Question

if non - zero numbers a,b,c are in harmonic progression then show that the equation x/a+y/b+1/c =0 represents a family of concurrent lines and find the point of con currency ​

Answer 1

SOLUTION

GIVEN

The non - zero numbers a,b,c are in harmonic progression

TO DETERMINE

The point of concurrency for the equation

$\displaystyle \sf{ \frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0 }$

represents a family of concurrent lines

EVALUATION

Here it is given that the non - zero numbers a,b,c are in harmonic progression

$\displaystyle \sf{ \frac{2}{b} = \frac{1}{a} + \frac{1}{c} }$

$\displaystyle \sf{ \implies \: \frac{1}{a} -\frac{2}{b} + \frac{1}{c} = 0 }$

$\displaystyle \sf{ \implies \: \frac{1}{a} + \frac{ - 2}{b} + \frac{1}{c} = 0 }$

Again the equation of the line is

$\displaystyle \sf{ \frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0 }$

Comparing we get x = 1 & y = - 2

Hence the required point of concurrency =

( 1, - 2)

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Answer 2

Answer:

The point of concurrence is (1,-2)