# Straight Lines Set 3

**Example – 5**

A rod *AB* of length *l* slides with its end on the coordinate axes. Let *O* be the origin. The rectangle *OAPB* is completed. Find the locus of the foot of the perpendicular drawn from *P* onto *AB*.

**Solution:**

In terms of the variable \(\theta ,\) *A* and *B*, and hence *P*, have coordinates

\[A \equiv (0,\,\,l\sin \theta ), \; \,\,\,B \equiv (l\cos \theta ,\,\,0),\,\,\,\,\,P \equiv (l\cos \theta ,\,l\sin \theta )\]

The slope of *PF* can be observed to be so that its equation is

\[y - l\sin \theta = \cot \theta (x - l\cos \theta )\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\]

The equation of *AB* is

\[x\sec \theta + y\,{\text{cosec }}\theta = l\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)\]

The intersection of (1) and (2) gives us the point \(F(h,\,k):\)

\[\begin{align}&h = l{\cos ^3}\theta , \; k = l{\sin ^3}\theta \\ \Rightarrow \qquad& \cos \theta = {\left( {\frac{h}{l}} \right)^{1/3}}, \; \sin \theta = {\left( {\frac{k}{l}} \right)^{1/3}} \\\end{align} \]

Eliminating \(\theta ,\) we have

\[{h^{2/3}} + {k^{2/3}} = {l^{2/3}}\]

Thus, the locus of *F* is

\[{x^{2/3}} + {y^{2/3}} = {l^{2/3}}\]

**Example – 6**

**(a)** Consider a line segment *AB* where \(A \equiv ({x_1},\,{y_1})\) and \(B \equiv ({x_2},\,{y_2}).\) In what ratio does a line \(L \equiv ax + by + c = 0\) divide *AB*?

**Solution: **Let the required ratio be \(\lambda :1\)

The coordinates of *C* are (from the internal division formula),

\[C \equiv \left( {\frac{{\lambda {x_2} + {x_1}}}{{\lambda + 1}},\,\,\frac{{\lambda {y_2} + {y_1}}}{{\lambda + 1}}} \right)\]

Since this lies on *L*, we have

\[\begin{align}& a\left( {\frac{{\lambda {x_2} + {x_1}}}{{\lambda + 1}}} \right) + b\left( {\frac{{\lambda {y_2} + {y_1}}}{{\lambda + 1}}} \right) + c = 0 \\ \Rightarrow \qquad &\lambda (a{x_2} + b{y_2} + c) + (a{x_1} + b{y_1} + c) = 0 \\ \Rightarrow \qquad &\lambda = \frac{{ - a{x_1} + b{y_1} + c}}{{a{x_2} + b{y_2} + c}} \\ \Rightarrow \qquad &\boxed{\lambda = - \frac{{L({x_1},\,{y_1})}}{{L({x_2},\,{y_2})}}} \\

\end{align} \]

This is a useful result (as we’ll see from part(b), the next example) and it would be worth memorizing it.

**(b)** A line intersects *BC*, *CA* and *AB* in at *P*, *Q* and *R* respectively. Show that

\[\frac{{BP}}{{PC}} \cdot \frac{{CQ}}{{QA}} \cdot \frac{{AR}}{{RB}} = - 1\]

**Solution:**

Using the result of the last example, we have

\[\begin{align} \frac{{AR}}{{RB}} = - \frac{{L({x_1},\,{y_1})}}{{L({x_2},\,{y_2})}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right) \\ \frac{{BP}}{{PC}} = - \frac{{L({x_2},\,{y_2})}}{{L({x_3},\,{y_3})}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right) \\ \frac{{CQ}}{{QA}} = - \frac{{L({x_3},\,{y_3})}}{{L({x_1},\,{y_1})}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 3 \right) \\ \end{align}\]

From (1), (2) and (3), it should be evident that the assertion stated in the question is valid.

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