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

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−2α−1=−1⇒ 2α=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+(2α−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 ∠BAC=90∘and area Δ(ABC)=55 sq. units then the abscissa of the vertex C is

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−2α−1=−1⇒ 2α=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+(2α−2)2=105⇒ 5|α−1|=10⇒ |α−1|=25⇒α−1=±25⇒α=1±25

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$1 + \sqrt{5}$

$2 \sqrt{5} - 1$

$1 + 2 \sqrt{5}$

$2 + \sqrt{5}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let the coordinates of vertex C be $(α, β)$ . It is given that $∠ B A C = 90^{\circ}$

$\Rightarrow$ Slope of AB x Slope of AC = -1

$\Rightarrow \frac{1 - 2}{3 - 1} \times \frac{β - 2}{α - 1} = - 1$

$\Rightarrow 2 α = β$ …(i)

Area of $Δ (A B C) = 5 \sqrt{5}$

$\begin{array}{l} \Rightarrow \frac{1}{2} A B \times A C = 5 \sqrt{5} \\ \Rightarrow \sqrt{(3 - 1)^{2} + (1 - 2)^{2}} \times \sqrt{(α - 1)^{2} + (β - 2)^{2}} = 10 \sqrt{5} \\ \Rightarrow \sqrt{5} \sqrt{(α - 1)^{2} + (2 α - 2)^{2}} = 10 \sqrt{5} \end{array}$

$\begin{array}{l} \Rightarrow \sqrt{5} | α - 1 | = 10 \\ \Rightarrow | α - 1 | = 2 \sqrt{5} \Rightarrow α - 1 = \pm 2 \sqrt{5} \Rightarrow α = 1 \pm 2 \sqrt{5} \end{array}$