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

Hint: 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
\[\left| \begin{gathered}
  0&\dfrac{h}{2}&\dfrac{{ - g}}{2} \\
  \dfrac{h}{2}&0&\dfrac{{ - f}}{2} \\
  \dfrac{{ - g}}{2}&\dfrac{{ - f}}{2}&c \\
\end{gathered} \right| = 0\]
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.