Use calculus to find the area of the triangle with the vertices (0, 5), (2, -2), and (5, 1).?
draw the triangle ABC on a graph paper.
Plot the point D(5,5)
From A, draw a horizontal line AD (parallel to x-axis)
Plot the point F(0, -2)
plot the point E(5, -2)
from B, draw a horizontal line (FBE)
area of the reactangle (AFED) =7*5=35
area of triangle (AFB) = .5 * 7*2 = 7
area of triangle (BCE) = .5 * 3*3 = 4.5
area of triangle (ADC) = .5 * 5*4 = 10
area triangle (ABC) = area (AFED) – area(AFB) – area (BCE) – area(ADC)
area triangle (ABC) = 35 – 7 – 4.5 – 10 = 13.5
Here’re some steps:
1. You’re given 3 vertices, call them A, B, and C.
2. Find the equations of the 3 lines AB, BC, and CA.
How do you do that?
If A is (x1, y1) and B is (x2, y2), equation of line going thru A and B is y – y1 = m * ( x – x1) , where m = (y2 – y1)/(x2 – x1)
3. Once you get the equations of AB, BC, and CA, plot the lines on a graph, you should get a triangle
4. At this point, you should figure out if your variable of integration is dx or dy and integrate accordingly to get the area covered by the lines. I am assuming you know how to carry forward from here.
Hope this helps 🙂
The points then become: A(0,7), B(2,0), C(5,3).
If you work out the equations of the lines relative to the new axes, you can then integrate these equations to find the areas:
P bounded by AB, the x axis, and the y axis;
Q bounded by BC, the x axis, and the line x = 5;
R bounded by AC, the x axis, the y axis, and the line x = 5.
The area of the triangle is then R – (P + Q).
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