Name: The Line Family
Uploaded: 2019-07-19
Duration: 4 min 33 s
Description: A family of lines is a set of lines that have something in common with each other. Learn more about the family of lines by watching this video.

Question

If bases and the sum of area of a number of triangles have a common vertex, then show that the locus of this vertex is a straight line.

Accepted Answer

Let the common vertex be $(h, k)$ .

Let the bases of triangles be $(b_{1}, b_{2}, b_{3}, . . . . . . . . . . . . . . b_{n})$ .

Assume normal form of line for $b_{1}$ , which is $x \cos α + y \sin α = p_{1}$ .

The perpendicular length $k_{1}$ from the common vertex $(h, k)$ on the base $b_{1}$ will be $k_{1} = \frac{|h \cos α + k \sin α - p_{1}|}{\sqrt{\cos^{2} α + \sin^{2} α}}$ .

Similarly assume normal form of line for $b_{2}$ , which is $x \cos β + y \sin β = p_{2}$ .

The perpendicular length $k_{2}$ from the common vertex $(h, k)$ on the base $b_{2}$ will be $k_{2} = \frac{|h \cos β + k \sin β - p_{2}|}{\sqrt{\cos^{2} β + \sin^{2} β}}$ .

$\Rightarrow k_{2} = |h \cos β + k \sin β - p_{2}|$

Similarly, for every base, there will be perpendiculars $k_{3}, k_{4}, . . . . . . . . . k_{n}$ .

Let's assume the area of triangle to be $A_{1}, A_{2}, . . . . . . . A_{n}$ .

Now we will come to sum of areas of the triangles $(A_{1,} A_{2}, . . . . . . . A_{n})$ .

$A_{1} + A_{2} + A_{3} . . . . . . + A_{n} = c$

$(\frac{1}{2} \times b_{1} \times k_{1} + \frac{1}{2} \times b_{2} \times k_{2} + \frac{1}{2} \times b_{3} \times k_{3} . . . . .) = c$ (constant)

Also replace $(h, k)$ by $(x, y)$ in all the perpendicular distances $k_{1}, k_{2}, k_{3}, . . . k_{n}$ .

We know distances are always positive, so take $+ v e$ sign.

$\Rightarrow (\frac{1}{2} \times b_{1} \times (x \cos α + y \sin α - p_{1}) + \frac{1}{2} \times b_{2} \times (x \cos β + y \sin β - p_{2}) + \frac{1}{2} \times b_{3} \times k_{3} + . . . . + \frac{1}{2} \times b_{n} \times k_{n}) = c$

$\Rightarrow (b_{1} \times (x \cos α + y \sin α - p_{1}) + b_{2} \times (x \cos β + y \sin β - p_{2}) + . . . . + b_{n} \times k_{n}) = 2 c$

$\Rightarrow (x (b_{1} \cos α + b_{2} \cos β + . . . + b_{n} \cos β) + y (b_{1} \sin α + b_{2} \sin β . . . . . + b_{n} \sin α) - (b_{1} p_{1} + b_{2} p_{2} + . . . + b_{n} p_{n})) = 2 c$

$\Rightarrow x (\sum b_{1} \cos α) + y (\sum b_{1} \sin α) - (\sum b_{1} p_{1}) - 2 c = 0$

This is the format of the equation of a straight line $a x + b y + k = 0$ .

Hence, proved.

S L Loney Solutions for Chapter: The Straight Line (Continued), Exercise 2: EXAMPLES XI

S L Loney Mathematics Solutions for Exercise - S L Loney Solutions for Chapter: The Straight Line (Continued), Exercise 2: EXAMPLES XI

Questions from S L Loney Solutions for Chapter: The Straight Line (Continued), Exercise 2: EXAMPLES XI with Hints & Solutions

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