Consider ABC whose vertices are A(2,1), B(3,3), and C(1,6). Use the segment AC as the base of the triangle. 1. Find the equation of the base segment AC in slope-intercept form. 2. Find the equation of the height segment in slope-intercept form by finding the equation of the line passes through B and is perpendicular to AC. 3.Graph the height segment using the equation you found in Part 2. Label the intersection of the height line and the base segment as point D. Then, find the exact coordinates of D using an appropriate technique for solving systems of equations. 4.Use the distance formula to find: the length of the base AC the height BD 5. Find the area of ABC using the triangle area formula A=1/2bh.

2. The point-slope form of the equation of a line can be used to find the perpendicular to AC through B. The slope will be the negative reciprocal of that of AC, so will be -1/-5 = 1/5. Then the equation is ...

y = m(x -h) +k

y = (1/5)(x -3) +3

y = 1/5x +12/5

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3. Equating expressions for y in these equations, we can find their intersection point.

-5x +11 = 1/5x +12/5

-25x +55 = x +12 . . . . . multiply by 5

43 = 26x . . . . . . . . . . . . add 2x -12

x = 43/26 ≈ 1.654

y = -5x +11 = -5(43/26) +11 = 71/26 ≈ 2.731

Point D is (43/26, 71/26).

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4. Using the distance formula we find the length of AC to be ...

d = √((1-2)^2 +(6-1)^2) = √(1 +25) = √26

Using the distance formula to find the length of BD, we get ...

d = √((3 -43/26)^2 +(3 -71/26)^2) = (1/26)√(35^2 +7^2) = (√1274)/26

= (7√26)/26

The length AC is √26; the height BD is (7√26)/26.

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5. Using the area formula, the area of the triangle is ...

A = (1/2)bh

A = (1/2)(√26)(7√26)/26 = 7/2 = 3.5

The triangle area is 3.5 square units.

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Comment on the method

While the method serves to make use of formulas for lines and distance, it may not be the easiest. A way to compute the area directly from coordinates is ...

area = (1/2)|Ax(By-Cy) +Bx(Cy-Ay) +Cx(Ay-By)|

= (1/2)|2(3-6) +3(6-1) +1(1-3)| = (1/2)|-6 +15 -2|

area = 7/2