Given A(a,2) B(3,4) and C(-2,-1). If A is at the same distance to the points B and C. Find a.
distance AB= sqrt((a-3)^2+(2-4)^2)
distance AC= sqrt(a+2)^2+(2+1)^2)
setting them equal
((a-3)^2+(2-4)^2)=(a+2)^2+(2+1)^2)
expand out the terms, combine, and solve for a.
(a-3)^2-(a+2)^2=9-4
so if the right side is five, then we have on the left a difference of two squares is five, and immediately comes to mind 9 and 4, and because math teachers often chose whole numbers, it is worth a try..
then if (a-3)^2=9, then a must be six, or zero. six wont work, but zero does.
the left side is also just a difference of two squares, so
((a-3)+(a-2))((a-3)-(a-2)) = (2a-5)(-1)
so, you have
5-2a = 5
a = 0
To find the value of "a" such that point A is equidistant from points B and C, we need to use the distance formula.
The distance formula between two points (x1, y1) and (x2, y2) is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Let's calculate the distances between points A and B, and between points A and C.
Distance between A and B:
d_AB = sqrt((3 - a)^2 + (4 - 2)^2)
d_AB = sqrt((3 - a)^2 + 2^2)
d_AB = sqrt((3 - a)^2 + 4)
Distance between A and C:
d_AC = sqrt((-2 - a)^2 + (-1 - 2)^2)
d_AC = sqrt((-2 - a)^2 + (-3)^2)
d_AC = sqrt((-2 - a)^2 + 9)
d_AC = sqrt((2 + a)^2 + 9)
Since A is equidistant from B and C, we can set the two distances equal to each other:
d_AB = d_AC
sqrt((3 - a)^2 + 4) = sqrt((2 + a)^2 + 9)
Squaring both sides of the equation to eliminate square roots, we get:
(3 - a)^2 + 4 = (2 + a)^2 + 9
Expanding and simplifying the equation, we have:
9 - 6a + a^2 + 4 = 4 + 4a + a^2 + 9
Rearranging the terms, we get:
3a = 0
Dividing both sides by 3, we have:
a = 0
Therefore, the value of "a" that makes point A equidistant from points B and C is 0.