Question

If the lines $$ax + ky + 10 = 0$$, $$bx + \left( {k + 1} \right)y + 10 = 0$$ and $$cx + \left( {k + 2} \right)y + 10 = 0$$ are concurrent, then
A.a, b, c are in GP
B.a, b, c are in HP
C.a, b, c are in AP
D.$${\left( {a + 1} \right)^2} = c$$

Answer 1

Hint: Here in this question given a linear equation of lines, we need to find if the lines are concurrent then the co-efficient of the lines are in which sequence. For this, first we need to convert a given linear equation in the form of determinant and equate a determinant with zero then on further simplification and by the nature of the final answer we get the required solution.

Complete step-by-step answer:
Consider the given linear equations
$$ax + ky + 10 = 0$$ ---------(1)
$$bx + \left( {k + 1} \right)y + 10 = 0$$ -----(2)
$$cx + \left( {k + 2} \right)y + 10 = 0$$ -----(3)
Now rearrange the linear equation into determinant form.
$$ \Rightarrow \,\,\,\,\left| \begin {matrix}
   a & k & {10} \\
   b & {k + 1} & {10} \\
   c & {k + 2} & {10} \\

 \end{matrix} \right|$$
If these three lines are concurrent, the determinant of the coefficients should be equal to zero.
$$ \Rightarrow \,\,\,\,\left| \begin{matrix}
   a & k & {10} \\
   b & {k + 1} & {10} \\
   c & {k + 2} & {10} \\

 \end{matrix} \right| = 0$$
Now apply a row reduced echelon form
$${R_2} \to {R_2} - {R_1}$$
$${R_3} \to {R_3} - {R_1}$$
$$ \Rightarrow \,\,\,\,\left| \begin{matrix}
   a & k & {10} \\
   {b - a} & {k + 1 - k} & {10 - 10} \\
   {c - a} & {k + 2 - k} & {10 - 10} \\

 \end{matrix} \right| = 0$$
$$ \Rightarrow \,\,\,\,\left| \begin{matrix}
   a & k & {10} \\
   {b - a} & 1 & 0 \\
   {c - a} & 2 & 0 \\

 \end{matrix} \right| = 0$$
Now expand a determinant
$$ \Rightarrow \,\,\,\,a\left[ {1\left( 0 \right) - 2\left( 0 \right)} \right] - k\left[ {\left( {b - a} \right)\left( 0 \right) - \left( {c - a} \right)\left( 0 \right)} \right] + 10\left[ {\left( {b - a} \right)\left( 2 \right) - \left( {c - a} \right)\left( 1 \right)} \right] = 0$$
$$ \Rightarrow \,\,\,\,a\left[ 0 \right] - k\left[ 0 \right] + 10\left[ {2b - 2a - c + a} \right] = 0$$
On simplification, we have
$$ \Rightarrow \,\,\,10\left[ {2b - a - c} \right] = 0$$
Divide both side by 10, then we have
$$ \Rightarrow \,\,\,2b - a - c = 0$$
Add ‘a’ and ‘c’ on both side, then we get
 $$ \Rightarrow \,\,\,2b = a + c$$
Or
$$\therefore \,\,\,\,a + c = 2b$$
By the common difference of A.P
If a, b, c are in A.P, the arithmetic mean is given by $$b = \dfrac{{a + c}}{2}$$ then ‘$$2b = a + c$$’.
So, a, b, c are in AP.
Therefore, option (3) is the correct answer.
So, the correct answer is “Option 3”.

Note: Three straight lines are said to be concurrent if they pass through a point i.eThey meet at a point then the determinant of the coefficients of lines should be equal to zero. The use of row reduced echelon form makes the simplification of determinants simple.
The general arithmetic progression is of the form $$a,a + d,a + 2d,...$$ where a is first term nth d is the common difference which is same between the distance on any two number in sequence otherwise the fixed number that must be added to any term of an AP to get the next term is known as the common difference of the AP.