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)^ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

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}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 + 2 \sqrt{5}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 \sqrt{5} - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 + \sqrt{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $A B C = 5 \sqrt{5}$
$\frac{1}{2} \times \sqrt{5} \times \sqrt{(a - 1)^{2} + (b - 2)^{2}} = 5 \sqrt{5}$
$\Rightarrow \sqrt{(a - 1)^{2} + (b - 2)^{2}} = 10$
$\Rightarrow \sqrt{(a - 1)^{2} + 4 (a - 1)^{2}} = 10$
$\Rightarrow \sqrt{5} (a - 1) = 10$
$∴ a = 1 + 2 \sqrt{5}$

Suggest Corrections

146

Similar questions

Q. If in a

Δ A B C

b = 12

units,

c = 5

units and

Δ = 30

sq. units,

then the distance between vertex

A

and incentre of the triangle is equal to?

Q. The area of an equilateral triangle whose two vertices are (1,0) and (3,0) and third vertex lying in the first quadrant is-

Q. In a triangle

A B C

, coordiantes of

A

are

(1, 2)

and the equation of the medians through

B

and

C

are respectively,

x + y = 5

and

x = 4

. Then area of

△ A B C

( in

s q u n i t s

) is

Q. The vertices

B

and

C

of a

△ A B C

lie on the line,

\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}

such that

B C = 5

units. Then the area (in sq. units) of this triangle, given that the point

A (1, - 1, 2)

is:

Q. In a triangle

A B C

, coordinates of

A

are

(1, 2)

and the equations of the medians through

B

and

C

are

x + y = 5

and

x = 4

respectively,Then area of

△ A B C

(in sq. units) is

Join BYJU'S Learning Program

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