Question

The lines ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ cuts the coordinate axis in co-cyclic points then $\left| {{a_1}{a_2}} \right| = \left| {{b_1}{b_2}} \right|$.
A.True
B.False

Answer 1

We know that if two lines intersect the coordinate axis in concyclic points then they are at the same distance from the origin of the circle. First find the $x$-intercepts by substituting $y = 0$ and $y$-intercepts by substituting $x = 0$ of both the lines.

Complete step-by-step answer:
Let us consider that the lines be $m = {a_1}x + {b_1}y + {c_1} = 0$ and $n = {a_2}x + {b_2}y + {c_2} = 0$.
Put $y = 0$ in the equation of line $m$ to find $x$-intercept.
${a_1}x + {b_1}\left( 0 \right) + {c_1} = 0$
$\Rightarrow$ ${a_1}x + {c_1} = 0$
$\Rightarrow$ $x = - \dfrac{{{c_1}}}{{{a_1}}}$
Put $x = 0$ in the equation of line $m$ to find $y$-intercept.
${a_1}\left( 0 \right) + {b_1}y + {c_1} = 0$
$\Rightarrow$ ${b_1}y + {c_1} = 0$
$\Rightarrow$ $y = - \dfrac{{{c_1}}}{{{b_1}}}$
Put $y = 0$ in the equation of line $n$ to find $x$-intercept.
${a_2}x + {b_2}\left( 0 \right) + {c_2} = 0$
$\Rightarrow$ ${a_2}x + {c_2} = 0$
$\Rightarrow$ $x = - \dfrac{{{c_2}}}{{{a_2}}}$
Put $x = 0$ in the equation of line $n$ to find $y$-intercept.
${a_2}\left( 0 \right) + {b_2}y + {c_2} = 0$
$\Rightarrow$ ${b_2}y + {c_2} = 0 = 0$
$\Rightarrow$ $y = - \dfrac{{{c_2}}}{{{b_2}}}$
We get the concyclic points that the cut the coordinate axis at $A\left( { - \dfrac{{{c_1}}}{{{a_1}}},0} \right)$, $B\left( {0, - \dfrac{{{c_1}}}{{{b_1}}}} \right)$, $C\left( { - \dfrac{{{c_2}}}{{{a_2}}},0} \right)$ and $D\left( {0, - \dfrac{{{c_2}}}{{{b_2}}}} \right)$.
Let the center of the circle be the origin $O\left( {0,0} \right)$. The distance of point $A$ from the origin is:
$\Rightarrow$ $OA = \dfrac{{{c_1}^2}}{{{a_1}^2}}$
The distance of point $B$ from the origin is:
$\Rightarrow$ $OB = \dfrac{{{c_1}^2}}{{{b_1}^2}}$
The distance of point $C$ from the origin is:
$\Rightarrow$ $OA = \dfrac{{{c_1}^2}}{{{a_1}^2}}$
The distance of point $D$ from the origin is:
$\Rightarrow$ $OD = \dfrac{{{c_2}^2}}{{{b_2}^2}}$
Now, solve for $OA \cdot OC = OB \cdot OD$.
$\Rightarrow \dfrac{{{c_1}^2}}{{{a_1}^2}} \cdot \dfrac{{{c_2}^2}}{{{a_2}^2}} = \dfrac{{{c_1}^2}}{{{b_1}^2}} \cdot \dfrac{{{c_2}^2}}{{{b_2}^2}}$
$\Rightarrow \dfrac{1}{{{a_1}^2}} \cdot \dfrac{1}{{{a_2}^2}} = \dfrac{1}{{{b_1}^2}} \cdot \dfrac{1}{{{b_2}^2}}$
$\Rightarrow {a_1}^2 \cdot {a_2}^2 = {b_1}^2 \cdot {b_2}^2$
$\Rightarrow \left| {{a_1}{a_2}} \right| = \left| {{b_1}{b_2}} \right|$
Therefore, the given statement is true.
So, the option (A) is correct.

Note: If a set of points lies on the same circle, they are said to be concyclic points. Also, the distance of points on the circle will be the same from its origin. Correctly find the $x$-intercept and $y$-intercept of both the given lines, otherwise this may lead to the incorrect answer.