Problem 16 - The Calculus with Analytic Geometry

Question

Given the two points \(A(-3,4)\) and \(B(2,5)\), find the coordinates of a point \(P\) on the line through \(A\) and \(B\) such that \(P\) is (a) twice as far from \(A\) as from \(B\), and (b) twice as far from \(B\) as from \(A\).

Step-by-Step Solution

Answer

For part (a), P is \( \left( \frac{1}{3}, \frac{14}{3} \right) \). For part (b), P is \( \left( \frac{-4}{3}, \frac{13}{3} \right) \).

Step 1 - Calculate the distance between points A and B

Use the distance formula to find the distance between points A(-3,4) and B(2,5). The distance formula is \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). Substituting the coordinates of A and B, we get \( d = \sqrt{(2 - (-3))^2 + (5 - 4)^2} = \sqrt{5^2 + 1^2} = \sqrt{26} \).

Step 2 - Determine the ratio of distances for part (a)

For part (a), we need point P such that it is twice as far from A as it is from B. This means the ratio of distances PA:PB is 2:1.

Step 3 - Calculate coordinates of point P for part (a)

Using the section formula, the coordinates of point P dividing the line segment AB in the given ratio m:n can be found using \( P(x, y) = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \). Here, m = 2, n = 1, \( x_1 = -3, y_1 = 4, x_2 = 2, y_2 = 5 \). Substituting these values, \( P(x, y) = \left( \frac{2(2) + 1(-3)}{2+1}, \frac{2(5) + 1(4)}{2+1} \right) = \left( \frac{4 - 3}{3}, \frac{10 + 4}{3} \right) = \left( \frac{1}{3}, \frac{14}{3} \right) \). Therefore, the coordinates of point P are \( \left( \frac{1}{3}, \frac{14}{3} \right) \).

Step 4 - Determine the ratio of distances for part (b)

For part (b), we need point P such that it is twice as far from B as it is from A. This means the ratio of distances PA:PB is 1:2.

Step 5 - Calculate coordinates of point P for part (b)

Using the section formula again with m = 1, n = 2, \( x_1 = -3, y_1 = 4, x_2 = 2, y_2 = 5 \), we get \( P(x, y) = \left( \frac{1(2) + 2(-3)}{1+2}, \frac{1(5) + 2(4)}{1+2} \right) = \left( \frac{2 - 6}{3}, \frac{5 + 8}{3} \right) = \left( \frac{-4}{3}, \frac{13}{3} \right) \). Therefore, the coordinates of point P are \( \left( \frac{-4}{3}, \frac{13}{3} \right) \).

Key Concepts

Distance Formula

Understanding how to calculate the distance between two points is essential to many geometry problems. The distance formula helps us determine how far apart two points are in a coordinate plane. This formula is derived from the Pythagorean theorem and is written as: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
To use the formula, we simply plug in the coordinates of our two points. For example, to find the distance between point A(-3,4) and point B(2,5), we substitute their coordinates into the formula:

• First calculate the differences: \( x_2 - x_1 = 2 - (-3) = 5 \) and \( y_2 - y_1 = 5 - 4 = 1 \).
• Next, square the differences: \( 5^2 = 25 \) and \( 1^2 = 1 \).
• Add the squares: \( 25 + 1 = 26 \).
• Finally, take the square root: \( \sqrt{26} \approx 5.1 \). This is the distance between points A and B.

Section Formula

The section formula is extremely useful when we need to find a point that divides a line segment between two points in a given ratio. It can be visualized as placing weights on a balance: the coordinates of the dividing point are the weighted averages of the coordinates of the two points. The section formula is given by:
\( P(x, y) = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \), where \( m \) and \( n \) are the weights, or the ratio parts.
Let's apply the section formula based on the exercise:

• For part (a): We need point P such that it is twice as far from A as it is from B. This translates to a ratio of \( m:n = 2:1 \).
• Coordinates of A are \( (-3,4) \) and B are \( (2,5) \).
• Substitute these values: \( P(x, y) = \left( \frac{2(2) + 1(-3)}{2+1}, \frac{2(5) + 1(4)}{2+1} \right) = \left( \frac{4 - 3}{3}, \frac{10 + 4}{3} \right) = \left( \frac{1}{3}, \frac{14}{3} \right) \).
This means the coordinates of P are \( \left( \frac{1}{3}, \frac{14}{3} \right) \).
• For part (b): We need point P such that it is twice as far from B as it is from A. This translates to a ratio of \( m:n = 1:2 \).
• Substitute these values: \( P(x, y) = \left( \frac{1(2) + 2(-3)}{1+2}, \frac{1(5) + 2(4)}{1+2} \right) = \left( \frac{2 - 6}{3}, \frac{13}{3} \right) \).
This means the coordinates of P are \( \left( \frac{-4}{3}, \frac{13}{3} \right) \).

Ratio of Distances

The ratio of distances helps us position points along a line segment based on how far we want them from the endpoints. Understanding the ratio can help solve more complex problems of segment division.
In our exercise, we dealt with two specific scenarios:

• Finding a point P such that it is twice as far from A as from B (part (a)).
• Finding a point P such that it is twice as far from B as from A (part (b)).

In both cases, the solution involves:

• Using the given ratio to determine the weights (m and n) for the section formula.
• Calculating the coordinates of the point P that divides the line segment AB using these weights.

For part (a), the ratio was 2:1 (PA:PB). For part (b), the ratio was 1:2 (PA:PB). By applying the section formula accordingly, we were able to find the exact coordinates of P in both scenarios. This understanding can be extended to other problems where the division ratios change.
With practice, these concepts become more intuitive and easier to apply to different problems.