Question

For \[a>b>c>0\] the distance between \[(1,1)\]and the point of intersection of lines\[ax+by+c=0\] and \[bx+ay+c=0\]is less than \[2\sqrt{2}\] then,
(a) \[a+b-c>0\]
(b) \[a-b+c<0\]
(c) \[a-b+c>0\]
(d) \[a+b-c<0\]

Answer 1

Hint: Find x from \[ax+by+c=0\] and substitute in the line , \[bx+ay+c=0\] to find the intersection point of both the lines. After finding the intersection point, apply the distance formula to the given condition.

Complete step by step solution:
The given line equations are:
\[ax+by+c=0......1\]
\[bx+ay+c=0.......2\]
Now to find the intersection point of them, let us solve equations (1) & (2) in the manner below.
From equation 1, we can have,
\[ax+by+c=0\]
Rearranging for finding \[x\], we have:
\[ax=-by-c\]
And therefore, \[\ x=\dfrac{-by-c}{a}\].
Substituting \[\ x=\dfrac{-by-c}{a}\] in equation 2, we get:
\[b\left( \dfrac{-c-by}{a} \right)+ay+c=0\]
\[\Rightarrow -bc-{{b}^{2}}y+{{a}^{2}}y+ac=0\]
\[\begin{align}
  & \Rightarrow y\left( {{a}^{2}}-{{b}^{2}} \right)=bc-ac \\
 & \Rightarrow y=\dfrac{-c(a-b)}{(a-b)(a+b)} \\
\end{align}\]
Since, \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\].
Therefore, \[y=\dfrac{-c}{a+b}\]
As, we have found out the y-coordinate of the intersection point, we shall also determine the x-coordinate accordingly.
Therefore, from:
\[x=\dfrac{-by-c}{a},\]
Substituting \[y=\dfrac{-c}{a+b}\], in the above equation, we have:
\[\begin{align}
  & x=\dfrac{-b\left( \dfrac{-c}{a+b} \right)-c}{a} \\
 & \Rightarrow x=\dfrac{bc-c(a+b)}{a(a+b)} \\
 & \Rightarrow x=\dfrac{c(b-a-b)}{a(a+b)} \\
 & \Rightarrow x=\dfrac{-ac}{a(a+b)} \\
\end{align}\] ,
Therefore, \[x=\dfrac{-c}{a+b}\].
Thus, the x-coordinate of the intersection point is found to be, \[x=\dfrac{-c}{a+b}\].
So, the point of intersection of the given lines,\[ax+by+c=0\] and \[bx+ay+c=0\] is found out to be \[\left( \dfrac{-c}{a+b},\dfrac{-c}{a+b} \right)\].
Given, that distance between \[(1,1)\] and point of intersection, \[\left( \dfrac{-c}{a+b},\dfrac{-c}{a+b} \right)\] is less than \[2\sqrt{2}\], then we have:
Applying the distance formula $\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}$, between \[(1,1)\] and \[\left( \dfrac{-c}{a+b},\dfrac{-c}{a+b} \right)\] according to the condition, we have
\[\begin{align}
  & \sqrt{{{\left( 1+\dfrac{c}{a+b} \right)}^{2}}+{{\left( 1+\dfrac{c}{a+b} \right)}^{2}}}<2\sqrt{2} \\
 & \sqrt{2{{\left( 1+\dfrac{c}{a+b} \right)}^{2}}}<2\sqrt{2} \\
\end{align}\]
Taking the LCM under the root, we get
\[\begin{align}
  & \sqrt{2{{\left( \dfrac{a+b+c}{a+b} \right)}^{2}}}<2\sqrt{2} \\
 & \left( \dfrac{a+b+c}{a+b} \right)\sqrt{2}<2\sqrt{2} \\
\end{align}\]
Cancelling the like terms and cross multiplying, we get
\[\begin{align}
  & a+b+c<2a+2b \\
 & \Rightarrow 0<2a-a+2b-b-c \\
 & \Rightarrow a+b-c>0 \\
\end{align}\]
Therefore, \[a+b-c>0\] is the required condition when the distance between \[(1,1)\] and the point of intersection of lines \[ax+by+c=0\] and \[bx+ay+c=0\] is less than \[2\sqrt{2}\], for \[a>b>c>0\].

Hence, option (a) is the correct answer.

Note: While solving for the conditions of intersections points for \[ax+by+c=0\] and \[bx+ay+c=0\], do not get confused with their coefficients. As the x-coordinate and the y-coordinate of the intersection point is the same, we can also say that the intersection of these two points lie on the line \[x=y\].