Question

Find the locus of a point so that its distance from the point (3,0) is three times its distance from (0,2).

Answer 1

Hint: Take the point P as (h, k) use the distance formula which is, $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ and take the two points as A and B. Then apply the condition that is PA=3PB as the point P is three times its distance from the other point and find the result.

Complete step-by-step answer:

Locus is a set of points making a curve or surface that satisfy conditions like equidistance. We have been given points A(3,0) and B(0,2). Let us assume a point P(h,k) which satisfies the given condition of being as three times distant from B(0,2) as distant from A(3,0).

Mathematically,
$PA=3PB..............(i)$

Using the distance formula for points,
$d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}...........(ii)$

Substituting the values of the coordinates from A(3,0) and P(h,k) in equation (ii), we have
$\begin{aligned} & PA=\sqrt{{{\left( h-3 \right)}^{2}}+{{\left( k-0 \right)}^{2}}} \\\ & \Rightarrow PA=\sqrt{{{\left( h-3 \right)}^{2}}+{{k}^{2}}}.........(iii) \\\ \end{aligned}$

Similarly, substituting the values of the coordinates from B(0, 2) and P(h, k) in equation (ii), we have
$\begin{aligned} & PB=\sqrt{{{\left( h-0 \right)}^{2}}+{{\left( k-2 \right)}^{2}}} \\\ & \Rightarrow PB=\sqrt{{{h}^{2}}+{{\left( k-2 \right)}^{2}}}.........(iv) \\\ \end{aligned}$

Substituting the values of PA and PB in equation (i), we get
$\sqrt{{{\left( h-3 \right)}^{2}}+{{k}^{2}}}=3\sqrt{{{h}^{2}}+{{\left( k-2 \right)}^{2}}}$

On squaring both sides, we get
$\begin{aligned} & {{\left( h-3 \right)}^{2}}+{{k}^{2}}={{\left( 3 \right)}^{2}}\left[ {{h}^{2}}+{{\left( k-2 \right)}^{2}} \right] \\\ & \Rightarrow {{\left( h-3 \right)}^{2}}+{{k}^{2}}=9\left[ {{h}^{2}}+{{\left( k-2 \right)}^{2}} \right].............(v) \\\ \end{aligned}$

Now we will use the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ we can write equation (v) as,
$\begin{aligned} & {{h}^{2}}-6h+9+{{k}^{2}}=9\left[ {{h}^{2}}+{{k}^{2}}-4k+4 \right] \\\ & \Rightarrow {{h}^{2}}-6h+9+{{k}^{2}}=9{{h}^{2}}+9{{k}^{2}}-36k+36 \\\ & \Rightarrow 9{{h}^{2}}+9{{k}^{2}}-36k+36-{{h}^{2}}+6h-9-{{k}^{2}}=0 \\\ & \Rightarrow 8{{h}^{2}}+8{{k}^{2}}+6h-36k+27=0..........(vi) \\\ \end{aligned}$

The equation (vi) satisfies a single locus point P for the given condition. Now for generalisation for all the points satisfying the given condition we can replace (h, k) with general variables (x, y). Therefore the required equation is
$8{{x}^{2}}+8{{y}^{2}}+6x-36y+27=0$

The required locus for the given condition is $8{{x}^{2}}+8{{y}^{2}}+6x-36y+27=0$.

Note: The chances of mistakes are if while using the distance formula you are only calculating the distance of AB. We need to find the locus of a point that satisfies the given condition for the points (3,0) and (2,0).