The Straight Line Joining the Points (Ap, 3) and B(1, -1) has a Length of 5. What are the Values of p?

In this article, we will explore how to find the values of p such that the distance between the points (Ap, 3) and B(1, -1) is 5. This involves a combination of geometric principles and algebraic techniques. Let's break down the problem step-by-step and demonstrate the calculations.

Understanding the Problem

We are given two points Ap, 3 and B(1, -1). We need to find the value(s) of p such that the distance between these points is 5. To solve this, we will use the distance formula, which is derived from the Pythagorean theorem.

Using the Distance Formula

The distance formula is:

d sqrt{x2 - x1^2 y2 - y1^2}

Given:

x1 p y1 3 x2 1 y2 -1

The distance d is given as 5. So we can set up the equation:

sqrt{1 - p^2 (-1 - 3^2)} 5

Working Through the Equation

First, let's calculate -1 - 3^2:

-1 - 3 -4
Therefore, -4^2 16

Now, substituting into the distance equation:

sqrt{1 - p^2 16} 5

To eliminate the square root, we square both sides:

1 - p^2 16 25

Isolate 1 - p^2:

1 - p^2 25 - 16
1 - p^2 9

Next, take the square root of both sides:

1 - p plusmn;3

Solving for p in each case:

1 - p 3
-p 3 - 1
-p 2
p -2 1 - p -3
-p -3 - 1
-p -4
p 4

Therefore, the values of p that satisfy the condition are:

boxed{-2 and 4}

Additionally, we can verify that the points (Ap, 3) and B(1, -1) are indeed 5 units apart on both the line and the circle centered at B(1, -1) with a radius of 5.

Verification:

The equation for the circle centered at B(1, -1) with radius 5 is:

(x - 1)^2 (y 1)^2 5^2
(x - 1)^2 (y 1)^2 25

Substituting y 3 into the circle's equation:

(x - 1)^2 (3 1)^2 25
(x - 1)^2 16 25
(x - 1)^2 9
x - 1 plusmn;3
x 4 or x -2

This confirms our solutions, as the values of x (i.e., p) are -2 and 4.

In conclusion, the values of p that satisfy the condition are:

boxed{-2 and 4}