Let the equations of two adjacent sides of a parallelogram ABCD be 2x - 3y = -23 and 5x + 4y = 23. I

Q: Wait, why do we need to find all the vertices? Can't we just use the slope of the sides to find the other vertices?

You absolutely could use slopes, but that would be a much longer and more tedious path! In coordinate geometry, the most elegant solution is often the one that uses the properties of the shape. The fact that diagonals bisect each other is a powerful shortcut that lets us find $D$ instantly once we have $A$, $B$, and $C$.

Q: How do I know which intersection corresponds to which vertex?

Look at the equations! The side $AB$ and the diagonal $AC$ both pass through vertex $A$. Therefore, their intersection point must be $A$. Similarly, $BC$ and $AC$ intersect at $C$, and $AB$ and $BC$ intersect at $B$. It is all about identifying the common points shared by the lines.

Q: Is there a faster way to find $d$ without finding the equation of $BD$?

Not really. To use the point-to-line distance formula, you need the equation of the line $BD$. Geometry is like a puzzle; you have to build the pieces step-by-step. Finding the equation of $BD$ is just one of those necessary pieces, and it is quite satisfying when it all comes together!

Question

Let the equations of two adjacent sides of a parallelogram $ABCD$ be $2x - 3y = -23$ and $5x + 4y = 23$. If the equation of its one diagonal $AC$ is $3x + 7y = 23$ and the distance of $A$ from the other diagonal is $d$, then $50d^2$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the coordinate plane. Today, we stand before a parallelogram, a shape that seems simple on the surface but hides a beautiful, rigid structure within its lines.

We are given two adjacent sides, $2x - 3y = -23$ and $5x + 4y = 23$, and one diagonal, $AC$, defined by $3x + 7y = 23$. Our mission is to find the perpendicular distance $d$ from vertex $A$ to the other diagonal, $BD$, and ultimately, to calculate $50d^2$.

### The Intersection Hunt

To understand our parallelogram, we must first know where its corners lie. Think of the vertices as the anchors of our shape.

Vertex $A$ is the meeting point of side $AB$ and diagonal $AC$. By solving the system of equations $2x - 3y = -23$ and $3x + 7y = 23$, we find the coordinates of $A$ to be $(-4, 5)$.

Vertex $C$ is the intersection of side $BC$ and diagonal $AC$. Solving $5x + 4y = 23$ and $3x + 7y = 23$ reveals $C$ to be $(3, 2)$.

Vertex $B$ is the intersection of the two given sides, $AB$ and $BC$. Solving $2x - 3y = -23$ and $5x + 4y = 23$ gives us $B(-1, 7)$. We have successfully anchored three of our four corners.

### The Midpoint Magic

Now, we need vertex $D$. This is where the magic of geometry comes in.

A fundamental property of any parallelogram is that its diagonals bisect each other. This means the midpoint $M$ of diagonal $AC$ is the exact same point as the midpoint of diagonal $BD$.

Let us calculate $M$ using $A(-4, 5)$ and $C(3, 2)$:

$$M = \left( \frac{-4+3}{2}, \frac{5+2}{2} \right) = \left( -\frac{1}{2}, \frac{7}{2} \right)$$

Since $M$ is also the midpoint of $BD$, and we know $B(-1, 7)$, we set up the following:

$$\left( \frac{-1 + x_D}{2}, \frac{7 + y_D}{2} \right) = \left( -\frac{1}{2}, \frac{7}{2} \right)$$

Solving this, we find $x_D = 0$ and $y_D = 0$. Vertex $D$ is beautifully located at the origin, $(0, 0)$.

### The Final Distance

With $B(-1, 7)$ and $D(0, 0)$ in hand, we find the equation of the diagonal $BD$. The slope $m$ is:

$$m = \frac{7 - 0}{-1 - 0} = -7$$

Since it passes through the origin, the equation is $y = -7x$, or $7x + y = 0$. Now, we calculate the perpendicular distance $d$ from $A(-4, 5)$ to this line using the formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$:

$$d = \frac{|7(-4) + 1(5)|}{\sqrt{7^2 + 1^2}} = \frac{|-28 + 5|}{\sqrt{49 + 1}} = \frac{23}{\sqrt{50}}$$

Finally, the question asks for $50d^2$. Squaring $d$ gives:

$$d^2 = \frac{529}{50}$$

Multiplying by $50$, the denominators cancel out, leaving us with the final result of **529**.

Enter Numerical Value:

Visualized Solution

Visualizing the Parallelogram

Finding Vertex $A$

Finding Vertex $C$

Finding Vertex $B$

Midpoint of Diagonals

Finding Vertex $D$

Equation of Diagonal $B D$

Calculating Distance $d$

Final Answer: $50 d^{2}$

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

The Intersection Hunt

The Midpoint Magic

The Final Distance

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Enter Numerical Value:

Visualized Solution

Visualizing the Parallelogram

Finding Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Finding Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Finding Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Midpoint of Diagonals

Finding Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Equation of Diagonal .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BD

Calculating Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Final Answer: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 50d2