A triangle ABC lying in the first quadrant has two vertices at $$A(1$$,$$2)$$ and $$B(3$$,$$1)$$. If ∠$$BAC=90$$∘and area Δ$$(ABC)=55$$ sq. units then the abscissa of the vertex C is

Question

A triangle ABC lying in the first quadrant has two vertices at $$A(1$$,$$2)$$ and $$B(3$$,$$1)$$. If ∠$$BAC=90$$∘and area Δ$$(ABC)=55$$ sq. units then the abscissa of the vertex C is

Infinity Learn · Accepted Answer

Let the coordinates of vertex C be $$($$α,β$$)$$. It is given that ∠$$BAC=90$$∘⇒ Slope of AB x Slope of AC $$=$$ $$-1$$⇒ $$1$$−$$23$$−$$1$$×β−$$2$$α−$$1=$$−$$1$$⇒ $$2$$α$$=$$β              …$$(i)Area$$ of Δ$$(ABC)=55$$⇒ $$12AB$$×$$AC=55$$⇒ $$(3$$−$$1)2+(1$$−$$2)2$$×$$($$α−$$1)2+($$β−$$2)2=105$$⇒ $$5($$α−$$1)2+(2$$α−$$2)2=105$$⇒ $$5$$|α−$$1$$|$$=10$$⇒ |α−$$1$$|$$=25$$⇒α−$$1=$$±$$25$$⇒α$$=1$$±$$25$$

A triangle ABC lying in the first quadrant has two vertices at $A (1, 2)$ and $B (3, 1)$ . If $∠ B A C = 90^{\circ}$ and area $Δ (A B C) = 5 \sqrt{5}$ sq. units then the abscissa of the vertex C is

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A triangle ABC lying in the first quadrant has two vertices at A(1,2) and B(3,1). If ∠BAC=90∘and area Δ(ABC)=55 sq. units then the abscissa of the vertex C is

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A triangle ABC lying in the first quadrant has two vertices at $A (1, 2)$ and $B (3, 1)$ . If $∠ B A C = 90^{\circ}$ and area $Δ (A B C) = 5 \sqrt{5}$ sq. units then the abscissa of the vertex C is