Prove that the area of the triangle whose vertices are (t, t-2), (t+2, t+2) and (t+3, t) independent of t.

Asked by Simran K | 23rd Feb, 2014, 01:28: PM

Expert Answer:

Let open parentheses x subscript 1 comma y subscript 1 close parentheses comma open parentheses x subscript 2 comma y subscript 2 close parentheses space a n d space open parentheses x subscript 3 comma y subscript 3 close parentheses are the vertices of the triangle.
 
Then the area of the triangle, A r e a equals 1 half open square brackets x subscript 1 open parentheses y subscript 2 minus y subscript 3 close parentheses plus x subscript 2 open parentheses y subscript 3 minus y subscript 1 close parentheses plus x subscript 3 open parentheses y subscript 1 minus y subscript 2 close parentheses close square brackets
Given the vertices of the triangle open parentheses t comma space t minus 2 close parentheses comma space open parentheses t plus 2 comma t plus 2 close parentheses space a n d space open parentheses t plus 3 comma t close parentheses
 
Thus the area of the triangle A r e a space equals space 1 half open vertical bar t open parentheses t plus 2 minus t close parentheses plus open parentheses t plus 2 close parentheses open parentheses t minus t plus 2 close parentheses plus open parentheses t plus 3 close parentheses open parentheses t minus 2 minus t minus 2 close parentheses close vertical bar equals 1 half open vertical bar 2 t plus 2 t plus 4 minus 4 t minus 12 close vertical bar equals 1 half open vertical bar minus 8 close vertical bar equals 4 space s q. space u n i t s

Answered by  | 25th Feb, 2014, 10:42: AM

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