Question

If the area of triangle is 12 sq. Units with vertices (a, −3), (3, a) (−1, 5) find ‘a’.

(a) – 1,2 (b) 1,3 (c) 2,1 (d) 3,1​

Answer 1

If the area of a triangle with vertices $(x_1,\ y_1),\ (x_2,\ y_2)$ and $(x_3,\ y_3)$ is A sq. units, then,

• $\dfrac{1}{2}\left|\begin{array}{cc}x_1-x_2&y_1-y_2\\x_2-x_3&y_2-y_3\end{array}\right|=\pm A$

In the question,

• $(x_1,\ y_1)=(a,\ -3)$

• $(x_2,\ y_2)=(3,\ a)$

• $(x_3,\ y_3)=(-1,\ 5)$

• $A=12$

Then,

$\longrightarrow\dfrac{1}{2}\left|\begin{array}{cc}a-3&-3-a\\4&a-5\end{array}\right|=\pm12$

$\longrightarrow\left|\begin{array}{cc}a-3&-(a+3)\\4&a-5\end{array}\right|=\pm24$

$\longrightarrow(a-3)(a-5)+4(a+3)=\pm24$

$\longrightarrow a^2-8a+15+4a+12=\pm24$

$\longrightarrow a^2-4a+27=\pm24$

Take,

$\longrightarrow a^2-4a+27=24$

$\longrightarrow a^2-4a+3=0$

$\longrightarrow(a-1)(a-3)=0$

$\longrightarrow a\in\{1,\ 3\}$

Take,

$\longrightarrow a^2-4a+27=-24$

$\longrightarrow a^2-4a+51=0$

The roots of this quadratic equation are not real since D < 0.

Hence the possible values of a are 1 and 3.

Answer 2

Given :-

Area = 12 sq. units

vertices (a, −3), (3, a) (−1, 5)

Solution :-

We know that

$\sf Area_{\triangle}=\dfrac{1}{2}\left[\begin{array}{ccc}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1}\end{array}\right]$

$\sf 12=\dfrac{1}{2}\left[\begin{array}{ccc}a&-3&1\\3&a&1\\ -1&5&1}\end{array}\right]$

$\sf 12\times 2=\left[\begin{array}{ccc}a&-3&1\\3&a&1\\ -1&5&1}\end{array}\right]$

$\sf 24=\left[\begin{array}{ccc}a&-3&1\\3&a&1\\ -1&5&1}\end{array}\right]$

$\sf 24=(a-3)(a-5)+4(a+3)$

$\sf 24 = (a-3)(a-5)+4a+12$

$\sf 24= a^2 - (8a - 15) + 4a+12$

$\sf 24 = a^2 - 8a+15+4a+12$

$\sf 24 = a^2- (8a - 4a)+(15+12)$

$\sf 24 = a^2+4a+27$

$\sf 27-24=a^2-4a$

$\sf 3= a^2-4a$

$\sf a =3 \;or \;a= 1$