Question

If a and b are two arbitrary constants, then the straight line $\left( a-2b \right)x+\left( a+3b \right)y+3a+4b=0$ will pass through:
(1) $\left( -1,-2 \right)$
(2) $\left( 1,2 \right)$
(3) $\left( -2,-3 \right)$
(4) $\left( 2,3 \right)$

Answer 1

Hint: From the question, we have been asked to find the points through which the given straight line $\left( a-2b \right)x+\left( a+3b \right)y+3a+4b=0$ passes consider that $a$ and $b$ are two arbitrary constants. Now for answering this question we will simplify the given equation of straight line in the form of ${{L}_{1}}+\lambda {{L}_{2}}=0$ and get the intersection point of ${{L}_{1}}$ and ${{L}_{2}}$ .

Complete step by step solution:
Now considering from the question we have been asked to find the point through which the given straight line $\left( a-2b \right)x+\left( a+3b \right)y+3a+4b=0$ passes, consider that $a$ and $b$ are two arbitrary constants.
Now for answering this question we will simplify the given equation of straight line in the form of ${{L}_{1}}+\lambda {{L}_{2}}=0$ and get the intersection point of ${{L}_{1}}$ and ${{L}_{2}}$ .
By simplifying the expression we will have
$\begin{align}
  & \left( a-2b \right)x+\left( a+3b \right)y+3a+4b=0 \\
 & \Rightarrow a\left( x+y+3 \right)+b\left( -2x+3y+4 \right)=0 \\
\end{align}$ .
Now we need to find the intersection point of $x+y+3=0$ and $-2x+3y+4=0$ .
For solving both the expressions we will multiply the first expression $x+y+3=0$ with 2 and then add both the expressions. By doing that we will have
$\begin{align}
  & \Rightarrow 2\left( x+y+3 \right)-2x+3y+4=0 \\
 & \Rightarrow 2y+3y+6+4=0 \\
 & \Rightarrow 5y+10=0 \\
 & \Rightarrow y=-2 \\
\end{align}$ .
Now by substituting $y=-2$ in the expression $x+y+3=0$ we will have
$\begin{align}
  & x+\left( -2 \right)+3=0 \\
 & \Rightarrow x=-1 \\
\end{align}$ .
Therefore we can conclude that the given straight line $\left( a-2b \right)x+\left( a+3b \right)y+3a+4b=0$ passes through the point $\left( -1,-2 \right)$, consider that $a$ and $b$ are two arbitrary constants.
Hence we will mark the option “1” as correct.

Note: During the process of answering questions of this type, we should be sure with the calculations that we are going to perform in between the steps. This question can also be answered by substituting the given options in the given equation of the straight line and verifying. For example, let us consider the point $\left( 1,2 \right)$ and verify if it passes through the given expression or not. By doing that we will have
$\begin{align}
  & \Rightarrow \left( a-2b \right)+2\left( a+3b \right)+3a+4b=0 \\
 & \Rightarrow a-2b+2a+6b+3a+4b=0 \\
 & \Rightarrow 6a+8b=0 \\
\end{align}$ .
This implies that the point is not satisfying the given condition.