Straight Lines Set 4

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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 C1:

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

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 C2: 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 m1 and m2 so that m1 and m2 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 m2 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

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

Download practice questions along with solutions for FREE:
Straight Lines
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Download practice questions along with solutions for FREE:
Straight Lines
grade 11 | Answers Set 2
Straight Lines
grade 11 | Questions Set 1
Straight Lines
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Straight Lines
grade 11 | Questions Set 2
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