Question

If the straight line \[ax + by + c = 0\] always passes through \[\left( {1, - 2} \right)\], then $a,b,c$ are:
1) In A.P.
2) In H.P.
3) In G.P.
4) None of these

Answer 1

Hint: In order to solve this question, we will substitute \[1\] in place of \[x\] and \[ - 2\] in place of $y$ in the equation of line \[ax + by + c = 0\]. After that, we will add $2b$ each side of the obtained expression and cancel out the equal-like term to simplify the obtained expression.

Complete step-by-step solution:
Since, the given equation of line is \[ax + by + c = 0\] and the given point is \[\left( {1, - 2} \right)\] from which the line passes.
If the line passes through any point, the point will satisfy the equation of the line.
Here, we will substitute \[1\] in place of \[x\] and \[ - 2\] in place of $y$ in the equation of line \[ax + by + c = 0\] as:
\[ \Rightarrow 1 \times a + \left( { - 2} \right)b + c = 0\]
Now, we will simplify the equation using multiplication with variables.
\[ \Rightarrow a - 2b + c = 0\]
Then, we will add $2b$ each side of the obtained expression and will simplify it.
\[ \Rightarrow a - 2b + c + 2b = 0 + 2b\]
Here, we will cancel out the term $2b$ from the left side of obtained expression and simplify it using addition and subtraction.
\[ \Rightarrow a + c = 2b\]
Now, we will divide by $2$ in the obtained expression and simplify it.
\[ \Rightarrow \dfrac{{a + c}}{2} = \dfrac{{2b}}{2}\]
After that, we will cancel out the term $2$ from the right side of the obtained expression and simplify it.
\[ \Rightarrow \dfrac{{a + c}}{2} = b\]
Since, the obtained expression denotes that $a,b,c$ are in A.P, the correct option is 1.

Note: A.P. is an order of a series of a number in which the mid term is the average of the first and last term. Let, $a,b,c$ are a series in A.P of three numbers in which $b$ is mid term. Then, $ \Rightarrow b = \dfrac{{a + c}}{2}$.