Let the lines $3x - 4y - \alpha = 0$, $8x - 11y - 33 = 0$, and $2x - 3y + \lambda = 0$ be concurrent. If the image of the point $(1, 2)$ in the line $2x - 3y + \lambda = 0$ is $\left(\frac{87}{13}, \frac{-60}{13}\right)$, then $|\alpha \lambda|$ is equal to

Step 1: Find the intersection point of the first two lines

We are given three concurrent lines. First, we find the intersection point of the two lines with known coefficients: $3x - 4y - \alpha = 0$ and $8x - 11y - 33 = 0$. We can rewrite these as $3x - 4y = \alpha$ and $8x - 11y = 33$. To solve this system of equations, we can use the elimination method. Multiply the first equation by 8 and the second by 3 to make the coefficients of $x$ equal.

Step 2: Solve for y and x in terms of alpha

Subtracting the modified second equation from the modified first equation eliminates $x$, allowing us to solve for $y$. Once $y$ is found, substitute it back into the first original equation to solve for $x$. This gives us the coordinates of the intersection point in terms of $\alpha$.

Step 3: Use concurrency to find alpha

Since all three lines are concurrent, the intersection point $(x, y)$ must also lie on the third line $2x - 3y + \lambda = 0$. Substitute the expressions for $x$ and $y$ (in terms of $\alpha$) into the third line's equation. This gives us a relationship between $\alpha$ and $\lambda$.

Step 4: Use the image point formula to find lambda

The problem states that the image of point $(1, 2)$ in the line $2x - 3y + \lambda = 0$ is $\left(\frac{87}{13}, \frac{-60}{13}\right)$. We use the formula for the image of a point $(x_1, y_1)$ in a line $Ax + By + C = 0$. Substitute the given point, image point, and coefficients of the line into the formula.

Step 5: Solve for lambda

From the image point formula, we can equate the first part of the expression to the third part to solve for $\lambda$. This simplifies to $37 = -2(\lambda - 4)$, which allows us to find the value of $\lambda$.

Step 6: Find alpha and calculate |alpha lambda|

Now that we have the value of $\lambda$, substitute it back into the relationship between $\alpha$ and $\lambda$ (from step 3) to find $\alpha$. Finally, calculate the absolute value of the product of $\alpha$ and $\lambda$.

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