Question

Locus of the perpendicular lines one belonging to family $\left( x+y-2 \right)+\lambda \left( 2x+3y-5 \right)=0$ and other belonging to family $\left( 2x+y-11 \right)+\lambda \left( x+2y-13 \right)=0$ is a
(a) Circle
(b) Straight line
(c) Pair of lines
(d) None

Answer 1

First we find the point of intersection of individual lines. That is, we find the intersection of two line equations given in the first line between $\left( x+y-2 \right)$ and $\left( 2x+3y-5 \right)$. Then we find the intersection of the second line between $\left( 2x+y-11 \right)$ and $\left( x+2y-13 \right)$. Then we assume that the point of intersection of two families is $\left( h,k \right)$. Since these lines are perpendicular, the product of slopes from $\left( h,k \right)$ to two points of intersections will be ‘-1’. Then we get required locus.

Complete step-by-step solution:
Let us consider the first family $\left( x+y-2 \right)+\lambda \left( 2x+3y-5 \right)=0$
Let us assume
$\left( x+y-2 \right)=0.......equation(i)$
$\left( 2x+3y-5 \right)=0.......equation(ii)$
Now let us solve equation (i) and equation (ii)
By multiplying equation (i) with ‘3’ and subtract equation (ii) from equation (i) we get

& \Rightarrow \left( 2x+3y-5 \right)-3\left( x+y-2 \right)=0 \\\ & \Rightarrow -x+1=0 \\\ & \Rightarrow x=1 \\\ \end{aligned}$$ By substituting value of $$x$$ in equation (i) we get $$\begin{aligned} & \Rightarrow 1+y-2=0 \\\ & \Rightarrow y=1 \\\ \end{aligned}$$ So, let us assume that first point of intersection as $$P=\left( 1,1 \right)$$ Now let us take second family $$\left( 2x+y-11 \right)+\lambda \left( x+2y-13 \right)=0$$ Let us assume that $$2x+y-11=0.......equation(iii)$$ $$x+2y-13=0........equation(iv)$$ By multiplying equation (iii) by ‘2’ and subtract equation (iv) from equation (iii) we get $$\begin{aligned} & \Rightarrow 2\left( 2x+y-11 \right)-\left( x+2y-13 \right)=0 \\\ & \Rightarrow 3x-9=0 \\\ & \Rightarrow x=3 \\\ \end{aligned}$$ By substituting value of $$x$$ in equation (iii) we get $$\begin{aligned} & \Rightarrow 2\left( 3 \right)+y-11=0 \\\ & \Rightarrow y=5 \\\ \end{aligned}$$ Now let us assume that the second point of intersection as $$Q=\left( 3,5 \right)$$ Now let us assume that the point of intersection of two families as $$R=\left( h,k \right)$$ Since the families are perpendicular to each other we can write $$\begin{aligned} & \Rightarrow \left( \text{slope of PR} \right)\left( \text{slope of QR} \right)=-1 \\\ & \Rightarrow {{m}_{1}}\times {{m}_{2}}=-1............equation(v) \\\ \end{aligned}$$ We know that if $$A\left( {{x}_{1}},{{y}_{1}} \right)$$ and $$B\left( {{x}_{2}},{{y}_{2}} \right)$$ are two points then slope of AB is $$m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$$ By using this slope formula let us find slope of PR as $$\Rightarrow {{m}_{1}}=\dfrac{k-1}{h-1}$$ Now let us find the slope of QR as $$\Rightarrow {{m}_{2}}=\dfrac{k-5}{h-3}$$ Now by substituting the slopes in equation (v) we get $$\begin{aligned} & \Rightarrow \left( \dfrac{k-1}{h-1} \right)\left( \dfrac{k-5}{h-3} \right)=-1 \\\ & \Rightarrow \dfrac{{{k}^{2}}-6k+5}{{{h}^{2}}-4h+3}=-1 \\\ \end{aligned}$$ By cross multiplying and rearranging the terms we get $$\Rightarrow \left( {{k}^{2}}-6k \right)+\left( {{h}^{2}}-4h \right)+8=0$$ Let us add ‘9’ to $$k$$ terms and ‘4’ to $$h$$ terms and add ‘13’ on other side to make perfect square then we get $$\begin{aligned} & \Rightarrow \left( {{k}^{2}}-6k+9 \right)+\left( {{h}^{2}}-4h+4 \right)=13-8 \\\ & \Rightarrow {{\left( k-3 \right)}^{2}}+{{\left( h-2 \right)}^{2}}=5 \\\ \end{aligned}$$ Therefore by replacing $$h$$ by $$x$$ and $$k$$ by $$y$$ we get the locus of $$R=\left( h,k \right)$$ as $${{\left( x-2 \right)}^{2}}+{{\left( y-3 \right)}^{2}}=5$$ We know that this equation is the general form of a circle. The locus of points of intersection of given two families is a circle. **Note:** There will be some students who can make mistakes in solving the two equations in finding the point of intersection. Students need to take care in the calculations part. Also, in the locus part leaving the equation at $${{\left( k-3 \right)}^{2}}+{{\left( h-2 \right)}^{2}}=5$$ is wrong. You need to replace $$h$$ by $$x$$ and $$k$$ by $$y$$ and write $${{\left( x-2 \right)}^{2}}+{{\left( y-3 \right)}^{2}}=5$$ as the locus. Leaving there is not a complete solution.