# Straight Lines Set 3

Go back to  'Straight Lines'

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.