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:
$b\left( \dfrac{-c-by}{a} \right)+ay+c=0$
$\Rightarrow -bc-{{b}^{2}}y+{{a}^{2}}y+ac=0$

& \Rightarrow y\left( {{a}^{2}}-{{b}^{2}} \right)=bc-ac \\\ & \Rightarrow y=\dfrac{-c(a-b)}{(a-b)(a+b)} \\\ \end{aligned}$$ Since, $${{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{aligned} & 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{aligned}$$ , Therefore, $$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{aligned} & \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{aligned}$$ Taking the LCM under the root, we get $$\begin{aligned} & \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{aligned}$$ Cancelling the like terms and cross multiplying, we get $$\begin{aligned} & a+b+c<2a+2b \\\ & \Rightarrow 0<2a-a+2b-b-c \\\ & \Rightarrow a+b-c>0 \\\ \end{aligned}$$ Therefore, $$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$$.