7.3 Q-7

Question Statement

Show that the midpoint of the hypotenuse of a right-angled triangle is equidistant from its vertices.

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Background and Explanation

In a right-angled triangle, the midpoint of the hypotenuse is a unique point, and it has an interesting property: it is equidistant from the two vertices of the triangle that form the hypotenuse. This property is a result of the perpendicular bisector theorem. We will show this property by calculating the distances from the midpoint to each of the two vertices and proving they are equal.

To solve this, we will use the distance formula and basic vector operations.

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Solution

Consider a right-angled triangle $OAB$ with a right angle at $O$. Let $AB$ be the hypotenuse, and $M$ be the midpoint of the hypotenuse. We need to show that:

• $A \vec{M} = M \vec{B}$, meaning that the distances from $M$ to $A$ and $M$ to $B$ are equal.
• $O \vec{M}$ is perpendicular to the hypotenuse.

Step 1: Find the vector $A \vec{M}$

Let the coordinates of $A$ be $(a, 0)$ and the coordinates of $B$ be $(0, b)$. The coordinates of the midpoint $M$ are given by:

$$M = \left(\frac{a}{2}, \frac{b}{2}\right)$$

Now, find the vector $A \vec{M}$, which is the difference between the coordinates of $M$ and $A$:

$$A \vec{M} = \left(\frac{a}{2}, \frac{b}{2}\right) - (a, 0)$$ $$A \vec{M} = \left(\frac{a}{2} - a, \frac{b}{2} - 0\right)$$ $$A \vec{M} = \left(-\frac{a}{2}, \frac{b}{2}\right)$$

The magnitude of $A \vec{M}$ is:

$$|A \vec{M}| = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2} = \sqrt{\frac{a^2}{4} + \frac{b^2}{4}} = \sqrt{\frac{a^2 + b^2}{4}} \tag{1}$$

Step 2: Find the vector $B \vec{M}$

Now, find the vector $B \vec{M}$, which is the difference between the coordinates of $M$ and $B$:

$$B \vec{M} = \left(\frac{a}{2}, \frac{b}{2}\right) - (0, b)$$ $$B \vec{M} = \left(\frac{a}{2} - 0, \frac{b}{2} - b\right)$$ $$B \vec{M} = \left(\frac{a}{2}, \frac{b}{2} - b\right) = \left(\frac{a}{2}, \frac{-b}{2}\right)$$

The magnitude of $B \vec{M}$ is:

$$|B \vec{M}| = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{-b}{2}\right)^2} = \sqrt{\frac{a^2}{4} + \frac{b^2}{4}} = \sqrt{\frac{a^2 + b^2}{4}} \tag{2}$$

Step 3: Find the vector $O \vec{M}$

Finally, find the vector $O \vec{M}$, which is the difference between the coordinates of $M$ and $O$:

$$O \vec{M} = \left(\frac{a}{2}, \frac{b}{2}\right) - (0, 0)$$ $$O \vec{M} = \left(\frac{a}{2}, \frac{b}{2}\right)$$

The magnitude of $O \vec{M}$ is:

$$|O \vec{M}| = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2} = \sqrt{\frac{a^2}{4} + \frac{b^2}{4}} = \sqrt{\frac{a^2 + b^2}{4}} \tag{3}$$

Step 4: Conclusion

From equations $(1)$, $(2)$, and $(3)$, we see that:

$$|A \vec{M}| = |B \vec{M}| = |O \vec{M}|$$

Therefore, the midpoint $M$ of the hypotenuse is equidistant from the vertices $A$, $B$, and $O$. This proves that the midpoint of the hypotenuse of a right-angled triangle is equidistant from its vertices.

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Key Formulas or Methods Used

• Distance between two points: The distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ in the plane is given by:

$$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

• Magnitude of a vector: The magnitude of a vector $\mathbf{v} = (v_1, v_2)$ is:

$$|\mathbf{v}| = \sqrt{v_1^2 + v_2^2}$$

• Midpoint formula: The midpoint $M$ of the segment joining points $A(x_1, y_1)$ and $B(x_2, y_2)$ is:

$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$

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Summary of Steps

1. Define the vectors $A \vec{M}$, $B \vec{M}$, and $O \vec{M}$ based on the coordinates of points $A$, $B$, and $O$.
2. Calculate the magnitudes of the vectors $A \vec{M}$, $B \vec{M}$, and $O \vec{M}$.
3. Compare the magnitudes to show that $A \vec{M} = B \vec{M} = O \vec{M}$.
4. Conclude that the midpoint of the hypotenuse is equidistant from the vertices of the right-angled triangle.