ammuseeru wrote:

Hi

Bunuel,

How to calculate area of triangle when coordinates of three vertices are given. For example, how to solve following question.

On the coordinate plane there are 3 points A(-1, 1), B(0, -2), and C(3, 4). What is the area of the triangle connecting the 3 points?

(A) 15/4

(B) 15/2

(C) 15

(D) 13/4

(E) 13/2

Regards,

Ammu

Attachment:

tri.area.grid.rectangle.jpg [ 37.89 KiB | Viewed 10363 times ]
Because this triangle has no vertical or horizontal leg, plot its three coordinates, connect them, then

draw a rectangle around the triangle (see diagram).

Then find area of rectangle and subtract the areas of the three yellow triangles, whose areas are easy to find because they are right triangles (their legs are base and height).

1. (Area of rectangle) - (area of OTHER triangles) = area of THIS triangle

Area of rectangle = (L * W)

L = 6 (count on the graph, or subtract y-coordinates at points B and C)

W = 4 (count, or subtract x-coordinates at points A and B)

Rectangle area = 242. Find area of three triangles (yellow) ABX, BCY, and DXY (coordinates are in red - find base and height for each by counting or subtracting x- and then y-coordinates)

Area of triangle ABX = \(\frac{1}{2} b*h =

\frac{(3*4)}{2}\) = 6

Area of triangle BCY = \(\frac{(6 * 3)}{2}\) = 9

Area of triangle DXY = \(\frac{(1*3)}{2} = \frac{3}{2}\)

Total area of yellow triangles = 6 + 9 + \(\frac{3}{2}\) =

\(\frac{33}{2}\)3.

(Rectangle area) - (total area of 3 yellow triangles) =

area of THIS triangle24 - \(\frac{33}{2}\) =

\(\frac{48}{2} - \frac{33}{2} = \frac{15}{2}\)

Answer B

Hope it helps.

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