Question

A point equidistant from the lines 4x+3y+10 = 0, 5x-12y+26 = 0 and 7x+24y-50 = 0 is
[a] (1,-1)
[b] (1,1)
[c] (0,0)
[d] (0,1)

Answer 1

Hint: Assume that the coordinates of the point be P (x,y). Find the distance of P from 4x+3y+10=0, 5x-12y+26=0 and 7x+24y-50 = 0. Let those distances be ${{d}_{1}},{{d}_{2}}$ and ${{d}_{3}}$.
Equate ${{d}_{1}}$ and ${{d}_{2}}$ and form an equation in x and y.
Again equate ${{d}_{2}}$ and ${{d}_{3}}$ and form an equation in x and y.
Solve the system of the equations for x and y.
The value of x and y gives the value of the coordinates of point P. Notice that there should exist four such points which are the centres of the three excircles and one incircle of the triangle formed by these lines.

Complete step-by-step answer:
Let the coordinates of the point be P(x,y).
We know that the distance of the point $P\left( {{x}_{1}},{{y}_{1}} \right)$ from the line $Ax+By+C=0$ is given by $\dfrac{\left| A{{x}_{1}}+B{{y}_{1}}+C \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$
Hence we have
The distance of P from 4x+3y+10 = 0 is ${{d}_{1}}=\dfrac{\left| 4x+3y+10 \right|}{\sqrt{{{3}^{2}}+{{4}^{2}}}}=\dfrac{\left| 4x+3y+10 \right|}{5}$
The distance of P from 5x-12y+26 = 0 is ${{d}_{2}}=\dfrac{\left| 5x-12y+26 \right|}{\sqrt{{{5}^{2}}+{{12}^{2}}}}=\dfrac{\left| 5x-12y+26 \right|}{13}$
The distance of P from 7x+24y-50 = 0 is ${{d}_{3}}=\dfrac{\left| 7x+24y-50 \right|}{\sqrt{{{7}^{2}}+{{24}^{2}}}}=\dfrac{\left| 7x+24y-50 \right|}{25}$
Now we have
${{d}_{1}}={{d}_{2}}$
Hence we get
$\begin{aligned} & \dfrac{\left| 4x+3y+10 \right|}{5}=\dfrac{\left| 5x-12y+26 \right|}{13} \\\ & \Rightarrow 13\left| 4x+3y+10 \right|=5\left| 5x-12y+26 \right|\text{ (i)} \\\ \end{aligned}$
Also, ${{d}_{1}}={{d}_{3}}$
Hence we get
$\begin{aligned} & \dfrac{\left| 4x+3y+10 \right|}{5}=\dfrac{\left| 7x+24y-50 \right|}{25} \\\ & \Rightarrow 5\left| 4x+3y+10 \right|=\left| 7x+24y-50 \right|\text{ (ii)} \\\ \end{aligned}$
Now we know that if $a\left| x \right|=b\left| y \right|,a,b>0$ then $ax=\pm by$
Hence equation (i) becomes $13\left( 4x+3y+10 \right)=\pm 5\left( 5x-12y+26 \right)\text{ }$
Taking with + sign, we get
$\begin{aligned} & 52x+39y+130=25x-60y+130 \\\ & \Rightarrow 27x+99y=0\text{ (A)} \\\ \end{aligned}$
Taking with the – sign, we get

