How do you find all points that have an x -coordinate of –4 and whose distance from point (4, 2) is 10?

Question

Avery Alexander · Accepted Answer

By using the distance formula. #D = root 2 ((x_1 - x_2)^2 + (y_1 - y_2)^2)# We want #D = 10# Let #(x_1, y_1)# be #(4, 2)#
Let #(x_2, y_2)# be #(-4, y_2)# #10 = root 2 ((4 - (-4))^2 + (2 - y_2)^2)# Squaring both sides we have #(10 = root 2 ((4 - (-4))^2 + (2 - y_2)^2))^2# #=> 100 = (4 - (-4))^2 + (2 - y_2)^2# #=> 100 = (8^2) + (2 - y_2)^2# #=> 100 = 64 + (4 - 4y_2 + (y_2)^2)# #=> 36 = (y_2)^2 + 4y_2 + 4# #=> (y_2)^2 + 4y_2 + 4 = 36# #=> (y_2)^2 + 4y_2 - 32 = 0# #=> (y_2 + 8)(y_2 - 4) = 0# y_2 = -8, y_2 = 4

Avery Alexander · Answer

undefined To find all points with an x-coordinate of -4 and a distance of 10 from point (4, 2), we can use the distance formula. The distance formula is given by:

d = √((x2 - x1)^2 + (y2 - y1)^2)

Substituting the given values, we have:

10 = √((-4 - 4)^2 + (y - 2)^2)

Simplifying the equation, we get:

100 = (-8)^2 + (y - 2)^2

100 = 64 + (y - 2)^2

36 = (y - 2)^2

Taking the square root of both sides, we have:

±6 = y - 2

Solving for y, we get:

y = 6 + 2 = 8 or y = -6 + 2 = -4

Therefore, the two points that satisfy the given conditions are (-4, 8) and (-4, -4).