Answer:
x -5y = -12
Explanation:
The slope of segment AC can be found using the slope formula:
m = (y2 -y1)/(x2 -x1)
m = (6 -1)/(1 -2) = -5
The desired perpendicular to AC will have a slope that is the opposite reciprocal of this value.
m' = -1/(-5) = 1/5
Using the point-slope form of the equation for a line, we can write the equation of the desired perpendicular through B as ...
y -k = m(x -h) . . . . . . . . line with slope m through point (h, k)
y -3 = 1/5(x -3) . . . . . . . line with slope 1/5 through (3, 3)
We can put this into standard form by multiplying by 5 and rearranging some terms.
5(y -3) = (x -3)
5y -15 = x -3 . . . . . eliminate parentheses
x -5y = -12 . . . . . add 3-5x
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Additional comment
You can find the distance from a vertex to the opposite line using the formula for the distance from a point to a line. You do not need to write the equation of the perpendicular line.
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You can find the area directly from the vertex coordinates using a formula like ...
A = (1/2)|x1(y2 -y3) +x2(y3 -y1) +x3(y1 -y2)|
Here, that is ...
A = 1/2|2(3-6) +3(6-1) +1(1-3)| = 1/2|-6 +15 -2| = 7/2
If the vertices are on grid points, then Pick's theorem is another way to find the area of the triangle. It tells you the area is ...
a = i +b/2 -1 . . . . . i = number of interior grid point; b = number of boundary grid points
The only places the boundary lines of this triangle are on grid points are the vertices, so b=3. There are 3 interior grid points, so the area from Pick's theorem is ...
A = 3 +3/2 -1 = 7/2 . . . . same as above.