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Chapter 38: Chapter 38

Problem 676

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Find the coordinates of the foot of the altitude to side \(\underline{\mathrm{AC}}\) of the triangle whose vertices are given by \(\mathrm{A}(-2,1), \mathrm{B}(4,7)\), and \(\mathrm{C}(6,-3)\). From this, find the length of the altitude and then the area of the triangle.

Short Answer

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The foot of the altitude to side AC is point D(5, -\(\frac{5}{2}\)). The length of the altitude is \(\sqrt{\frac{365}{4}}\). Thus, the area of the triangle is \(\frac{1}{2} * 80^\frac{1}{2} * 365^\frac{1}{4}\) square units.
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TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Find the equation of line AC

To find the equation of line AC, first, we need to find the slope of AC which is given by \(m = \frac{y_2 - y_1}{x_2 - x_1}\), where \((x_1, y_1) = A(-2, 1)\) and \((x_2, y_2) = C(6,-3)\). Calculating the slope: \[m = \frac{-3-1}{6-(-2)} = \frac{-4}{8} = -\frac{1}{2}\] Now use point-slope form to get the equation of line AC: \[y - y_1 = m(x - x_1)\] Plug in the coordinates of point A and the calculated slope: \[y - 1 = -\frac{1}{2}(x - (-2))\] Simplify the equation: \[y - 1 = -\frac{1}{2}x -1\] Add 1 to both sides: \[y = -\frac{1}{2}x\] So, the equation of line AC is \(y = -\frac{1}{2}x\).

Step 2: Find the equation of the altitude from B

The altitude from B is perpendicular to AC. So the slope of the altitude can be found using the negative reciprocal of the slope of AC: \[m_\perp = \frac{1}{2}\] Now use the point-slope form to find the equation of the altitude passing through B (4,7) and having a slope of 1/2: \[y - 7 = \frac{1}{2}(x - 4)\] Simplify the equation: \[y - 7 = \frac{1}{2}x - 2\] Add 7 to both sides: \[y = \frac{1}{2}x + 5\] So, the equation of the altitude from B is \(y = \frac{1}{2}x + 5\).

Step 3: Find the coordinates of the foot of the altitude (point D)

To find point D, we solve the equations of lines AC and the altitude simultaneously: \[-\frac{1}{2}x = \frac{1}{2}x + 5\] Add \(\frac{1}{2}x\) to both sides: \[x = 5\] Now substitute this x-value into the equation of AC to find the y-coordinate: \[y = -\frac{1}{2}(5)\] \[y = -\frac{5}{2}\] So, the coordinates of point D are (5, -\(\frac{5}{2}\)).

Step 4: Find the length of the altitude from B

Now we can find the length of the altitude (BD) using the distance formula: \[BE = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\] Plugging in the coordinates B(4, 7) and D(5, -\(\frac{5}{2}\)): \[BE= \sqrt{(5-4)^2 + (-\frac{5}{2} - 7)^2} = \sqrt{1^2 + (-\frac{19}{2})^2}\] \[BD = \sqrt{1 + \frac{361}{4}} = \sqrt{\frac{365}{4}}\] So, the length of the altitude from B is \(\sqrt{\frac{365}{4}}\).

Step 5: Find the area of the triangle

To find the area of the triangle, we can use the formula: Area = \(\frac{1}{2}\) * base * height, where the base is the length of side AC and height is the length of the altitude from B. First, we will find the length of AC using the distance formula: \[AC = \sqrt{(6 - (-2))^2 + (-3 - 1)^2} = \sqrt{8^2 + (-4)^2} = \sqrt{64 + 16} = \sqrt{80}\] Now, the area of the triangle ABC is: Area = \(\frac{1}{2} * 80^\frac{1}{2} * 365^\frac{1}{4}\) So, the area of the triangle ABC is \(\frac{1}{2} * 80^\frac{1}{2} * 365^\frac{1}{4}\) square units.

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