A triangle ABC lying in the first quadrant has two vertices as A 1-2 and B3-1- If - BAC-90- and ar- ABC-5-5 sq- units- then the abscissa of the vertex C is

Since nbsp;∠ BAC=90^∘ ∴ m_AC· m_AB= 1 ⇒b 2a 1·2 11 3= 1 ⇒b 2a 1=2 ⇒ b=2a ⋯(1) Area of triangle ABC=5√(5) 12×√(5)×√((a 1)^2+(b 2)^2)=5√(5) ⇒ nbsp;√((a 1)^ ...

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Byju's Answer

Standard XII

Mathematics

Area Method to Find Condition for Co-Linearity

A triangle AB...

Question

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

1 + \sqrt{5}

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1 + 2 \sqrt{5}

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2 \sqrt{5} - 1

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2 + \sqrt{5}

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Solution

The correct option is B $1 + 2 \sqrt{5}$

Since $∠ B A C = 90^{\circ}$
$∴ m_{A C} \cdot m_{A B} = - 1$
$\Rightarrow \frac{b - 2}{a - 1} \cdot \frac{2 - 1}{1 - 3} = - 1$
$\Rightarrow \frac{b - 2}{a - 1} = 2$
$\Rightarrow b = 2 a \dots (1)$
Area of triangle $A B C = 5 \sqrt{5}$
$\frac{1}{2} \times \sqrt{5} \times \sqrt{(a - 1)^{2} + (b - 2)^{2}} = 5 \sqrt{5}$
$\Rightarrow \sqrt{(a - 1)^{2} + (b - 2)^{2}} = 10$
$\Rightarrow \sqrt{(a - 1)^{2} + 4 (a - 1)^{2}} = 10$
$\Rightarrow \sqrt{5} (a - 1) = 10$
$∴ a = 1 + 2 \sqrt{5}$

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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)=5√5 sq. units, then the abscissa of the vertex C is

The correct option is B 1+2√5 Since ∠BAC=90∘ ∴mAC⋅mAB=−1 ⇒b−2a−1⋅2−11−3=−1 ⇒b−2a−1=2 ⇒b=2a ⋯(1) Area of triangle ABC=5√5 12×√5×√(a−1)2+(b−2)2=5√5 ⇒√(a−1)2+(b−2)2=10 ⇒√(a−1)2+4(a−1)2=10 ⇒√5(a−1)=10 ∴a=1+2√5

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