The point (0,1) has the distance ____, from the line defined by y=-2x-1
How do you solve this?
I tried the pythagorean theorem but I don't think I did it right as I came out with the square root of two. Can someone please explain this to me.
The distance of a point (x0,y0) from a line L: Ax+By+C=0 is given by
Ax0+By0+C/√(A²+B²).
Here (0,1) ⇒ x0=0, y0=1.
L: y=-2x-1 ⇒ L: 2x+y+1 = 0
⇒ A=2, B=1, C=1.
So distance,
d = (2(0)+1+1)/√(2²+1²)
= 2/√(5)
= 2√(5)/5
To find the distance between a point and a line, you can use the formula for the distance between a point and a line in coordinate geometry. Here's how you can solve this problem step by step:
1. Start with the equation of the line: y = -2x - 1.
2. Let the given point be (x₁, y₁) = (0, 1). Plug in these values into the equation of the line to find the y-coordinate of the corresponding point on the line: y = -2(0) - 1 = -1.
3. Now, you have two points: (0, 1) and (0, -1), which form a right-angled triangle. The distance between these points is the vertical distance between them, which is simply the absolute difference between their y-coordinates: |-1 - 1| = 2.
4. Therefore, the distance between the point (0, 1) and the line y = -2x - 1 is 2 units.
The Pythagorean theorem is not necessary in this case since the line is vertical, and the distance between the given point and the line is purely vertical.