A triangle has two of its vertices at (0 , 1) and (2 , 2) in the Cartesian plane. Its third vertex lies on the x -axis. If the area of the triangle is 2 square units, then the sum of the square of the possible values of abscissa of the third vertex is

Question

A triangle has two of its vertices at (0 , 1) and (2 , 2) in the Cartesian plane. Its third vertex lies on the x -axis. If the area of the triangle is 2 square units, then the sum of the square of the possible values of abscissa of the third vertex is (A) 40 (B) 10 (C) 52 (D) 61

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-mlqm54uepygl.png" /> Let, the third vertex be (a , 0) Area of triangle, Δ =(1/2)| 0 1 1 2 2 1 a 0 1 |=2 ∴ | 0 0 1 2 1 1 a -1 1 |=4 ⇒ |a + 2|=4 i.e. a=2 or a=-6 Hence, the sum of the square of the possible values of abscissa =40

Q. A triangle has two of its vertices at $(0, 1)$ and $(2, 2)$ in the Cartesian plane. Its third vertex lies on the $x$ -axis. If the area of the triangle is $2$ square units, then the sum of the square of the possible values of abscissa of the third vertex is

$40$

$10$

$52$

$61$

Let, the third vertex be $(a, 0)$
Area of triangle, $Δ = \frac{1}{2} ∣ ∣ 02 a 120111 ∣ ∣ = 2$
$∴ ∣ ∣ 02 a 01 - 1 111 ∣ ∣ = 4$
$\Rightarrow ∣ a + 2 ∣ = 4$
i.e. $a = 2$ or $a = - 6$
Hence, the sum of the square of the possible values of abscissa $= 40$

Q. A triangle has two of its vertices at (0,1) and (2,2) in the Cartesian plane. Its third vertex lies on the x -axis. If the area of the triangle is 2 square units, then the sum of the square of the possible values of abscissa of the third vertex is

40

10

52

61