Finding the Equation of a Perpendicular Line

Let's find the equation of a line that meets two conditions: it's perpendicular to a given line and forms a specific area with the coordinate axes.

Step 1: Determine the Slope of the Given Line

The given line is \(5{\rm{x}} - {\rm{y}} = 0\). We can rewrite this equation in the slope-intercept form, \({\rm{y}} = {\rm{mx}} + {\rm{c}}\), where \({\rm{m}}\) is the slope.

Rearranging \(5{\rm{x}} - {\rm{y}} = 0\), we get:

The slope of this line is \({\rm{m}}_1 = 5\).

Step 2: Determine the Slope of the Perpendicular Line

If a line is perpendicular to another line with slope \({\rm{m}}_1\), its slope \({\rm{m}}_2\) is the negative reciprocal of \({\rm{m}}_1\). The relationship is \({\rm{m}}_1 \cdot {\rm{m}}_2 = -1\).

So, the slope of the line perpendicular to \(5{\rm{x}} - {\rm{y}} = 0\) is:

Step 3: Write the General Equation of the Perpendicular Line

The equation of a line with slope \({\rm{m}}_2 = -1/5\) can be written in the form \({\rm{y}} = {\rm{m}}_2{\rm{x}} + {\rm{c}}\) or \({\rm{Ax}} + {\rm{By}} + {\rm{C}} = 0\). Using the slope-intercept form:

Multiplying by 5 to clear the fraction:

Rearranging into the general form \({\rm{Ax}} + {\rm{By}} + {\rm{C}} = 0\):

Let's replace the constant term \(-5{\rm{c}}\) with a new constant, say \({\rm{K}}\). The equation of the perpendicular line is of the form:

Step 4: Find the Intercepts of the Perpendicular Line

The line \({\rm{x}} + 5{\rm{y}} + {\rm{K}} = 0\) forms a triangle with the coordinate axes. The vertices of this triangle are the origin (0,0), the x-intercept, and the y-intercept.

• To find the x-intercept, set \({\rm{y}} = 0\) in the equation: \[{\rm{x}} + 5(0) + {\rm{K}} = 0 \implies {\rm{x}} = -{\rm{K}}\] The x-intercept is \((-{\rm{K}}, 0)\). Let \({\rm{a}} = -{\rm{K}}\).
• To find the y-intercept, set \({\rm{x}} = 0\) in the equation: \[0 + 5{\rm{y}} + {\rm{K}} = 0 \implies 5{\rm{y}} = -{\rm{K}} \implies {\rm{y}} = -\frac{{\rm{K}}}{5}\] The y-intercept is \((0, -{\rm{K}}/5)\). Let \({\rm{b}} = -{\rm{K}}/5\).

Step 5: Use the Area of the Triangle Formed by Intercepts

The triangle formed by the line \({\rm{x}} + 5{\rm{y}} + {\rm{K}} = 0\) and the coordinate axes has base \(|{\rm{a}}|\) along the x-axis and height \(|{\rm{b}}|\) along the y-axis. The area of this triangle is given by:

We are given that the area is 5 square units. Substituting the values of \({\rm{a}}\) and \({\rm{b}}\):

Multiplying both sides by 10:

This means \({\rm{K}}^2 = 50\). Taking the square root of both sides:

We can simplify \(\sqrt{50}\):

So, the possible values for \({\rm{K}}\) are \({\rm{K}} = \pm 5\sqrt{2}\).

Step 6: Write the Final Equation of the Line

Substitute the values of \({\rm{K}}\) back into the equation \({\rm{x}} + 5{\rm{y}} + {\rm{K}} = 0\):

• If \({\rm{K}} = 5\sqrt{2}\), the equation is \({\rm{x}} + 5{\rm{y}} + 5\sqrt{2} = 0\).
• If \({\rm{K}} = -5\sqrt{2}\), the equation is \({\rm{x}} + 5{\rm{y}} - 5\sqrt{2} = 0\).

Combining these two possibilities, the equation of the line is \({\rm{x}} + 5{\rm{y}} \pm 5\sqrt{2} = 0\).

Let's quickly check the options provided.

The derived equation \({\rm{x}} + 5{\rm{y}} \pm 5\sqrt 2 = 0\) matches option 1.

Revision Table: Key Concepts for Line Equations and Area

Additional Information: Alternative Formulations

The equation of a line perpendicular to \({\rm{Ax}} + {\rm{By}} + {\rm{C}} = 0\) is of the form \({\rm{Bx}} - {\rm{Ay}} + {\rm{K}} = 0\) (or \({\rm{Bx}} - {\rm{Ay}} = {\rm{K}}'\) where \({\rm{K}}' = -{\rm{K}}\)).

In our case, the given line is \(5{\rm{x}} - 1{\rm{y}} = 0\). Here \({\rm{A}}=5\) and \({\rm{B}}=-1\). A line perpendicular to this would be of the form \(-1{\rm{x}} - 5{\rm{y}} + {\rm{K}} = 0\), which is \(-{\rm{x}} - 5{\rm{y}} + {\rm{K}} = 0\), or \({\rm{x}} + 5{\rm{y}} - {\rm{K}} = 0\). This matches the form \({\rm{x}} + 5{\rm{y}} + {\rm{K}} = 0\) we used, just with a sign change for the constant term.

Let the equation be \({\rm{x}} + 5{\rm{y}} = {\rm{C}}\). To find the intercepts, set \({\rm{x}}=0\) to get \(5{\rm{y}} = {\rm{C}}\), so \({\rm{y}} = {\rm{C}}/5\) (y-intercept). Set \({\rm{y}}=0\) to get \({\rm{x}} = {\rm{C}}\) (x-intercept).

The area is \(\frac{1}{2} |{\rm{C}}| \left|\frac{{\rm{C}}}{5}\right| = \frac{|{\rm{C}}|^2}{10}\). Setting this equal to 5:

Substituting \({\rm{C}} = \pm 5\sqrt{2}\) back into \({\rm{x}} + 5{\rm{y}} = {\rm{C}}\), we get \({\rm{x}} + 5{\rm{y}} = \pm 5\sqrt{2}\).

Rearranging into the form \({\rm{Ax}} + {\rm{By}} + {\rm{C}} = 0\):

Note that \( \pm \) and \( \mp \) just indicate both possibilities, so \({\rm{x}} + 5{\rm{y}} \mp 5\sqrt{2} = 0\) is equivalent to \({\rm{x}} + 5{\rm{y}} \pm 5\sqrt{2} = 0\). Both forms represent the same pair of lines.