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Show that the points (2,0), (-2,0) and (0,2) are the vertices of a triangle. Also state with reason the type of the triangle . |
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Answer» Let A(2,0), B(-2,0) and C(0,2) By distance formula `AB = sqrt((x_(2) - x_(1))^(2) + (Y_(2) - y_(2))^(2))` ` = sqrt((-2-2)^(2) + (0- 0)^(2))` ` = sqrt((-4)^(2) + (0)^(2))` ` sqrt(16)` ` = 4 ` Similary , ` BC = 2sqrt(2)" and " AC = 2sqrt(2)` For any triangle Sum of two sides will always be greter then the third side ` therefore 2sqrt(2) + 2sqrt(2) gt 4 ` i.e. ` 4sqrt(2) gt 4 ` and also ` 4 + 2 sqrt(2) gt 2sqrt(2)` ` therefore A ,B C ` are the vertices of a triangle Now, ` BC^(2) + AC^(2) = (2sqrt(2)^(2)+ 2sqrt(2))^(2)` ` therefore BC^(2) + AC^(2) = 16` ...(1) `AB^(2) = 4^(2)` ` therefore AB^(2) = 16` ...(2) ` therefore BC^(2) + AC^(2) = AB^(2)" "` [from (1) and (2)] ` therefore ` by converse, of Pythagoras theorem ` Delta ABC ` is a right angled triangle and also BC = AC ` therefore Delta ABC ` is an isosceles right angled triangle. |
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