$$A($$−$$5$$,$$0)$$ and $$B(3$$,$$0)$$ are two of the vertices of a triangle ABC. Its area is $$20$$ square ems. The vertex C lies on the line x−$$y=2$$. The coordinates of C, are

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$$A($$−$$5$$,$$0)$$ and $$B(3$$,$$0)$$ are two of the vertices of a triangle ABC. Its area is $$20$$ square ems. The vertex C lies on the line x−$$y=2$$. The coordinates of C, are

Infinity Learn · Accepted Answer

Let the coordinates of the third vertex C be (h$$, $$k). Then, Area of ABC $$=$$ $$20$$ sq. units ⇒ $$12h$$    k    $$1$$−$$5$$    $$0$$    $$13$$    $$0$$    $$1=$$±$$20$$⇒$$k=$$±$$5$$.             …(i)Since$$, $$(h$$, $$k) lies on $$x-y=2$$ Therefore, h−$$k=2$$                                                        …(ii)Solving$$ $$(i) and (ii), we find that the coordinates of the third vertex are $$(-3$$,$$-5)$$ or,(7$$, $$5).

$A (- 5, 0)$ and $B (3, 0)$ are two of the vertices of a triangle $A B C .$ Its area is 20 square ems. The vertex $C$ lies on the line $x - y = 2 .$ The coordinates of $C,$ are

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A(−5,0) and B(3,0) are two of the vertices of a triangle ABC. Its area is 20 square ems. The vertex C lies on the line x−y=2. The coordinates of C, are

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$A (- 5, 0)$ and $B (3, 0)$ are two of the vertices of a triangle $A B C .$ Its area is 20 square ems. The vertex $C$ lies on the line $x - y = 2 .$ The coordinates of $C,$ are