**Example - 19**

Find the area of the triangle formed by the lines \({y^2} - 9xy + 18{x^2} = 0\;{\rm{and}}\;y = 9.\)

**Solution:** The joint equation can be factorized to obtained

\[(y - 3x)(y - 6x) = 0\]

Thus, the three lines forming the sides of the triangle are

\[y = 3x, \; y = 6x, \; y = 9\]

The three intersection points can easily be seen to be

\[(0,\,0),\;(3,\,9),\;\left( {\frac{3}{2},\,9} \right)\]

Thus, the area of the triangle is

\[\begin{align} \Delta & = \frac{1}{2}\left| {\begin{array}{*{20}{c}} 0&3&{\frac{3}{2}}\\ 0&9&9\\ 1&1&1 \end{array}} \right|\\ & = \frac{{27}}{4}{\rm{sq}}{\rm{. units}} \end{align}\]

**Example - 20**

The slope of one of the two lines represented by \(a{x^2} + 2hxy + b{y^2} = 0\) is the square of the other. Prove that

\[\frac{{a + b}}{h} + \frac{{8{h^2}}}{{ab}} = 6\]

**Solution:** Let the two slopes be \(m{\text{ and }}{m_2};\) these are the roots of the quadratic

\[b{M^2} + 2hM = a = 0\]

so that

\[m + {m^2} = \frac{{ - 2h}}{b}, \qquad {m^3} = \frac{a}{b}\]

Cubing the first relation, we have

\[\begin{align} & {m^3} + {m^6} + 3{m^3}(m + {m^2}) = \frac{{ - 8{h^3}}}{{{b^3}}}\\ \Rightarrow \qquad & \frac{a}{b} + \frac{{{a^2}}}{{{b^2}}} + \frac{{3a}}{b}\left( {\frac{{ - 2h}}{b}} \right) = \frac{{ - 8{h^3}}}{{{b^3}}}\\ \Rightarrow \qquad & b{a^2} + a{b^2} + 8{h^3} = 6hab\\ \Rightarrow \qquad & \frac{{a + b}}{h} + \frac{{8{h^2}}}{{ab}} = 6 \end{align}\]