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Question
The sides of a triangle are given by the equations y - 2 = 0; y + 1 = 3 (x - 2) and x + 2y = 0.
Find, graphically :
(i) the area of a triangle;
(ii) the coordinates of the vertices of the triangle.
Solution
y - 2 = 0
⇒ y = 2
y + 1 = 3(x - 2)
⇒ y + 1 = 3x - 6
⇒ y = 3x - 6 - 1
⇒ y = 3x - 7
The table for y + 1 = 3(x - 2) is
| X | 1 | 2 | 3 |
| Y | - 4 | - 1 | 2 |
Also we have
x + 2y = 0
⇒ x = - 2y
The table for x + y = 0 is
| X | - 4 | 4 | - 6 |
| Y | 2 | - 2 | 3 |
Plotting the above points we get the folllowing required graph:
(i) The area of the triangle ABC = `(1)/(2) xx "AB" xx "CD"`
= `(1)/(2) xx 7 xx 3`
= `(21)/(2)`
= 10.5 sq.units
(ii) The coordinates of the verticles of the triangle are ( - 4, 2), (3, 2) and (2, -1).
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