Locus of a point which moves such that its distance from the $X-axis$ is twice its distance from the line $x - y = 0$ is
- $x^2 + 4xy - y^2 = 0$
- $2x^2 - 4xy + y^2 = 0$
- $x^2 - 4xy + y^2 = 0$
- $x^2 - 4xy - y^2 = 0$
The Correct Option is B
Solution and Explanation
$P_2 =$ length of perpendicular from p to y = x line.
$P_1 = 2P_2$
$\left|k\right| = 2.\frac{\left|h-k\right|}{\sqrt{2}}$
Squaring on both sides,
$k^2 =2(h-k)^2$
$k^{2}= 2h^{2} + 2k^{2} - 4hk$
$\Rightarrow 2h^{2} - 4hk + k^{2} = 0$
So, the locus of a point P is,
$2x^{2} - 4xy + y^{2} = 0$
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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