Question

The equation of straight line passing through $(-a,0)$ and making the triangle with axes of area $T$ is

• A. $2Tx+a^{2}y+2aT=0$
• B. $2Tx-a^{2}y+2aT=0$
• C. $2Tx-a^{2}y-2aT=0$
• D. None of these

Answer 1

Answer: (A), (B)

$2Tx-a^{2}y+2aT=0$

Explanation for correct answer:

Since it is given that the line passes through $(-a,0)$ , thus the $x$ intercept =$-a$

Now we know the formula for

$$\begin{gathered} Areaoftriangle=\frac{1}{2}\times |xintercept|\times |yintercept| \\ \Rightarrow T=\frac{1}{2}a\times |yintercept| \\ \Rightarrow y=\pm \frac{2T}{a} \end{gathered}$$

So the equation of the line will be calculated by

$$\begin{gathered} \frac{x}{A}+\frac{y}{B}=1where \\ A=-a \\ B=\pm \frac{2T}{a} \end{gathered}$$

$$\begin{gathered} \frac{x}{-a}+\frac{y}{\left({\displaystyle \frac{2T}{a}}\right)}=1or\frac{x}{-a}+\frac{y}{\left({\displaystyle \frac{-2T}{a}}\right)}=1 \\ \Rightarrow -2Tx+a^{2}y=2aTor-2Tx–a^{2}y=2aT \\ \Rightarrow 2Tx–a^{2}y+2aT=0or2Tx+a^{2}y+2aT=0 \end{gathered}$$

Hence, the correct answer is option (A) and (B).