The straight line l1 has equation 3 x – 2 y = 1.
Find an equation of the straight l2 which is perpendicular to l1 and passes through the point A(4,-1), giving the answer in the form ax + by + c = 0 where a, b and c are integers.
The straight line l1 meets the coordinate axes at the points P and Q .
The point O represents the origin.
Show that the area of the triangle OPQ is 1 /12 of a square unit.

Let's go through the solution step by step.

### Part 1: Equation of the Perpendicular Line \( l_2 \)

The equation of the line \( l_1 \) is given by:

\[
3x - 2y = 1
\]

To find the slope of \( l_1 \), rewrite the equation in the slope-intercept form \( y = mx + c \):

\[
-2y = -3x + 1
\]
\[
y = \frac{3}{2}x - \frac{1}{2}
\]

So, the slope \( m_1 \) of \( l_1 \) is \( \frac{3}{2} \).

Since \( l_2 \) is perpendicular to \( l_1 \), its slope \( m_2 \) will be the negative reciprocal of \( m_1 \):

\[
m_2 = -\frac{1}{m_1} = -\frac{2}{3}
\]

We know that line \( l_2 \) passes through the point \( A(4, -1) \). Using the point-slope form of the equation of a line:

\[
y - y_1 = m_2(x - x_1)
\]

Substituting \( (x_1, y_1) = (4, -1) \) and \( m_2 = -\frac{2}{3} \):

\[
y + 1 = -\frac{2}{3}(x - 4)
\]

Simplify this equation:

\[
y + 1 = -\frac{2}{3}x + \frac{8}{3}
\]

To eliminate the fractions, multiply the entire equation by 3:

\[
3y + 3 = -2x + 8
\]

Rearranging terms to get the equation in the form \( ax + by + c = 0 \):

\[
2x + 3y - 5 = 0
\]

So, the equation of the line \( l_2 \) is:

\[
2x + 3y - 5 = 0
\]

### Part 2: Area of Triangle \( OPQ \)

To find the area of triangle \( OPQ \), we need to determine the coordinates of points \( P \) and \( Q \), where \( P \) is the x-intercept and \( Q \) is the y-intercept of line \( l_1 \).

**Finding Point \( P \) (x-intercept):**

At the x-intercept, \( y = 0 \):

\[
3x - 2(0) = 1 \implies 3x = 1 \implies x = \frac{1}{3}
\]

So, the coordinates of \( P \) are \( \left(\frac{1}{3}, 0\right) \).

**Finding Point \( Q \) (y-intercept):**

At the y-intercept, \( x = 0 \):

\[
3(0) - 2y = 1 \implies -2y = 1 \implies y = -\frac{1}{2}
\]

So, the coordinates of \( Q \) are \( \left(0, -\frac{1}{2}\right) \).

**Area of Triangle \( OPQ \):**

The area \( A \) of triangle \( OPQ \) formed by the points \( O(0, 0) \), \( P\left(\frac{1}{3}, 0\right) \), and \( Q\left(0, -\frac{1}{2}\right) \) can be calculated using the formula:

\[
A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]

Substituting the coordinates \( (x_1, y_1) = (0, 0) \), \( (x_2, y_2) = \left(\frac{1}{3}, 0\right) \), and \( (x_3, y_3) = \left(0, -\frac{1}{2}\right) \):

\[
A = \frac{1}{2} \left| 0\left(0 - \left(-\frac{1}{2}\right)\right) + \frac{1}{3}\left(-\frac{1}{2} - 0\right) + 0\left(0 - 0\right) \right|
\]

\[
A = \frac{1}{2} \left| \frac{1}{3} \times -\frac{1}{2} \right|
\]

\[
A = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}
\]

Thus, the area of triangle \( OPQ \) is \( \frac{1}{12} \) square units.

---

Would you like to explore any specific part of the solution further, or do you have any questions? Here are some related questions to consider:

1. What is the equation of a line parallel to \( l_1 \) that passes through a different point?
2. Calculate the distance between point \( A(4, -1) \) and the line \( l_1 \).
3. Determine the coordinates of the point where lines \( l_1 \) and \( l_2 \) intersect.
4. Find the midpoint of the line segment \( PQ \).
5. What is the slope of a line that bisects the angle between \( l_1 \) and \( l_2 \)?

**Tip:** When calculating the area of a triangle using its vertices, always double-check the coordinates to avoid mistakes!

Learn how to find the equation of a line perpendicular to 3x - 2y = 1 passing through point A(4,-1) and calculate the area of triangle OPQ formed by this line and the coordinate axes. Suitable for high school students in Grades 10-12.

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