Find all points having an x-coordinate of 1 whose distance from the point is (-2,-3) is 5.

Answer 1

(-2,-3) is the center of a circle with radius = 5

(x+2)^2 + (y+3)^2 = 25

let x = 1

3^2 + (y+3)^2 = 25

(y+3)^2 = 25-9 = 16

y+3 = + or - 4

y = -7 or 1

points are (1,1) or (1,-7)

Answer 2

Ok, so we know that the Pythagorean Theorem is a²+b²=c².

From this, we know that the distance between two points is (x₂-x₁)²+(y₂-y₁)²=distance² since you can make a right triangle with any line segment as the hypotenuse by taking the change in x and change in y as the legs of the triangle.

In this case, we know that the distance is 5, (x₁,y₁)=(-2,-3), and x₂=1. It really doesn't matter which point we call point 1 and which we call point 2, but the assignment I gave is convenient.

So now we plug these numbers in: (1- -2)²+(y₂- -3)² = 5²

and we can solve for y₂. For simplicity, I'll just call that variable y.

(3)²+(y+3)² = 25

(y+3)² = 25-9

(y+3)² = 16

Taking the square root of both sides,

y+3 = 4 or y+3 = -4 since both 4 and -4 will square to 16.

So now, y = 1 or y = -7.

One way to visualize why this question has two answers is to imagine a circle with radius 5, centered at (-2,-3). Now add a line at x=1, and you can see that it passes through the circle not once, but twice!

Answer 3

The set of all points that are 5 units from the point (-2,-3) is the circle with center (-2,-3) and radius 5.

An equation of the circle is (x+2)² + (y+3)² = 25.

If (x, y) is on the circle and if x = 1, then we have 9 + (y+3)² = 25

(y+3)² = 16

y + 3 = 4 or -4

y = 1 or -7

The points are (1,1) and (1,-7).