Question

Show that the path of a moving point, such that its distance from two lines \[3x - 2y = 5\] and \[3x + 2y = 5\] are equal is a straight line.

Answer 1

Hint: In the above question, we are given two straight lines \[3x - 2y = 5\] and \[3x + 2y = 5\] . We have to find the equation of a straight line that is equidistant from both the above-given lines. We can use the following formula to find the distance of a point on the new straight line from the given two lines.

The distance from a point \[(h,k)\;\] to a straight line \[ax + by + c = 0\] is given by the formula :
\[ \Rightarrow d = \dfrac{{\left| {ah + bk + c} \right|}}{{\sqrt {{a^2} + {b^{^2}}} }}\]

Complete step-by-step solution:
Given two straight lines are \[3x - 2y = 5\] and \[3x + 2y = 5\]
We can also write is as,
\[ \Rightarrow 3x - 2y - 5 = 0\] and \[ \Rightarrow 3x + 2y - 5 = 0\]
Let the point \[(h,k)\;\] be any point which is equidistant from both the lines \[3x - 2y - 5 = 0\] and \[3x + 2y - 5 = 0\] .
Now using the formula of distance between a point and a straight line, that is
\[ \Rightarrow d = \dfrac{{\left| {ah + bk + c} \right|}}{{\sqrt {{a^2} + {b^{^2}}} }}\]
Hence, the distance of point \[(h,k)\;\] from the straight line \[3x - 2y - 5 = 0\] is given by
\[ \Rightarrow {d_1} = \dfrac{{\left| {3h - 2k - 5} \right|}}{{\sqrt {{3^2} + {{\left( { - 2} \right)}^{^2}}} }}\]
Or,
\[ \Rightarrow {d_1} = \dfrac{{\left| {3h - 2k - 5} \right|}}{{\sqrt {9 + 4} }}\]
So,
\[ \Rightarrow {d_1} = \dfrac{{\left| {3h - 2k - 5} \right|}}{{\sqrt {13} }}\]
Similarly, the distance of point \[(h,k)\;\] from the straight line \[3x + 2y - 5 = 0\] is given by,
\[ \Rightarrow {d_2} = \dfrac{{\left| {3h + 2k - 5} \right|}}{{\sqrt {{3^2} + {2^{^2}}} }}\]
Or,
\[ \Rightarrow {d_2} = \dfrac{{\left| {3h + 2k - 5} \right|}}{{\sqrt {9 + 4} }}\]
So,
\[ \Rightarrow {d_2} = \dfrac{{\left| {3h + 2k - 5} \right|}}{{\sqrt {13} }}\]
Now since, \[(h,k)\;\] is equidistant from both the lines \[3x - 2y - 5 = 0\] and \[3x + 2y - 5 = 0\] ,
Therefore we can write,
\[ \Rightarrow {d_1} = {d_2}\]
That gives,
\[ \Rightarrow \dfrac{{\left| {3h - 2k - 5} \right|}}{{\sqrt {13} }} = \dfrac{{\left| {3h + 2k - 5} \right|}}{{\sqrt {13} }}\]
Cancelling the common denominators, we get
\[ \Rightarrow \left| {3h - 2k - 5} \right| = \left| {3h + 2k - 5} \right|\]
Removing the modulus sign, we can write
\[ \Rightarrow 3h - 2k - 5 = \pm \left( {3h + 2k - 5} \right)\]
Case 1: If \[3h - 2k - 5 = 3h + 2k - 5\]
Then, we get
\[ \Rightarrow 3h - 2k - 5 = 3h + 2k - 5\]
Removing the common terms, that gives
\[ \Rightarrow - 2k = 2k\]
Or,
\[ \Rightarrow - k = k\]
Therefore,
\[ \Rightarrow k = 0\]
Taking locus of \[k = 0\] we can write the equation of the straight line as
\[ \Rightarrow y = 0\]
Case 2: If \[3h - 2k - 5 = - \left( {3h + 2k - 5} \right)\]
Then we get,
\[ \Rightarrow 3h - 2k - 5 = - \left( {3h + 2k - 5} \right)\]
Or,
\[ \Rightarrow 3h - 2k - 5 = - 3h - 2k + 5\]
Removing the common term, we can write
\[ \Rightarrow 3h - 5 = - 3h + 5\]
That gives,
\[ \Rightarrow 6h = 10\]
Or,
\[ \Rightarrow h = \dfrac{{10}}{6}\]
Hence,
\[ \Rightarrow h = \dfrac{5}{3}\]
Taking locus of \[h = \dfrac{5}{3}\] we can write the equation of the straight line as,
\[ \Rightarrow x = \dfrac{5}{3}\]
Hence, the two required straight lines, which are equidistant from the straight lines \[3x - 2y = 5\] and \[3x + 2y = 5\] are \[y = 0\] and \[x = \dfrac{5}{3}\] .

Therefore, the path of a moving point, such that its distance from two lines \[3x - 2y = 5\] and \[3x + 2y = 5\] are equal is a straight line.

Note: We can also plot the graph of the above lines as,