4.1 Q-15

Question Statement

Find the point on the line joining $A(1, 4)$ and $B(5, 6)$ that is twice as far from $A$ as $B$ is from $A$. Solve for:

1. The point on the same side of $A$ as $B$.
2. The point on the opposite side of $A$ as $B$.

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Background and Explanation

To solve this problem, we use the section formula. This formula is used to find the coordinates of a point dividing a line segment in a specific ratio. The given ratio in this problem is either 1:1 or 2:1, depending on whether the point lies on the same or opposite side of $A$.

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Solution

Part 1: Point on the Same Side of $A$ as $B$

We are asked to find the point that is twice as far from $A$ as $B$ is, but on the same side of $A$ as $B$. This means the ratio of distances between points $A$, $B$, and $C$ is 1:1.

Let $P(x_1, y_1)$ be the required point. Using the section formula, we divide the segment $AB$ in the ratio 1:1:

$$\frac{1 + x_1}{2} = 5, \quad \frac{4 + y_1}{2} = 6$$

Now, solving for $x_1$ and $y_1$:

$$1 + x_1 = 10 \quad \Rightarrow \quad x_1 = 9$$ $$4 + y_1 = 12 \quad \Rightarrow \quad y_1 = 8$$

Thus, the coordinates of the point on the same side of $A$ as $B$ are $C(9, 8)$.

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Part 2: Point on the Opposite Side of $A$ as $B$

Next, we need to find the point that divides the line segment $AB$ in the ratio 2:1, but on the opposite side of $A$ as $B$.

Let $P(x_2, y_2)$ be the required point. Using the section formula, we divide the segment in the ratio 2:1:

For the $x$-coordinate:

$$1 = \frac{2(5) + 1(x_2)}{2 + 1}$$ $$1 = \frac{10 + x_2}{3} \quad \Rightarrow \quad 10 + x_2 = 3 \quad \Rightarrow \quad x_2 = -7$$

For the $y$-coordinate:

$$4 = \frac{2(6) + 1(y_2)}{2 + 1}$$ $$4 = \frac{12 + y_2}{3} \quad \Rightarrow \quad 12 + y_2 = 12 \quad \Rightarrow \quad y_2 = 0$$

Thus, the coordinates of the point on the opposite side of $A$ as $B$ are $P(-7, 0)$.

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Key Formulas or Methods Used

• Section Formula: To find the coordinates of a point dividing a line segment in the ratio $m:n$:

$$x = \frac{n x_1 + m x_2}{m + n}, \quad y = \frac{n y_1 + m y_2}{m + n}$$

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Summary of Steps

1. For the point on the same side of $A$ as $B$:
   • Use the section formula with a ratio of 1:1.
   • Solve for $x_1$ and $y_1$ to get the coordinates $(9, 8)$.
2. For the point on the opposite side of $A$ as $B$:
   • Use the section formula with a ratio of 2:1.
   • Solve for $x_2$ and $y_2$ to get the coordinates $(-7, 0)$.