Final answer:
To locate the circumcenter of ΔABC, the perpendicular bisectors of the triangle's sides are used.
For ΔABC with given vertices, the circumcenter's coordinates are (7 / 3, −5).
Step-by-step explanation:
To find the coordinates of the circumcenter of ΔABC with vertices A (1, 0), B (7, −6), and C (7, −4), we need to use the perpendicular bisectors of the sides of the triangle. The circumcenter is the point where these bisectors intersect.
However, since vertices B and C have the same x-coordinate, we know that the perpendicular bisector of BC would be a horizontal line and since the triangle is not a right angle triangle, the circumcenter will lie on this line.
To find its precise location, we need to find the midpoint of BC and then create a perpendicular line through A which will intersect our horizontal line.
Midpoint of BC: ((Bx + Cx) / 2, (By + Cy) / 2) = ((7 + 7) / 2, (−6 + −4) / 2) = (7, −5).
The perpendicular bisector through A will have a slope that is the negative reciprocal of the slope of AC since line AC is not vertical.
Slope of AC = (Cy − Ay) / (Cx − Ax)
= (−4 − 0) / (7 − 1)
= −4 / 6
= −2 / 3.
The negative reciprocal is 3 / 2. Using the point-slope form of a line: y − y1 = m(x − x1), we plug in the coordinates of A and the slope to find the equation of the bisector.
y − 0 = (3 / 2)(x − 1), simplifying we get y = (3 / 2)x − (3 / 2).
Setting this equal to the y-coordinate of our horizontal line (which is −5) gives us the x-coordinate of the circumcenter.
−5 = (3 / 2)x − (3 / 2)
3 / 2x = −3.5
x = −3.5 / (3 / 2)
= −3.5 × 2 / 3
= −7 / 3
Thus, the coordinates of the circumcenter are (7 / 3, −5).