A straight line AB of length 10 units is free to move with its end on the axes.find the locus of the point p(x,y) on the line at the distance of 3 units from the end on the x-axis
The line segment touches bot the x- and y-axes at all times. So, the point p is 3 units away from the x-axis, along the line.
So, let the line make an angle θ with the x-axis (measured clockwise).
Then picture point P as being at the end of a rod of length 7, which slides down the y-axis.
x = 7cosθ
y = 3sinθ
The graph is an ellipse.
JISKHA ORU THEVIDIA PUNDA
I am wondering where is the end of an axis, all those I draw have no ends. x axis is not a line segement.
What do you mean by the end of the axis?
To find the locus of the point P(x, y) on the line at a distance of 3 units from the end on the x-axis, we can start by visualizing the problem.
Let's assume that the line AB is parallel to the y-axis and its end is on the positive x-axis. We need to find the locus of the point P(x, y) such that the distance between P and the end of the line on the x-axis is 3 units.
The distance between two points (x1, y1) and (x2, y2) in a coordinate plane can be found using the distance formula:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In our case, we want the distance between P(x, y) and the end of the line on the x-axis to be 3 units. Let's call the end of the line on the x-axis point A(10, 0). The coordinates of P(x, y) are not known yet, so we can assume them to be (x, y).
Using the distance formula, the equation becomes:
3 = sqrt((10 - x)^2 + (0 - y)^2)
Squaring both sides of the equation to remove the square root, we get:
9 = (10 - x)^2 + y^2
Expanding and simplifying the equation, we have:
9 = 100 - 20x + x^2 + y^2
Rearranging the equation, we get:
x^2 - 20x + y^2 - 91 = 0
This equation represents a circle with the center at (10, 0) and a radius of sqrt(91). Therefore, the locus of the point P(x, y) on the line at a distance of 3 units from the end on the x-axis is a circle centered at (10, 0) with a radius of sqrt(91).