Finding Parallelogram Diagonal Equation Using Side Lines and Center

This solution explains how to find the equation of a specific diagonal of a parallelogram, given the equations of two adjacent sides and the point where the diagonals intersect (the center).

Understanding the Problem Setup

We are given:

• Equation of side 1: $x + y = 3$
• Equation of side 2: $x - y + 3 = 0$
• Intersection point of diagonals (center): $C = (2, 4)$

We need to find the equation of the diagonal that does *not* go through the vertex formed by the intersection of the two given side lines. Let's call this vertex V.

Step 1: Find the Vertex V

The vertex V is the intersection point of the two given sides. We can find it by solving the system of linear equations:

1. $x + y = 3$
2. $x - y = -3$ (Rearranging the second equation)

Add the two equations:

$ (x + y) + (x - y) = 3 + (-3) $

$ 2x = 0 $

$ x = 0 $

Substitute $x = 0$ back into the first equation ($x + y = 3$):

$ 0 + y = 3 \implies y = 3 $

So, the vertex V is at $(0, 3)$.

Step 2: Understand the Diagonals

A parallelogram has two diagonals which bisect each other at the center. The center is given as $C = (2, 4)$.

One diagonal passes through the vertex V (where the given sides intersect) and the opposite vertex. This diagonal must also pass through the center C.

The other diagonal connects the remaining two vertices and also passes through the center C.

The question asks for the diagonal that does *not* pass through vertex V = (0, 3).

Step 3: Calculate the Equation of the Diagonal Through V

One diagonal passes through vertex $V = (0, 3)$ and the center $C = (2, 4)$. Let's find its equation.

First, find the slope ($m$) of this diagonal:

$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 3}{2 - 0} = \frac{1}{2} $

Now, use the point-slope form of a linear equation ($y - y_1 = m(x - x_1)$) with point V=(0, 3) and slope $m = 1/2$:

$ y - 3 = \frac{1}{2}(x - 0) $

$ 2(y - 3) = x $

$ 2y - 6 = x $

$ x - 2y + 6 = 0 $

This is the equation of the diagonal that passes through the vertex V=(0, 3).

Step 4: Identify the Required Diagonal

We are looking for the equation of the diagonal that does *not* pass through vertex $V = (0, 3)$. This must be the *other* diagonal.

We know this other diagonal must also pass through the center $C = (2, 4)$. We need to check the given options to find the equation of a line that passes through $C = (2, 4)$ and is *not* $x - 2y + 6 = 0$.

Step 5: Test the Options

Let's check which options satisfy the center point $(2, 4)$ and are not $x - 2y + 6 = 0$. The options are:

• Option 1: $3x - y - 2 = 0$
• Option 2: $x - 2y + 6 = 0$ (This is the diagonal passing through V, so we exclude it based on the question).
• Option 3: $x + y - 4 = 0$
• Option 4: $2x - y = 0$

Check Option 1: Substitute $(2, 4)$

$ 3(2) - 4 - 2 = 6 - 4 - 2 = 0 $

Option 1 passes through the center $(2, 4)$.

Check Option 3: Substitute $(2, 4)$

$ 2 + 4 - 4 = 2 $

Option 3 does *not* pass through the center $(2, 4)$.

Check Option 4: Substitute $(2, 4)$

$ 2(2) - 4 = 4 - 4 = 0 $

Option 4 passes through the center $(2, 4)$.

We have two potential candidates for the other diagonal: Option 1 ($3x - y - 2 = 0$) and Option 4 ($2x - y = 0$).

Step 6: Confirm the Correct Diagonal

We found the vertex $V = (0, 3)$. The diagonal passing through V has equation $x - 2y + 6 = 0$. The other diagonal must pass through the center $C = (2, 4)$.

Let's consider the properties of a parallelogram. The lines containing the sides are $x+y-3=0$ and $x-y+3=0$. The center is $(2,4)$. The lines containing the other two sides must be parallel to these and pass through the vertices opposite to $V$. The opposite vertex $V'$ can be found using the center as the midpoint: $V' = (2 \times 2 - 0, 2 \times 4 - 3) = (4, 5)$.

The lines containing the other sides are:

• Parallel to $x+y-3=0$ through $(4,5)$: $x+y = 4+5 \implies x+y=9$
• Parallel to $x-y+3=0$ through $(4,5)$: $x-y = 4-5 \implies x-y=-1$

The vertices of the parallelogram are the intersections of these four lines:

• $V = (x+y=3) \cap (x-y=-3) = (0, 3)$
• Vertex B = $(x+y=3) \cap (x-y=-1) = (1, 2)$
• $V' = (x+y=9) \cap (x-y=-1) = (4, 5)$
• Vertex D = $(x+y=9) \cap (x-y=-3) = (3, 6)$

The diagonals connect opposite vertices: $VV'$ and $BD$.

Diagonal $VV'$ passes through $V=(0,3)$ and $C=(2,4)$. Its equation is $x - 2y + 6 = 0$.

Diagonal $BD$ passes through $B=(1,2)$ and $D=(3,6)$. Let's find its equation:

Slope $m = \frac{6 - 2}{3 - 1} = \frac{4}{2} = 2$.

Using point-slope form with $B=(1,2)$:

$ y - 2 = 2(x - 1) $

$ y - 2 = 2x - 2 $

$ 2x - y = 0 $

This is the equation of the diagonal $BD$. Does it pass through the center $C=(2,4)$? $2(2) - 4 = 4 - 4 = 0$. Yes, it does.

This diagonal $BD$ (equation $2x - y = 0$) does not pass through vertex $V=(0,3)$ because $2(0) - 3 = -3 \ne 0$. This matches Option 4.

Conclusion

The vertex formed by the intersection of the given sides $x + y = 3$ and $x - y + 3 = 0$ is $V=(0, 3)$.

The diagonal passing through this vertex V and the center $(2, 4)$ has the equation $x - 2y + 6 = 0$.

The other diagonal passes through the center $(2, 4)$ but does not pass through vertex $V=(0, 3)$. Its equation was calculated to be $2x - y = 0$.

Therefore, the equation of the required diagonal is $2x - y = 0$. This corresponds to Option 4.