Straight Lines Set 3

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