Question

The lines $(a+2b)x+(a-3b)y=a-b$for different values of $a$ and $b$ pass through the fixed point, whose coordinates are

• A. $\left(\frac{2}{5},\frac{3}{5}\right)$
• B. $\left(\frac{3}{5},\frac{3}{5}\right)$
• C. $\left(\frac{1}{5},\frac{1}{5}\right)$
• D. $\left(\frac{3}{5},\frac{2}{5}\right)$

Answer 1

The correct option is A

$\left(\frac{2}{5},\frac{3}{5}\right)$

Explanation for the correct options:

Find the coordinates of the fixed points

Given, the equation of the lines $=(a+2b)x+(a-3b)y=a-b$

$$\begin{gathered} \Rightarrow ax+2bx+ay-3by-a+b=0 \\ \Rightarrow a(x+y-1)+b(2x-3y+1)=0 \end{gathered}$$

$a{L}_1+b{L}_2=0$ form a family of straight lines,all lines always pass through the intersection point of $L_{1}and{L}_2$.

The two equation

$$\begin{gathered} \Rightarrow x+y-1=0.....\left(1\right) \\ \Rightarrow 2x-3y+1=0.....\left(2\right) \end{gathered}$$

Solving equations $\left(1\right)$ and $\left(2\right)$, we get

From $\left(1\right)$

$\Rightarrow y=1-x.......\left(3\right)$

Putting the above value in $\left(2\right)$

$$\begin{gathered} \Rightarrow 2x-3(1-x)+1=0 \\ \Rightarrow 5x-2=0 \\ \Rightarrow x=\frac{2}{5} \end{gathered}$$

Putting the value of $x$ in $\left(3\right)$

$$\begin{gathered} \Rightarrow y=1-\frac{2}{5} \\ \Rightarrow y=\frac{3}{5} \end{gathered}$$

$\Rightarrow x=\frac{2}{5}\&y=\frac{3}{5}$

So, the coordinates of the fixed point is $\left(\frac{2}{5},\frac{3}{5}\right)$

Hence, the correct option is option (A).