Find all points on the x-axis that are 14 units from the point (6,-7)

Answer 1

Let these points is (x, 0).Then (x - 6)² + (0 + 7)² = 14², or (x - 6)² = 196 - 49; x - 6 = ± 7√3

x = 6 ± √3.

So we have 2 points:(6 + 7√3, 0) and (6 - 7√3, 0)

Answer 2

All points that are 14 units from the point (6,-7) would be a circle of radius 14 with a center at (6,-7). The equation would be as follows:

r² = (x - h)² + (y - k)²

center = (h,k) = (6,-7)

radius = r = 14

14² = (x - 6)² + (y + 7)²

All points on the x-axis that are 14 units from the point (6,-7) would be the points along the circle that cross through the x-axis, or when y = 0. Set y = 0 to find the corresponding x-values:

14² = (x - 6)² + (0 + 7)²

196 = x² - 12x + 36 + 49

0 = x² - 12x - 111

Use the quadratic formula to solve for x:

x = [12 ± √((-12)² - 4(1)(-111)] / 2(1) = [12 ± √(588)] / 2 = -6.124, 18.124

All points on the x-axis that are 14 units from the point (6,-7) are (-6.124,0) and (18.124,0).