A point moves in such a way that it remains equidistant from each of the lines \(3x±2y= 5\). Then the path along which the point moves is?
A point moves in such a way that it remains equidistant from each of the lines \(3x±2y= 5\). Then the path along which the point moves is?
\(x=\dfrac{5}{3}\)
\(y=\dfrac{5}{3}\)
\(x=\dfrac{-5}{3}\)
\(x=0\)
\(y=\dfrac{-5}{3}\)
The Correct Option is A
Approach Solution - 1
Given that
The point is equidistant from the given two lines
i.e. ; 3x−2y−5=0 and 3x+2y−5=0
Now,
Let P(h, k) be a moving point such that it is equidistant from the lines 3x−2y−5=0 then
\(∣\dfrac{3h−2k−5}{(√9+4)}∣=∣\dfrac{3h+2k−5}{(√9+4)}∣\)
\(|3h−2k−5|=|3h+2k−5|\)
\(4k=0\) or \(6h−10=0\)
\(k=0\) or \(3h=5\)
therefore, \(h=\dfrac{5}{3}\)
Hence, the path of the moving points are \(y=0\) or \(3x=5\) (\(x=\dfrac{5}{3}\)), which are straight lines (Ans.)
Approach Solution -2
To find the path of a point that remains equidistant from the lines \(3x + 2y = 5\) and \(3x - 2y = 5\), we equate the distances from the point \((x, y)\) to both lines.
The distance from \((x, y)\) to \(3x + 2y - 5 = 0\) is:
\[ \frac{|3x + 2y - 5|}{\sqrt{3^2 + 2^2}} \]The distance to \(3x - 2y - 5 = 0\) is:
\[ \frac{|3x - 2y - 5|}{\sqrt{3^2 + (-2)^2}} \]Setting them equal:
\[ |3x + 2y - 5| = |3x - 2y - 5| \]This gives two cases:
- \(3x + 2y - 5 = 3x - 2y - 5\) leads to \(y = 0\).
- \(3x + 2y - 5 = - (3x - 2y - 5)\) simplifies to \(x = \frac{5}{3}\).
Only \(x = \frac{5}{3}\) appears in the options. Verifying:
For \(x = \frac{5}{3}\), the distances to both lines are:
\[ \frac{|2y|}{\sqrt{13}} \quad \text{and} \quad \frac{|2y|}{\sqrt{13}} \]Thus, the correct path is \(x = \frac{5}{3}\), corresponding to option A.
Approach Solution -3
Given lines:
\(L_1: 3x + 2y - 5 = 0\)
\(L_2: 3x - 2y - 5 = 0\)
Condition for equidistance:
The distance from a point \((x, y)\) to \(L_1\) and \(L_2\) must be equal. The distance formula gives:
\[\frac{|3x + 2y - 5|}{\sqrt{3^2 + 2^2}} = \frac{|3x - 2y - 5|}{\sqrt{3^2 + (-2)^2}}\]
Since the denominators are equal (\(\sqrt{13}\)), we simplify to:
\[|3x + 2y - 5| = |3x - 2y - 5|\]
Solving the absolute value equation:
This implies two cases:
Case 1: Positive equality
\[3x + 2y - 5 = 3x - 2y - 5\]
Simplify:
\[2y = -2y \quad \Rightarrow \quad 4y = 0 \quad \Rightarrow \quad y = 0\]
Case 2: Negative equality
\[3x + 2y - 5 = -(3x - 2y - 5)\]
Simplify:
\[3x + 2y - 5 = -3x + 2y + 5\]
\[6x = 10 \quad \Rightarrow \quad x = \frac{5}{3}\]
Conclusion:
The point moves along the line \(x = \frac{5}{3}\), which corresponds to option (A).
Final Answer: \(\boxed{\text{(A) } x = \frac{5}{3}}\)
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Concepts Used:
Straight lines
A straight line is a line having the shortest distance between two points.
A straight line can be represented as an equation in various forms, as show in the image below:
The following are the many forms of the equation of the line that are presented in straight line-
1. Slope – Point Form
Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.
y – y0 = m (x – x0)
2. Two – Point Form
Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2) are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes
The slope of P2P = The slope of P1P2 , i.e.
\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)
Hence, the equation becomes:
y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)
3. Slope-Intercept Form
Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by
y – c =m( x - 0 )
As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if
y = m x +c