Key Concepts and Formulas

• Intersection of two lines: The coordinates of the intersection point of two lines $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$ are found by solving these two linear equations simultaneously.
• Properties of a Parallelogram: Opposite sides are parallel and equal in length, and diagonals bisect each other.
• Distance of a point from a line: The perpendicular distance $d$ of a point $P(x_0, y_0)$ from a line $Ax+By+C=0$ is given by: $d = \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$

Step-by-Step Solution

We are given the equations of two adjacent sides of a parallelogram $\mathrm{ABCD}$ and the equation of one of its diagonals $\mathrm{AC}$. Let's denote the given equations:

1. Equation of side $\mathrm{AB}$: $2x - 3y = -23 \quad \ldots(1)$
2. Equation of side $\mathrm{BC}$: $5x + 4y = 23 \quad \ldots(2)$
3. Equation of diagonal $\mathrm{AC}$: $3x + 7y = 23 \quad \ldots(3)$

Step 1: Determine the coordinates of the vertices A, B, and C.

• Finding Vertex B: Vertex B is the intersection point of sides AB and BC. To find its coordinates, we solve equations (1) and (2) simultaneously. Multiply equation (1) by 4: $8x - 12y = -92 \quad \ldots(1')$ Multiply equation (2) by 3: $15x + 12y = 69 \quad \ldots(2')$ Adding (1') and (2'): $23x = -23$ $x = -1$ Substitute $x = -1$ into (1): $2(-1) - 3y = -23$ $-2 - 3y = -23$ $-3y = -21$ $y = 7$ Thus, the coordinates of Vertex B are $(-1, 7)$.

• Finding Vertex C: Vertex C is the intersection point of side BC and diagonal AC. To find its coordinates, we solve equations (2) and (3) simultaneously. Multiply equation (2) by 7: $35x + 28y = 161 \quad \ldots(2'')$ Multiply equation (3) by 4: $12x + 28y = 92 \quad \ldots(3')$ Subtracting (3') from (2''): $23x = 69$ $x = 3$ Substitute $x = 3$ into (2): $5(3) + 4y = 23$ $15 + 4y = 23$ $4y = 8$ $y = 2$ Thus, the coordinates of Vertex C are $(3, 2)$.

• Finding Vertex A: Vertex A is the intersection point of side AB and diagonal AC. To find its coordinates, we solve equations (1) and (3) simultaneously. Multiply equation (1) by 7: $14x - 21y = -161 \quad \ldots(1'')$ Multiply equation (3) by 3: $9x + 21y = 69 \quad \ldots(3'')$ Adding (1'') and (3''): $23x = -92$ $x = -4$ Substitute $x = -4$ into (1): $2(-4) - 3y = -23$ $-8 - 3y = -23$ $-3y = -15$ $y = 5$ Thus, the coordinates of Vertex A are $(-4, 5)$.

Step 2: Find the equation of the other diagonal BD.

Since the diagonals of a parallelogram bisect each other, the midpoint of AC is also the midpoint of BD. Let's find the midpoint of AC, which we'll call M. $M = \left(\frac{-4+3}{2}, \frac{5+2}{2}\right) = \left(-\frac{1}{2}, \frac{7}{2}\right)$ Now we know that M is the midpoint of BD, and we know B = (-1, 7). Let D = (x, y). Then, $\frac{x - 1}{2} = -\frac{1}{2} \implies x - 1 = -1 \implies x = 0$ $\frac{y + 7}{2} = \frac{7}{2} \implies y + 7 = 7 \implies y = 0$ So, D = (0, 0). The equation of the diagonal BD is given by the line passing through B(-1, 7) and D(0, 0). The slope of the line is $m = \frac{0 - 7}{0 - (-1)} = \frac{-7}{1} = -7$ The equation of the line is $y - 0 = -7(x - 0)$ $y = -7x$ $7x + y = 0$

Step 3: Calculate the distance 'd' of vertex A from diagonal BD.

We need to find the distance $d$ of point A$(-4, 5)$ from the line $7x + y = 0$. Using the distance formula $d = \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$: Here, $(x_0, y_0) = (-4, 5)$ and the line is $7x + 1y + 0 = 0$, so $A=7$, $B=1$, $C=0$. $d = \frac{|7(-4) + 1(5) + 0|}{\sqrt{7^2 + 1^2}}$ $d = \frac{|-28 + 5|}{\sqrt{49 + 1}}$ $d = \frac{|-23|}{\sqrt{50}}$ $d = \frac{23}{\sqrt{50}}$

Step 4: Calculate the final required value $50d^2$.

We have $d = \frac{23}{\sqrt{50}}$. $d^2 = \left(\frac{23}{\sqrt{50}}\right)^2 = \frac{23^2}{50} = \frac{529}{50}$ Now, we need to find $50d^2$: $50d^2 = 50 \times \frac{529}{50}$ $50d^2 = 529$

Common Mistakes & Tips

• Solving Linear Equations: Be extra careful when solving the systems of linear equations to find the intersection points. Small algebraic errors can lead to incorrect vertex coordinates.
• Using Parallelogram Properties: Remember that the diagonals of a parallelogram bisect each other. This property is crucial for finding the equation of the other diagonal.
• Distance Formula: Double-check the signs and values when applying the point-to-line distance formula.

Summary

This problem combines concepts from coordinate geometry, requiring the student to find intersection points, utilize parallelogram properties to deduce information about diagonals, and apply the point-to-line distance formula. We found the vertices A, B, and C, then used the midpoint of AC to find the equation of diagonal BD. Finally, we calculated the distance of point A from line BD and squared the result to get the final answer.

The final answer is \boxed{529}.