A(4,7) and B(-3,2) are points on a coordinate plane.Find the coordinates of a point C on the x-axis such that AC=BC .
If you make a sketch it should be easy to see that point C must be on the right-bisector of AB
midpoint of AB = ( (4-3)/2 , (7+2)/2) = (1/2 , 9/2)
slope o AB = (2-7)/(-3-4) = 5/7
so the slope of the right-bisector is -7/5
equation of right-bisector:
y - 9/2 = (-7/5)(x - 1/2)
for the x-intercept, let y = 0
-9/2 = (-7/5)(x-1/2)
-9/2 = -7x/5 + 7/10
times 10
-45 = -14x + 7
14x = 52
x = 52/14 = 26/7
point C is (26/7, 0)
or
let the point be C(x,0)
AC = BC
√( (x-4)^2 + 7^2) = √( (x+3)^2 + 4^2)
square both sides
(x-4)^2 + 49 = (x+3)^2 + 4
x^2 - 8x + 16 + 49 = x^2 + 6x + 9 + 4
-14x = -52
x = -52/-14 = 26/7 , just like before
THANKS A LOT !!:)
To find the coordinates of point C, we need to first find the midpoint of AB and then find its reflection across the x-axis.
Let's find the midpoint of AB:
Midpoint formula:
Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Given point A: (4,7)
Given point B: (-3,2)
Midpoint = ((4 + -3)/2, (7 + 2)/2)
Midpoint = (1/2, 9/2)
Midpoint = (0.5, 4.5)
Now, to find the reflection of the midpoint across the x-axis, we keep the x-coordinate the same and reverse the sign of the y-coordinate.
Therefore, the coordinates of point C on the x-axis such that AC = BC are:
C = (0.5, -4.5)
To find the point C on the x-axis such that AC=BC, we need to first calculate the distances AC and BC.
Step 1: Calculate the distance AC
The distance between two points A(x1, y1) and C(x2, y2) is given by the formula:
AC = √((x2 - x1)^2 + (y2 - y1)^2)
Given A(4, 7), we substitute the values to get:
AC = √((x2 - 4)^2 + (y2 - 7)^2)
Since point C lies on the x-axis, its y-coordinate is 0. So, we have:
AC = √((x2 - 4)^2 + (0 - 7)^2)
Simplifying, we get:
AC = √((x2 - 4)^2 + 49)
Step 2: Calculate the distance BC
Similarly, we calculate the distance BC using the formula:
BC = √((x2 - x3)^2 + (y2 - y3)^2)
Given B(-3, 2), we substitute the values to get:
BC = √((x2 - (-3))^2 + (y2 - 2)^2)
Since point C lies on the x-axis, its y-coordinate is 0. So, we have:
BC = √((x2 + 3)^2 + (0 - 2)^2)
Simplifying, we get:
BC = √((x2 + 3)^2 + 4)
Step 3: Set up the equation AC = BC
Now, we set up the equation AC = BC and solve for x2:
√((x2 - 4)^2 + 49) = √((x2 + 3)^2 + 4)
To eliminate the square roots, we square both sides of the equation:
((x2 - 4)^2 + 49) = ((x2 + 3)^2 + 4)
Expanding and simplifying, we get:
x2^2 - 8x2 + 16 + 49 = x2^2 + 6x2 + 9 + 4
Simplifying further, we get:
-14x2 + 42 = 0
Finally, solving for x2, we have:
x2 = 3
So, the coordinate of point C on the x-axis such that AC = BC is (3, 0).