Find all y such that the distance between the points (4, 3) and (8, y) is 11.
you need
√((8-4)^2 + (y-3)^2) = 11
√(16+(y-3)^2) = 11
16+(y-3)^2 = 121
(y-3)^2 = 105
y-3 = ±√105
y = 3±√105
you can check this by noting that the locus of all points a distance of 11 from (4,3) is a circle:
(x-4)^2 + (y-3)^2 = 121
Now pick x=8, and see where y is
You end up doing exactly the same steps as shown above.
To find all the values of y that satisfy the given condition, we need to calculate the distance between the two points and set it equal to 11.
The distance formula between two points (x1, y1) and (x2, y2) is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Substituting the given points (4, 3) and (8, y) into the formula, we have:
11 = sqrt((8 - 4)^2 + (y - 3)^2)
Simplifying the equation, we get:
11 = sqrt(16 + (y - 3)^2)
Squaring both sides of the equation to eliminate the square root, we have:
(11)^2 = 16 + (y - 3)^2
121 = 16 + (y - 3)^2
Rearranging the equation, we get:
(y - 3)^2 = 121 - 16
(y - 3)^2 = 105
Taking the square root of both sides and considering both positive and negative values, we have:
y - 3 = ±√105
To find the solutions for y, we need to solve for y in both cases separately:
Case 1: y - 3 = √105
Solving for y:
y = √105 + 3
Case 2: y - 3 = -√105
Solving for y:
y = -√105 + 3
Therefore, the solutions for y are:
y = √105 + 3
y = -√105 + 3