A line L intersects the lines $3 x-2 y-1=0$ and $x+2 y+$ $1=0$ at the points A and B . If the point $(1,2)$ bisects the line segment AB and $\frac{x}{a}+\frac{y}{b}=1$ is the equation of the line L. then $a+2 b+1=$

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Aicharya · Accepted Answer

Let point of intersection $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$

$\therefore \frac{x_1+x_2}{2}=1 \Rightarrow x_2=2-x_1 \Rightarrow \frac{y_1+y_2}{2}=2 \Rightarrow y_2=4-y_1$ $\because$ A lies on $3 x-2 y-1=0$ $\therefore 3 x_1-2 y_1-1=0$ ....(i) and $B$ lies on $x+2 y+1=0$ $\therefore\left(2-x_1\right)+2\left(4-y_1\right)+1=0 \Rightarrow x_1+2 y_1-11=0$ .....(ii) On solving equation (i) and (ii), we get $x_1=3, y_1=4 \Rightarrow x_2=-1, y_2=0$ So, point $\mathrm{A}(3,4)$ and $\mathrm{B}(-1,0)$ $\therefore$ Equation of line L i.e., AB : $y-4=\frac{4-0}{3+1}(x-3) \Rightarrow x-y+1=0 \Rightarrow \frac{x}{-1}+\frac{y}{1}=1$, Compare with $\frac{x}{a}+\frac{y}{b}=1$ we get $a=-1$ and $b=1 \Rightarrow$ Now, $a+2 b+1=-1+2+1=2$

A line L intersects the lines $3 x-2 y-1=0$ and $x+2 y+$ $1=0$ at the points A and B . If the point $(1,2)$…

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