Question

The number of possible straight lines, passing through $(2,3)$ and forming a triangle with co-ordinate axis, whose area is $12$ sq. units is______.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

($3$)

Equation of line in intercept form is $\frac{x}{a}+\frac{y}{b}=1$ through (2,3)

$\Rightarrow \frac{2}{a}+\frac{3}{b}=1$ $\Rightarrow 2b+3a=ab$-----------(1)

Intercepts of straight line forms a triangle with co-ordinates axis

Area of triangle = $\frac{1}{2}|ab|=12$

$\Rightarrow ab=\pm 24$

Case 1: When $ab=24,2b+3a=24$

$\Rightarrow 2ab+3a^2=24a$ (Multiplying $a$ on both sides)

$\Rightarrow 3a^2-24a+48=0$ [Since, $2\left(ab\right)=2\times 24=48$]

$\Rightarrow a^2-8a+16=0$

$\Rightarrow a^2-2\left(a\right)\left(4\right)+4^2=0$

$\Rightarrow (a-4{)}^2=0$ [Since, $x^2-2xy+y^2=(x-y{)}^2$]

$\Rightarrow a-4=0$

$\therefore a=4$

Substituting $a=4$ in equation (1)

$\Rightarrow 2b+3\left(4\right)=4\left(b\right)$

$\Rightarrow 4b-2b=12$

$\Rightarrow 2b=12$

$\therefore b=6$

Case 2:

When $ab=-24$

$\Rightarrow 2b+3a=-24$

$\Rightarrow 2ab+3a^2=-24a$ [Multiplying $a$ on both sides]

$\Rightarrow 3a^2+24a+2(-24)=0$

$\Rightarrow a^2+8a-16=0$

$a=\frac{-8\pm \sqrt{8^2-4\times 1\times (-16)}}{2\times 1}$
$=\frac{-8\pm \sqrt{64+64}}{2}$
$=\frac{-8\pm 8\sqrt{2}}{2}$
$=4\pm 4\sqrt{2}$

So, Two value of $a$ and $b$

Thus, two straight lines for case $2$.

Hence, $3$ straight lines (one straight line for case $1$ and Two straight lines for case $2$)

Hence the correct option is (c)