Finding the Diagonal Equation Through the Origin

The problem asks for the equation of the diagonal that passes through the origin of a quadrilateral formed by four given lines: \(x = 0\), \(y = 0\), \(x + y = 1\), and \(6x + y = 3\).

First, we need to find the vertices of the quadrilateral. The vertices are the intersection points of these lines that form the boundary of the enclosed region.

Let's find the intersection points of pairs of lines:

1. Intersection of \(x = 0\) and \(y = 0\): Substituting \(x=0\) into \(y=0\) gives the point \( (0,0) \). This is the origin.
2. Intersection of \(x = 0\) and \(x + y = 1\): Substituting \(x=0\) into \(x + y = 1\) gives \(0 + y = 1\), so \(y = 1\). The point is \( (0,1) \).
3. Intersection of \(x = 0\) and \(6x + y = 3\): Substituting \(x=0\) into \(6x + y = 3\) gives \(6(0) + y = 3\), so \(y = 3\). The point is \( (0,3) \).
4. Intersection of \(y = 0\) and \(x + y = 1\): Substituting \(y=0\) into \(x + y = 1\) gives \(x + 0 = 1\), so \(x = 1\). The point is \( (1,0) \).
5. Intersection of \(y = 0\) and \(6x + y = 3\): Substituting \(y=0\) into \(6x + y = 3\) gives \(6x + 0 = 3\), so \(6x = 3\), which means \(x = \frac{1}{2}\). The point is \( (\frac{1}{2},0) \).
6. Intersection of \(x + y = 1\) and \(6x + y = 3\): We can solve this system of equations. $$x + y = 1 \quad \text{(Eq 1)}$$ $$6x + y = 3 \quad \text{(Eq 2)}$$ Subtract (Eq 1) from (Eq 2): $$(6x + y) - (x + y) = 3 - 1$$ $$5x = 2$$ $$x = \frac{2}{5}$$ Substitute \(x = \frac{2}{5}\) into (Eq 1): $$\frac{2}{5} + y = 1$$ $$y = 1 - \frac{2}{5} = \frac{5 - 2}{5} = \frac{3}{5}$$ The point is \( (\frac{2}{5}, \frac{3}{5}) \).

The intersection points we found are \( (0,0) \), \( (0,1) \), \( (0,3) \), \( (1,0) \), \( (\frac{1}{2},0) \), and \( (\frac{2}{5}, \frac{3}{5}) \). These are potential vertices or points on the boundary lines.

The quadrilateral formed by these lines is the region bounded by them. By sketching or visualizing, the vertices of the quadrilateral are the points where the boundary lines intersect to form the enclosed shape. These vertices are:

• \( (0,0) \): Intersection of \(x=0\) and \(y=0\).
• \( (0,1) \): Intersection of \(x=0\) and \(x+y=1\).
• \( (\frac{2}{5}, \frac{3}{5}) \): Intersection of \(x+y=1\) and \(6x+y=3\).
• \( (\frac{1}{2},0) \): Intersection of \(y=0\) and \(6x+y=3\).

Let the vertices be \(A=(0,0)\), \(B=(0,1)\), \(C=(\frac{2}{5}, \frac{3}{5})\), and \(D=(\frac{1}{2},0)\). These vertices form a quadrilateral. The diagonals of this quadrilateral connect opposite vertices.

The diagonals are:

• Diagonal 1: Connects \(A=(0,0)\) and \(C=(\frac{2}{5}, \frac{3}{5})\).
• Diagonal 2: Connects \(B=(0,1)\) and \(D=(\frac{1}{2},0)\).

We are looking for the equation of the diagonal that passes through the origin \( (0,0) \). This is Diagonal 1, connecting \( (0,0) \) and \( (\frac{2}{5}, \frac{3}{5}) \).

The equation of a line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) can be found using the two-point form or by finding the slope and using the point-slope form.

Using point \( (0,0) \) and \( (\frac{2}{5}, \frac{3}{5}) \):

The slope \(m\) of the line is given by:

Since the line passes through the origin \( (0,0) \), its equation is of the form \( y = mx \). Substituting the slope \(m = \frac{3}{2}\):

To write this in the general form \(Ax + By + C = 0\) or a form similar to the options, we can multiply by 2:

Rearranging the terms to one side:

This is the equation of the diagonal passing through the origin.

Comparing this equation with the given options, we find that it matches one of the options.

Thus, the equation of the diagonal through the origin is \(3x - 2y = 0\).

Revision Table: Quadrilateral Vertices and Diagonals

Additional Information: Understanding Quadrilaterals and Diagonals

A quadrilateral is a polygon with four vertices and four sides. The vertices are points where the sides meet. In coordinate geometry, lines can intersect to form closed shapes like quadrilaterals.

A diagonal of a quadrilateral is a line segment connecting two non-adjacent vertices. A quadrilateral always has exactly two diagonals.

To find the equation of a line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \), you can use the formula:

If one of the points is the origin \( (0,0) \), the formula simplifies to:

This is the slope-intercept form \(y = mx\), where \(m = \frac{y_2}{x_2}\) is the slope, provided \(x_2 \neq 0\). If \(x_2 = 0\), the line is vertical (if \(y_2 \neq 0\)) and its equation is \(x = x_1\).

In this problem, one point is the origin \( (0,0) \) and the other is \( (\frac{2}{5}, \frac{3}{5}) \), where \(x_2 = \frac{2}{5} \neq 0\). So we can use the formula \(y = \frac{y_2}{x_2}x\) directly.

Equation of the other diagonal:

The other diagonal connects \(B=(0,1)\) and \(D=(\frac{1}{2},0)\).

Slope \(m_{BD} = \frac{0 - 1}{\frac{1}{2} - 0} = \frac{-1}{\frac{1}{2}} = -2\).

Using point \(B=(0,1)\) and slope \(m=-2\): \(y - y_1 = m(x - x_1)\)

This is the equation of the other diagonal.