Question

The line parallel to the x-axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx - 2ay - 3a = 0, where (a, b) is not origin.
\[\begin{align}
  & \left( \text{a} \right)\text{Above the X-axis at a distance of }\dfrac{\text{3}}{\text{2}}\text{ from it} \\
 & \left( \text{b} \right)\text{Above the X-axis at a distance of }\dfrac{2}{3}\text{ from it} \\
 & \left( \text{c} \right)\text{Below the X-axis at a distance of }\dfrac{\text{3}}{\text{2}}\text{ from it} \\
 & \left( \text{d} \right)\text{Below the X-axis at a distance of }\dfrac{2}{3}\text{ from it} \\
\end{align}\]

Answer 1

Hint: If a line passes through the intersection of two other lines then they are called a family of lines. So find the value of the variable from the equation below. By substituting the variable back you will get the equation of line. In the equation, when you substitute the x value to be 0, the y value you get will be the distance of the line below or above the x axis. By looking at the sign of y, negative sign indicates below the axis and positive sign indicates above the axis. However it is given that the line is parallel to the x axis so you won’t get any x in the equation. The value of y you get will itself become the distance from the axis. If a line passes through intersection of lines A and B, then every line of the family can be written as: \[A+\lambda .B\].

Complete step-by-step answer:
We require the equation of a line which passes through the intersection of the lines ax + 2by + 3b = 0 and bx - 2ay - 3a = 0, where (a, b) is not origin.
Let A be the straight line with line equation ax + 2by + 3b = 0
B be the straight line with line equation bx - 2ay - 3a = 0.
 So the required equation is also in the family of A, B.
Let the L be the required line equation.
\[L\equiv A+\lambda .B=0\]
By substituting values of A, B in the equation, we get:
\[L\equiv \left( ax+2by+3b \right)+\lambda \left( bx\text{ }-\text{ }2ay\text{ }-\text{ }3a \right)=0.....\left( 1 \right)\]
By simplifying, we get:
\[L\equiv (a+b\lambda )x+(2b-2a\lambda )y+3b-3\lambda a=0\]
As the line is parallel to X-axis the coefficient of x must be 0.
Equating the x-co efficient to 0, we get:
\[\begin{align}
  & a+b\lambda =0~ \\
 & ~\lambda =\left( -\dfrac{a}{b} \right)\text{ }.....\left( 2 \right) \\
\end{align}\]
By substituting equation (2) in equation (1), we get:
\[ax+2by+3b+\left( -\dfrac{a}{b} \right)\left( bx-2ay-3a \right)=0\]
By simplifying, we get:
\[\left( b \right)\left( ax+2by+3b \right)+\left( -a \right)\left( bx-2ay-3a \right)=0\]
\[abx+2{{b}^{2}}y+3{{b}^{2}}-abx+2{{a}^{2}}y+3{{a}^{2}}=0\]
By cancelling common terms and simplifying, we get:
\[\begin{align}
  & 2\left( {{a}^{2}}+{{b}^{2}} \right)y+3{{b}^{2}}+3{{a}^{2}}=0 \\
 & y=-\dfrac{3\left( {{a}^{2}}+{{b}^{2}} \right)}{2\left( {{a}^{2}}+{{b}^{2}} \right)}=-\dfrac{3}{2} \\
\end{align}\]
So as the y value is negative it lies below X-axis.
And the magnitude of y is:
\[\therefore \text{The required line is below x-axis by }\dfrac{3}{2}\text{ units}\]
So option (c) is correct.

Note: Alternate method is to find the intersection point of two lines in terms of a, b, then use the condition that the line is parallel to the x axis to find slope and then solve the question.