A line L intersects the lines 3x - 2y - 1 = 0 | vedclass

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(D) Let the points of intersection be $A(x_1, y_1)$ and $B(x_2, y_2)$.Since the point $(1, 2)$ bisects the line segment $AB$,we have:$\frac{x_1 + x_2}

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$2$

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(D) Let the points of intersection be $A(x_1, y_1)$ and $B(x_2, y_2)$.Since the point $(1, 2)$ bisects the line segment $AB$,we have:$\frac{x_1 + x_2}{2} = 1 \Rightarrow x_2 = 2 - x_1$$\frac{y_1 + y_2}{2} = 2 \Rightarrow y_2 = 4 - y_1$Since $A$ lies on $3x - 2y - 1 = 0$,we have:$3x_1 - 2y_1 - 1 = 0$  --- $(i)$Since $B$ lies on $x + 2y + 1 = 0$,we substitute $x_2$ and $y_2$:$(2 - x_1) + 2(4 - y_1) + 1 = 0$$2 - x_1 + 8 - 2y_1 + 1 = 0$$x_1 + 2y_1 - 11 = 0$  --- (ii)Solving equations $(i)$ and (ii) by adding them:$(3x_1 - 2y_1 - 1) + (x_1 + 2y_1 - 11) = 0$$4x_1 - 12 = 0 \Rightarrow x_1 = 3$Substituting $x_1 = 3$ in $(i)$:$3(3) - 2y_1 - 1 = 0 \Rightarrow 8 = 2y_1 \Rightarrow y_1 = 4$Thus,$A = (3, 4)$.Then $x_2 = 2 - 3 = -1$ and $y_2 = 4 - 4 = 0$,so $B = (-1, 0)$.The equation of line $L$ passing through $(3, 4)$ and $(-1, 0)$ is:$y - 0 = \frac{4 - 0}{3 - (-1)}(x - (-1))$$y = \frac{4}{4}(x + 1) \Rightarrow y = x + 1 \Rightarrow x - y = -1$Dividing by $-1$,we get $\frac{x}{-1} + \frac{y}{1} = 1$.Comparing with $\frac{x}{a} + \frac{y}{b} = 1$,we get $a = -1$ and $b = 1$.Therefore,$a + 2b + 1 = -1 + 2(1) + 1 = -1 + 2 + 1 = 2$.

$A$ line $L$ intersects the lines $3x - 2y - 1 = 0$ and $x + 2y + 1 = 0$ 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 + 2b + 1 = $

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