# Straight Lines Set 4

**Example – 7**

The curves

\[\begin{align} {C_1}:{a_1}{x^2} + 2{h_1}xy + {b_1}{y^2} + 2{g_1}x = 0 \\ {C_2}:{a_2}{x^2} + 2{h_2}xy + {b_2}{y^2} + 2{g_2}x = 0 \\ \end{align} \]

intersect at two points *A* and *B* other than the origin. Find the condition for *OA* and *OB* to be perpendicular.

**Solution:** Assume the equation of *AB* to be \(y = mx + c.\) Thus, using the homogenizing technique, we can write the joint equation of *OA* and *OB*:

**Homogenizing C**

_{1}

**:**

\[\begin{align}&{a_1}{x^2} + 2{h_1}xy + {b_1}{y^2} + 2{g_1}x\left( {\frac{{y - mx}}{c}} \right) = 0 \\ \Rightarrow \qquad &\left( {{a_1} - \frac{{2m{g_1}}}{c}} \right){x^2} + 2{h_1}xy + {b_1}{y^2} + \frac{{2{g_1}xy}}{c} = 0 \\ \end{align} \]

This is the joint equation of *OA* and *OB*. *OA* and *OB* are perpendicular if

\[\begin{align}&{a_1} - \frac{{2m{g_1}}}{c} + {b_1} = 0 \\ \Rightarrow \qquad &\frac{m}{c} = \frac{{{a_1} + {b_1}}}{{2{g_1}}}\qquad\qquad\qquad\qquad\dots(1) \\ \end{align}\]

**Homogenizing C**

_{2}

**:**Similarly, we can again evaluate the joint equation of

*OA*and

*OB*by homogenizing the equation of \({C_2}:\)

\[\left( {{a_2} - \frac{{2m{g_2}}}{c}} \right){x^2} + 2{h_2}xy + {b_2}{y^2} + \frac{{2{g_2}xy}}{c} = 0\]

The perpendicularity condition gives

\[\begin{align} &{a_2} - \frac{{2m{g_2}}}{c} + {b_2} = 0 \\ \Rightarrow \qquad &\frac{m}{c} = \frac{{{a_2} + {b_2}}}{{2{g_2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right) \\ \end{align} \]

From (1) and (2), the necessary required condition is

\[\frac{{{a_1} + {b_1}}}{{{g_1}}} = \frac{{{a_2} + {b_2}}}{{{g_2}}}\]

**Example – 8**

Find the orthocentre of the triangle formed by the lines \(a{x^2} + 2hxy + b{y^2} = 0\) and \(px + qy = 1.\)

**Solution:** The two lines given by the joint equation pass through the origin. Assume their slopes to be *m*_{1} and *m*_{2} so that *m*_{1} and *m*_{2} are the roots of

\[\begin{align}& b{m^2} + 2hm + a = 0 \\ \Rightarrow\qquad &{m_1} + {m_2} = - \frac{{2h}}{b}, \;\;{m_1}{m_2} = \frac{a}{b}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\\ \end{align} \]

To evaluate the orthocentre, we need two altitudes. We take one of them to be the one dropped from *O* onto *MN*.

\[qx - py = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)\]

Let us now find the altitude from *M* onto *ON*. The coordinates of *M* are, by solving \(y = {m_1}x\) and \(px + qy = 1\) simultaneously,

\[M \equiv \left( {\frac{1}{{p + q{m_1}}},\,\,\frac{{{m_1}}}{{p + q{m_1}}}} \right)\]

The slope of *ON* is *m*2 so that the slope of the altitude through *M* is \(- \frac{1}{{{m_2}}};\) thus, its equation is

\[\begin{align}&y - \frac{{{m_1}}}{{p + q{m_1}}} = \frac{{ - 1}}{{{m_2}}}\left( {x - \frac{1}{{p + q{m_1}}}} \right) \\ \Rightarrow \qquad& \frac{x}{{{m_2}}} + y = \frac{{1 + {m_1}{m_2}}}{{{m_2}(p + q{m_1})}} \\ \Rightarrow \qquad& x + {m_2}y = \frac{{1 + {m_1}{m_2}}}{{p + q{m_1}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 3 \right) \\ \end{align} \]

The intersection of (2) and (3) yields the orthocentre

\[\begin{align}h& = \frac{{p(1 + {m_1}{m_2})}}{{(p + q{m_1})(p + q{m_2})}} \\ &= \frac{{p(1 + {m_1}{m_2})}}{{{p^2} + pq({m_1} + {m_2}) + {q^2}{m_1}{m_2}}} \\ &= \frac{{p(a + b)}}{{b{p^2} - 2hpq + a{q^2}}} \\ k &= \frac{q}{p}.h\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {from\left( 1 \right)} \right) \\& = \frac{{q(a + b)}}{{b{p^2} - 2hpq + a{q^2}}} \\ \end{align} \]

Thus, the orthocentre has the coordinates

\[\left( {\frac{{p(a + b)}}{{b{p^2} - 2hpq + a{q^2}}},\,\,\,\frac{{q(a + b)}}{{b{p^2} - 2hpq + a{q^2}}}} \right)\]

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