A triangle ABC lying in the first quadrant has two vertices as $\mathrm{A}(1,2)$ and $\mathrm{B}(3,1)$. If $\angle BAC=90^{\circ}$, and $\operatorname{ar}(\triangle ABC)=5 \sqrt{5}$ sq. units, then the abscissa of the vertex $C$ is:

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Straight Lines

A triangle ABC lying in the first quadrant has two vertices as

A (1, 2)

and

B (3, 1)

. If

∠ B A C = 9 0^{\circ}

, and

ar (△ A B C) = 55

sq. units, then the abscissa of the vertex

C

is:

(A) $1 + 25$ ✓ Correct
(B) $1 + 5$
(C) $25 - 1$
(D) $2 + 5$

Answer: (A)

Solution

Let

C = (h, k)

. Since

∠ B A C = 9 0^{\circ}

, the vectors

A B = (2, - 1)

and

A C = (h - 1, k - 2)

are perpendicular, so

2 (h - 1) - (k - 2) = 0

, giving

k = 2 h

. The area is

\frac{1}{2} ∣ A B ∣∣ A C ∣ = 55

with

∣ A B ∣ = 5

, so

∣ A C ∣ = 10

and

(h - 1)^{2} + (k - 2)^{2} = 100

. Since

k - 2 = 2 (h - 1)

, this becomes

5 (h - 1)^{2} = 100

, so

(h - 1)^{2} = 20

. As the triangle lies in the first quadrant

h > 1

, giving

h - 1 = 25

, so the abscissa of

C

1 + 25

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option

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