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?
\(x=\dfrac{5}{3}\)
\(y=\dfrac{5}{3}\)
\(x=\dfrac{-5}{3}\)
\(x=0\)
\(y=\dfrac{-5}{3}\)
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.)
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-
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)
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 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)
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