So I take it that the coordinates of C are (x,0) if I let the x-coordinate of C be x.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

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!!