Two vertices of a triangle are (2$$, –$$1) and (3$$, $$2) and third vertex lies on the line x $$+$$ y $$=$$ $$5$$. If the area of the triangle is $$4$$ units then third vertex is

Question

Infinity Learn · Accepted Answer

Let A ≡ (2$$, –$$1) and B ≡ (3$$, $$2).Let the third vertex be Cα,βThen, α$$+$$β$$=5$$ (given)                                          (1)Area$$ of  △$$ABC=122-11321$$αβ$$1=$$±$$4$$ $$(given)⇒β$$-3$$α$$=1$$                                                        (2)or$$  β$$-3$$α$$=-15$$                                                   $$(3)Solving$$ $$(1) and (2), we get, α $$=$$ $$1$$, β $$=$$ $$4Solving$$ (1) and (3), we get, α $$=$$ $$5$$, β $$=$$ $$0Thus$$, the third vertex is either (5$$, $$0) or (1$$, $$4).

Two vertices of a triangle are (2, –1) and (3, 2) and third vertex lies on the line x + y = 5. If the area of the triangle is 4 units then third vertex is

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