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Chapter 40: Chapter 40

Problem 696

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Show that the points \(\mathrm{A}(2,-2), \mathrm{B}(-8,4)\), and \(\mathrm{C}(5,3)\) are the vertices of a right triangle and find its area.

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The given points $\mathrm{A}(2,-2), \mathrm{B}(-8,4)$, and $\mathrm{C}(5,3)$ form a right triangle with the right angle at point $\mathrm{B}$. The area of the triangle is \(34\) square units.
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TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Find side lengths AB, BC, and AC using the distance formula.

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) in a plane can be found using the distance formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Let's find the side lengths for the triangle formed by points \(\mathrm{A}(2,-2), \mathrm{B}(-8,4)\), and \(\mathrm{C}(5,3)\). Length of AB: \[ AB = \sqrt{(-8 - 2)^2 + (4 - (-2))^2} = \sqrt{(-10)^2 + (6)^2} = \sqrt{136} \] Length of BC: \[ BC = \sqrt{(5 - (-8))^2 + (3 - 4)^2} = \sqrt{(13)^2 + (-1)^2} = \sqrt{170} \] Length of AC: \[ AC = \sqrt{(5 - 2)^2 + (3 - (-2))^2} = \sqrt{(3)^2 + (5)^2} = \sqrt{34} \]

Step 2: Check if the Pythagorean theorem holds true for the side lengths.

We will check if any permutation of the side lengths satisfies the Pythagorean theorem, which states that for a right triangle with sides of length a, b, and c, where c is the longest side, then \(a^2 + b^2 = c^2\). Check the sum of the squares of the side lengths: \[ AB^2 + BC^2 = 136 + 170 = 306\\ AB^2 + AC^2 = 136 + 34 = 170\\ BC^2 + AC^2 = 170 + 34 = 204 \] We can see that \(AB^2 + AC^2 = 170\), which is equal to the square of side BC. Therefore, the given points form a right triangle, with the right angle at point \(\mathrm{B}\).

Step 3: Find the area of the right triangle.

Since we've determined that the triangle is indeed a right triangle, with the right angle at point \(\mathrm{B}\), we can now find its area using the following formula: Area = \(\frac{1}{2}\times\) base \(\times\) height. In our case, the base is the side AB and the height is the side AC: Area = \(\frac{1}{2} \times \sqrt{136} \times \sqrt{34}\). To simplify, we can factor out the square root of 2 from both sides: Area = \(\frac{1}{2} \times \sqrt{2^2 \cdot 34} \times \sqrt{2 \cdot 17}\). Area = \(\frac{1}{2} \times 2\sqrt{34} \times \sqrt{34}\). Thus, the area of the right triangle is \(34\) square units.

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