$.
Also, the separate equations of lines are $lx + my = 0$ and $px + qy = 0$.
As a consequence of this formula, we can conclude that
1. The lines are real and distinct, if ${h^2} - ab > 0$
2. The lines are real and coincident, if ${h^2} - ab = 0$
3. The lines are not real (imaginary), if ${h^2} - ab < 0$
The give equation $hxy + gx + fy + c = 0$will represent a pair of straight lines if

0&\dfrac{h}{2}&\dfrac{{ - g}}{2} \\\ \dfrac{h}{2}&0&\dfrac{{ - f}}{2} \\\ \dfrac{{ - g}}{2}&\dfrac{{ - f}}{2}&c; \\\ \end{gathered} \right| = 0$$ Expanding along $${R_1}$$ $$0.\left( {0 - \dfrac{{{f^2}}}{2}} \right) - \dfrac{h}{2}\left( {\dfrac{{hc}}{2} - \dfrac{{gf}}{4}} \right) - \dfrac{g}{2}\left( {\dfrac{{ - fh}}{4} - 0} \right) = 0$$ Hence we get , $$ - \dfrac{{{h^2}c}}{4} + \dfrac{{ghf}}{8} = \dfrac{{ - gfh}}{8}$$ On simplification we get , $$ - \dfrac{{{h^2}c}}{4} = \dfrac{{ - gfh}}{4}$$ Hence we get $$hc = gf$$ or $$fg = ch$$ **Therefore option(2) is the correct answer.** **Note:** Two lines are coincident if $$\tan \theta = 0$$ i.e. if $${h^2} - ab = 0$$. Two lines are perpendicular if $$\tan \theta = \infty $$ i.e. if $$a + b = 0$$. If two pairs of straight lines are equally inclined to one another, then both must have the same pair of bisectors. If the lines given by the equation $$a{x^2} + 2hxy + b{y^2} = 0$$ are equally inclined to axes, then the coordinate axes are the bisectors, i.e. the equation of pair of bisector must be$$xy = 0$$. Therefore$$h = 0$$. The two bisectors are always perpendicular.