Two vertices of a triangle are (4$$,$$-3) and $$(-2$$,$$5)$$. If the orthocenter of the triangle is at (1$$,$$2), then the third vertex is

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Infinity Learn · Accepted Answer

Let the third vertex be (h$$,$$k) Now, the slope of AD is (k$$−$$2)/(h$$−$$1), the slope of BC is (5+3)/($$−$$2$$−$$4)=$$−$$4/3$$,the slope of BE is $$($$−$$3$$−$$2)/(4$$−$$1)=$$−$$5/3$$, and the slope of AC is $$(k$$−$$5)/(h+2). Since AD⊥BC, we have k−$$2h$$−$$1$$×−$$43=$$−$$1or$$ $$3h$$−$$5k+31=0_______(i)Again$$, since BE⊥AC,, we have −$$53$$×k−$$5h+2=$$−$$1or$$ $$3h$$−$$5k+31=0______(ii)On$$ solving (i) and (ii), we get $$h=33$$, $$k=26$$. Hence, the third vertex $$is(33$$,$$26)$$

Two vertices of a triangle are (4,-3) and (-2,5). If the orthocenter of the triangle is at (1,2), then the third vertex is

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