Question

Let $ax + by + c = 0$ be a variable in a straight line, where a, b, and c are 1st, 3rd, and 7th terms of some increasing A.P. Then the variable straight line always passes through a fixed point which lies on
A) ${x^2} + {y^2} = 13$
B) ${x^2} + {y^2} = 5$
C) ${y^2} = 9x$
D) $3x + 4y = 9$

Answer 1

It is given in the question that $ax + by + c = 0$ be a variable in a straight line, where a, b, and c are $1^{\text{st}}, 3^{\text{rd}} \text{ and } 7^{\text{th}}$ terms of some increasing A.P.
Then, we will find that the variable straight line always passes through a fixed point which lies on?
Firstly, using formula ${T_n} = a + \left( {n - 1} \right)d$ find the value of b and c. Then put the value of b and c in $ax + \left( {a + 2} \right)y + \left( {a + 6} \right) = 0$
After that by assuming the value of x and y we will find if the equation is satisfied or not.
Then, if the equation is satisfied, we will find the value of r and by applying the equation of circle we will get our answer.

Complete step by step solution:
It is given in the question that $ax + by + c = 0$ be a variable in a straight line, where a, b, and c are $1^{\text{st}}, 3^{\text{rd}} \text{ and } 7^{\text{th}}$ terms of some increasing A.P.
Then, we have to find that the variable straight line always passes through a fixed point which lies on?
$ax + by + c = 0$ (I)
The above equation is given in the question
Here, a, b, and c are $1^{\text{st}}, 3^{\text{rd}} \text{ and } 7^{\text{th}}$ terms of increasing A.P.
$\therefore$By using formula ${T_n} = a + \left( {n - 1} \right)d$
$b = a + 2$ and $c = a + 6$
Now, put the values of b and c in the (I) equation
$\therefore ax + \left( {a + 2} \right)y + \left( {a + 6} \right) = 0$
Now, Assume $x = 2$ and $y = - 3$
put the value of x and y in the above equation.
$2a + \left( {a + 2} \right)\left( { - 3} \right) + a + 6 = 0 \\\ 2a + \left( { - 3a - 6} \right) + a + 6 = 0 \\\ 2a - 3a - 6 + a + 6 = 0 \\\ 2a - 2a = 0 \\\ 0 = 0 \\\$
$\therefore$On putting the values of x and y the above equation is satisfied.
Hence, the fixed point is (2,3) .
Here, one point is fixed and the other point is variable. So, a circle has formed.
$\because$ Radius of the circle is equal to the distance between the points (0,0) and (2,3).
$\therefore r = \sqrt {{2^2} + {3^2}} \\\ \therefore r = \sqrt {13}$
Now, the equation of the circle with Centre (0,0)and radius $= \sqrt {13}$
$\because$Equation of circle is ${x^2} + {y^2} = {r^2}$
Now, put the value of r in the equation of the circle.
$\therefore {x^2} + {y^2} = {\left( {\sqrt {13} } \right)^2}$
$\therefore {x^2} + {y^2} = 13$

$\therefore$ The variable straight line always passes through a fixed point which lies on ${x^2} + {y^2} = 13$.

Note:
The above question can solved with alternate method as follows:
It is given in the question that $ax + by + c = 0$ be a variable in a straight line, where a, b, and c are $1^{\text{st}}, 3^{\text{rd}} \text{ and } 7^{\text{th}}$ terms of some increasing A.P.
Then, we have to find that the variable straight line always passes through a fixed point which lies on?
a, b, c are $1^{\text{st}}, 3^{\text{rd}} \text{ and } 7^{\text{th}}$ terms of increasing A.P.
So, $b = a + \left( {3 - 1} \right)d = a + 2d$
$c = a + \left( {7 - 1} \right)d = a + 6d$
Then,
$3b - c = 2a$ (I)
$2a - 3b + c = 0$ (II)
Now,
$ax + by + c = 0$
It always passes through $\left( {2, - 3} \right)$
Then, $\left( {2, - 3} \right)$ lies on ${x^2} + {y^2} = 13$ .
The variable straight line always passes through a fixed point which lies on ${x^2} + {y^2} = 13$ .