Question

If the family 3a+2b+6c of straight lines ax+by+c=0 passes through a fixed point. The coordinates of fixed point are

Answer 1

Dear student

3a+2b+4c = 0

Dividing the equation by 4

3a/4+2b/4+4c/4 = 0

=> 3a/4 + b/2+c = 0

x = ¾ and y = ½ satisfied this relation

Hence the set of lines are concurrent at (3/4, ½)

Answer 2

Answer:

The co-ordinates of fixed point through which the family of straight lines passes are $(\frac{1}{2},\frac{1}{3})$

Step-by-step explanation:

Given that,

The family of straight lines $3a+2b+6c=0$ of $ax+by+c=0$ passes through a fixed point and we are required to find the co-ordinates of the fixed point.

The equation of family of straight lines is

$3a+2b+6c=0$

Dividing the relation with '6' on both sides,we get

$\frac{3a+2b+6c}{6}=0= > \frac{a}{2}+\frac{b}{3}+c=0\\= > a(\frac{1}{2})+b(\frac{1}{3})+c=0$

Comparing this with the line $ax+by+c=0$ we get the co-ordinates as

$x=\frac{1}{2},y=\frac{1}{3}$

Therefore,

The co-ordinates of fixed point through which the family of straight lines passes are $(\frac{1}{2},\frac{1}{3})$

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