Introduction to Perpendicular Lines and Calculating the Area of the Triangle Formed

In this article, we will explore the methods for finding the equation of a line perpendicular to a given line, and then derive the specific equations when the area of the triangle formed by this line with the coordinate axes is given.

Step-by-Step Solution

Let's begin with the given line equation: (3x - 2y - 4 0).

Step 1: Determine the Slope of the Given Line

The equation can be rearranged into the slope-intercept form (y mx b).

(3x - 2y - 4 0 implies -2y -3x - 4 implies y frac{3}{2}x - 2)

The slope of the given line, (m_1), is (frac{3}{2}).

Step 2: Determine the Slope of the Perpendicular Line

Since the lines are perpendicular, the slope of the new line, (m_2), is the negative reciprocal of (m_1).

(m_2 -frac{2}{3})

Step 3: Equation of the Perpendicular Line

The equation of the line (l) in point-slope form, passing through the origin, is:

(y -frac{2}{3}x)

However, we need to find where this line intersects the coordinate axes to form a triangle with an area of 27 square units.

Step 4: Finding the X- and Y-Intercepts

To find the x-intercept, set (y 0):

(0 -frac{2}{3}x implies x 0)

To find the y-intercept, set (x 0):

(y -frac{2}{3}(0) implies y 0)

These intercepts alone do not form a triangle with area 27, so we need a more general form.

Step 5: General Equation of the Perpendicular Line

Let's express the equation of line (l) in the form (y -frac{2}{3}x c), where (c) is the y-intercept.

To find the x-intercept, set (y 0):

(0 -frac{2}{3}x c implies x frac{3c}{2})

To find the y-intercept, set (x 0):

(y c)

Step 6: Area of the Triangle

The area (A) of the triangle formed by the intercepts on the axes is given by:

(A frac{1}{2} times text{X-intercept} times text{Y-intercept} frac{1}{2} times frac{3c}{2} times c frac{3c^2}{4})

Setting this area equal to 27:

(frac{3c^2}{4} 27)

Step 7: Solve for (c)

Multiply both sides by 4:

(3c^2 108 implies c^2 36 implies c 6 text{ or } c -6)

Step 8: Form the Equations of Line (l)

Using (c 6):

(y -frac{2}{3}x 6)

Using (c -6):

(y -frac{2}{3}x - 6)

Conclusion

The equations of the line (l) are:

(boxed{y -frac{2}{3}x 6}) and (boxed{y -frac{2}{3}x - 6})