& 52x+39y+130=-25x+60y-130 \\\ & \Rightarrow 77x-21y+260=0\text{ (B)} \\\ \end{aligned}$$ Also from equation (ii), we have $$5\left( 4x+3y+10 \right)=\pm \left( 7x+24y-50 \right)$$ Taking with + sign, we get $\begin{aligned} & 20x+15y+50=7x+24y-50 \\\ & \Rightarrow 13x-9y+100=0\text{ (C)} \\\ \end{aligned}$ Taking with the – sign, we get $\begin{aligned} & 20x+15y+50=-7x-24y+50 \\\ & \Rightarrow 27x+39y=0\text{ (D)} \\\ \end{aligned}$  Solving system A and C $\begin{aligned} & 27x+99y=0\text{ } \\\ & 13x-9y+100=0\text{ } \\\ \end{aligned}$ Multiply equation B by 11 and adding to equation B, we get $\begin{aligned} & 27x+143x+99y-99y+1100=0 \\\ & 170x+1100=0 \\\ & \Rightarrow x=\dfrac{-110}{17} \\\ \end{aligned}$ Substituting the value of x in equation A, we get $\begin{aligned} & 27\left( \dfrac{-110}{17} \right)+99y=0 \\\ & \Rightarrow y=\dfrac{30}{17} \\\ \end{aligned}$ Hence one point is $\left( \dfrac{-110}{17},\dfrac{30}{17} \right)$. Similarly solving system A and D, we get $\begin{aligned} & 27x+99y=0 \\\ & 27x+39y=0 \\\ \end{aligned}$ Subtracting equation D from equation A, we get $\begin{aligned} & 27x-27x+99y-39y=0 \\\ & \Rightarrow 60y=0 \\\ & \Rightarrow y=0 \\\ \end{aligned}$ Substituting the value of y in equation A, we get $\begin{aligned} & 27x+0=0 \\\ & \Rightarrow x=0 \\\ \end{aligned}$ Hence another point is (0,0) Solving the system B and C, we get $\begin{aligned} & 77x-21y+260=0\text{ } \\\ & 13x-9y+100=0\text{ } \\\ \end{aligned}$ Multiplying equation B by 3 and equation C by 7 and adding the two equations, we get $\begin{aligned} & 231x-91x-63x+63x+780-700=0 \\\ & \Rightarrow 140x+80=0 \\\ & \Rightarrow x=\dfrac{-80}{140}=\dfrac{-4}{7} \\\ \end{aligned}$ Substituting the value of x in equation B, we get $\begin{aligned} & 77\left( \dfrac{-4}{7} \right)-21y+260=0 \\\ & \Rightarrow y=\dfrac{72}{7} \\\ \end{aligned}$ Hence another point is $\left( \dfrac{-4}{7},\dfrac{72}{7} \right)$ Solving system B and D, we get $$\begin{aligned} & 77x-21y+260=0\text{ } \\\ & 27x+39y=0 \\\ \end{aligned}$$ Multiplying equation B by 13 and equation D by 7 and adding the two equations, we get $\begin{aligned} & 1001x+189x-273y+273y+3380=0 \\\ & \Rightarrow 1190x+3380=0 \\\ & \Rightarrow x=\dfrac{-3380}{1190}=-\dfrac{338}{119} \\\ \end{aligned}$ Substituting the value of x in equation D, we get $\begin{aligned} & 27\left( \dfrac{-338}{119} \right)+39y=0 \\\ & \Rightarrow y=\dfrac{234}{119} \\\ \end{aligned}$ Hence another point is $\left( \dfrac{-338}{119},\dfrac{234}{119} \right)$ Hence the points equidistant from the given lines are $\left( \dfrac{-110}{17},\dfrac{30}{17} \right),\left( 0,0 \right),\left( \dfrac{-4}{7},\dfrac{72}{7} \right)$ and $\left( \dfrac{-338}{119},\dfrac{234}{119} \right)$ Hence option [c] is correct. Note: Alternative Solution Find the coordinates of points of intersection of the lines. Hence find the lengths of the sides of the triangle formed by these lines. Find the coordinates of incentre by using the formula $$\left( \dfrac{a{{x}_{1}}+b{{x}_{2}}+c{{x}_{3}}}{a+b+c},\dfrac{a{{y}_{1}}+b{{y}_{2}}+c{{y}_{3}}}{a+b+c} \right)$$ and find the coordinates of excentres using the formula ${{I}_{1}}\equiv \left( \dfrac{-a{{x}_{1}}+b{{x}_{2}}+c{{x}_{3}}}{-a+b+c},\dfrac{-a{{y}_{1}}+b{{y}_{2}}+c{{y}_{3}}}{-a+b+c} \right),{{I}_{2}}\equiv \left( \dfrac{a{{x}_{1}}-b{{x}_{2}}+c{{x}_{3}}}{a-b+c},\dfrac{a{{y}_{1}}-b{{y}_{2}}+c{{y}_{3}}}{a-b+c} \right)$ and ${{I}_{3}}\equiv \left( \dfrac{a{{x}_{1}}+b{{x}_{2}}-c{{x}_{3}}}{a+b-c},\dfrac{a{{y}_{1}}+b{{y}_{2}}-c{{y}_{3}}}{a+b-c} \right)$, where a is the length of the side opposite to $A\left( {{x}_{1}},{{y}_{1}} \right)$, b is the length of the side opposite to $B\left( {{x}_{2}},{{y}_{2}} \right)$ and c is the length of the side opposite to $C\left( {{x}_{3}},{{y}_{3}} \right)$.