A triangle ABC lying in the first quadrant has two vertices as A (1,2) and B (3,1). If angle BAC =90°, and ar (Δ ABC )=5 √5 sq. units, then the abscissa of the vertex C is

Question

A triangle ABC lying in the first quadrant has two vertices as A (1,2) and B (3,1). If angle BAC =90°, and ar (Δ ABC )=5 √5 sq. units, then the abscissa of the vertex C is (A) 2+√5 (B) 1+√5 (C) 1+2 √5 (D) 2 √5-1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/ae9adc64f5df31a46721cd00875ed6b2-.png" /> (( K -2/ h -1))((1-2/3-1))=-1 ⇒ K =2 h dots(1) √5| h -1|=10 ∵[Δ ABC ]=5 √5 ⇒ (1/2)(√5) √( h -1)2+( K -2)2=5 √5 ldots(2) ⇒ h =2 √5+1( h >0)

Q. A triangle $A BC$ lying in the first quadrant has two vertices as $A (1, 2)$ and $B (3, 1)$ . If $∠ B A C = 9 0^{\circ},$ and $a r (Δ A BC) = 55$ sq. units, then the abscissa of the vertex $C$ is

$2 + 5$

$1 + 5$

$1 + 25$

$25 - 1$

$(\frac{K - 2}{h - 1}) (\frac{1 - 2}{3 - 1}) = - 1$
$\Rightarrow K = 2 h \dots$ (1)
$5 ∣ h - 1∣ = 10$
$∵ [Δ A BC] = 55$
$\Rightarrow \frac{1}{2} (5) (h - 1)^{2} + (K - 2)^{2} = 55 \dots$ (2)
$\Rightarrow h = 25 + 1 (h > 0)$

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

2+5​

1+5​

1+25​

25​−1