analytic geometrycoordinate systemsgeometry
So, I'm given a line , $y=-1$ , and a point $(1,0)$ , and I have to find the points that have the same distance between these two.
Now I know that the distance formula requires two points, so the question is how I can find a point on the line $y=-1$? So I can find the set of points?
Geometrically speaking, all points that are equidistance from points, p and q, all lie on the perdendicular bisector of pq.
So if $p = (x1, y1)$ and $q =(x2, y2)$ are the points, the line pq has slope, $\frac {y2 - y1}{x2 - x1}$. So the slope of a perpendicular line will have slope m = $- \frac {x2 - x1}{y2 - y1}$ As the line is a bisector, it will contain the midpoint $mid = (\frac{x1+x2}{2},\frac{y1+y2}{2})$. So the perpendicular bisector is the line with the equation $y - \frac{y1+y2}{2} = m(x - \frac{x1+x2}{2}) = y - \frac{y1+y2}{2} = - \frac {x2 - x1}{y2 - y1}(x - \frac{x1+x2}{2})$.
You want to find the points (x,y) where distance((x1,y1),(x,y)) = distance((x2,y2)(x,y)) = distance((x1,y1),(x2, y2)). Where (x,y) are on the line $y - \frac{y1+y2}{2} = - \frac {x2 - x1}{y2 - y1}(x - \frac{x1+x2}{2})$.
So you have three equations:
$y - \frac{y1+y2}{2} = - \frac {x2 - x1}{y2 - y1}(x - \frac{x1+x2}{2})$
$\sqrt{(x1 - x)^2 + (y1 - y)^2} = \sqrt{(x1 - x2)^2 + (y1 - y2)^2}$
and $\sqrt{(x2 - x)^2 + (y2 - y)^2} = \sqrt{(x1 - x2)^2 + (y1 - y2)^2}$
There will be two possible points that satisfy the the equations.
Okay, that's hard. BUT if you are given one of the two points the other will be symmetrically placed on the other side of the midpoint.
Ex: A=(7,3), B=(6,0), and C=(14,−1).
So the midpoint of A and B is M = (6.5, 1.5). C is 14 - 6.5 = 7.5 away from M in the x direction. C is -1 - 1.5 = -2.5 away in the y directions. So the 4th point X will also be 7.5 away in the x direction and -2.5 away in the y direction. So X = (6.5 - 7.5, 1.5 - (-2.5)) = (-1, 4).
Hint:- Let $P_2$ be (h, 0).
1) $x_1(h - x_1) = y_1^2$. Therefore, $(h - x_1)^2 = \dfrac {y_1^4}{x_1^2}$.
2) Use Pythagoras theorem. Then, $d^2 = ... = \dfrac {y_1^2}{x_1^2}(x_1^2 + y_1^2) = \dfrac {y_2^2}{x_2^2}a^2$.
The last equality shows the use of "a".
Best Answer
We know that the distance of a point $(x_0, y_0) $ from a line $ax+by+c=0$ is given by $\frac {|ax_0+ by_0 +c|}{\sqrt {a^2+b^2}} $.
Using this formula we get $$ y_0 +1 = \sqrt {(x_0-1)^2 +y_0^2} $$ $$\Rightarrow x_0^2 =2 (x_0 +y_0) $$ is the required locus of the points satisfying the condition. Hope it helps.