1. ## Coordinate Geometry

A(4,7) and B(-3,2) are points on a straight line. Find the coordinates of a point C on the x-axis such that AC=BC
So I take it that the coordinates of C are (x,0) if I let the x-coordinate of C be x.
Since the gradient of AB is:
$\displaystyle \frac{7-2}{4-(-3)} = \frac{5}{7}$
The y-intercept will be: (taking y-intercept as c)
$\displaystyle 7 = \frac{5}{7} (4) + c$
$\displaystyle c = 4\frac{1}{7}$
So equation of line AB is$\displaystyle y = \frac{5}{7}x + 4\frac{1}{7}$
I think lines AC and BC will meet at a point perpendicular to the mid-point of AB. But I do not know how to find where exactly is this point (not the mid point of AB, but the point where AC and BC meet)
Any help would be greatly appreciated, thank you very much in advance!!

2. Originally Posted by caramelcake
A(4,7) and B(-3,2) are points on a straight line. Find the coordinates of a point C on the x-axis such that AC=BC
the three points may not be collinear.

use the distance formula.

let $\displaystyle (x,0)$ be the point on the x-axis

$\displaystyle \sqrt{(x-4)^2 + (0-7)^2} = \sqrt{(x+3)^2 + (0-2)^2}$

$\displaystyle (x-4)^2 + 49 = (x+3)^2 + 4$

can you finish?

3. Thank you skeeter.
Hmm so $\displaystyle (3\frac{5}{7}, 0)$ is the answer?

4. Originally Posted by caramelcake
Thank you skeeter.
Hmm so $\displaystyle (3\frac{5}{7}, 0)$ is the answer?
$\displaystyle \displaystyle \left(\frac{26}{7} , 0\right)$

get used to using improper fractions rather than mixed numbers.

5. Well for the exams here it is a must to convert the final answer to mixed numbers that's why my answer is not in improper fractions. But thank you for your input anyway