Question

If \[a,b,c\] are in AP, then the straight line \[ax + by + c = 0\] will always pass through the point
A. \[\left( { - 1, - 2} \right)\]
B. \[\left( {1, - 2} \right)\]
C. \[\left( { - 1,2} \right)\]
D. \[\left( {1,2} \right)\]

Answer 1

Hint: In this question, we first convert \[a,b,c\] into general AP terms then find a relation between them to get a relation between \[a,b,c\] and then convert the relation to a straight line with algebraic coefficients.

Complete step-by-step solution:
 We know that a sequence is said to be an AP if the difference between any two successive numbers is a constant.
Now we have \[a,b,c\] that are in AP. So, we have \[a\] as the first term of the sequence.
Now let us assume that \[d\] is a common difference.
So, from above the value of \[b\] becomes:
\[b = a + d\,...\left( 1 \right)\]
Now we need the value of \[c\]:
\[c = a + 2d\,...\left( 2 \right)\]
Now by simplifying equation \[\left( 1 \right)\] and subtracting \[a\] on both sides:
\[b - a = a + d - a\]
By simplifying, we get
\[
  b - a = d \\
  d = b - a\,...\left( 3 \right)
 \]
Now, substituting the value of \[d\] from the equation \[\left( 3 \right)\] into the equation \[\left( 2 \right)\], we get
\[c = a + 2\left( {b - a} \right)\]
By simplifying, we get
\[
  c = a + 2b - 2a \\
   = 2b - a
 \]
Now, by adding \[a\] on both sides, we get
\[c + a = 2b - a + a\]
By cancelling the common terms, we get
\[c + a = 2b\]
Now, by subtracting \[2b\] on both sides, we get
\[
  c + a - 2b = 2b - 2b \\
   = 0
 \]
Now, by arranging the above equation, we get
\[a - 2b + c = 0\,\,...\left( 4 \right)\]
Now the given equation is
 \[ax + by + c = 0\,...\left( 5 \right)\]
Now by comparing equation \[\left( 4 \right)\] and equation \[\left( 5 \right)\], we get
\[x = 1\,\] and \[y = - 2\]
Therefore, the point \[\left( {x,y} \right)\] through which it passes is \[\left( {1, - 2} \right)\].

Hence, option (B) is correct

Note: There is an alternate method of the above solution in which we have to convert them \[a,b,c\] into another form of AP which is
\[a,b,c\, = \left( {b - d} \right)\,,b,\,\left( {b + d} \right)\] and this method will give us our result with less calculation.