To solve this problem, we will follow these steps:
### Step 1: Determine the equation of Line 1.
The given equation for Line 1 is:
[tex]\[ 2x + 3y = -24 \][/tex]
To convert this into slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[ 3y = -2x - 24 \][/tex]
[tex]\[ y = -\frac{2}{3}x - 8 \][/tex]
Thus, the slope ([tex]\( m_1 \)[/tex]) of Line 1 is [tex]\( -\frac{2}{3} \)[/tex].
### Step 2: Find the slope of Line 2.
Line 2 is perpendicular to Line 1. The slope of a perpendicular line is the negative reciprocal. Thus, the slope ([tex]\( m_2 \)[/tex]) of Line 2 is:
[tex]\[ m_2 = \frac{3}{2} \][/tex]
### Step 3: Determine the equation of Line 2.
Line 2 goes through the point [tex]\( R(3, -10) \)[/tex]. The equation of Line 2 in slope-intercept form is [tex]\( y = mx + b \)[/tex]. Substituting [tex]\( m = \frac{3}{2} \)[/tex] and the point [tex]\( (3, -10) \)[/tex]:
[tex]\[ -10 = \frac{3}{2}(3) + b \][/tex]
[tex]\[ -10 = \frac{9}{2} + b \][/tex]
[tex]\[ b = -10 - \frac{9}{2} \][/tex]
[tex]\[ b = -10 - 4.5 \][/tex]
[tex]\[ b = -14.5 \][/tex]
So, the equation of Line 2 is:
[tex]\[ y = \frac{3}{2}x - 14.5 \][/tex]
### Step 4: Find where Line 1 intersects the x-axis (Point P).
To find the x-intercept, set [tex]\( y = 0 \)[/tex] in the equation of Line 1:
[tex]\[ 2x + 3(0) = -24 \][/tex]
[tex]\[ 2x = -24 \][/tex]
[tex]\[ x = -12 \][/tex]
Thus, Point [tex]\( P \)[/tex] is [tex]\( (-12, 0) \)[/tex].
### Step 5: Find where Line 2 intersects the x-axis (Point Q).
To find the x-intercept, set [tex]\( y = 0 \)[/tex] in the equation of Line 2:
[tex]\[ 0 = \frac{3}{2}x - 14.5 \][/tex]
[tex]\[ \frac{3}{2}x = 14.5 \][/tex]
[tex]\[ x = \frac{14.5 \times 2}{3} \][/tex]
[tex]\[ x = \frac{29}{3} \][/tex]
Thus, Point [tex]\( Q \)[/tex] is [tex]\( \left(\frac{29}{3}, 0\right) \)[/tex].
### Step 6: Calculate the area of Triangle PQR.
The vertices of the triangle are:
- [tex]\( P(-12, 0) \)[/tex]
- [tex]\( Q\left(\frac{29}{3}, 0\right) \)[/tex]
- [tex]\( R(3, -10) \)[/tex]
We use the formula for the area of a triangle given by three points [tex]\((x_1, y_1)\)[/tex], [tex]\((x_2, y_2)\)[/tex], and [tex]\((x_3, y_3)\)[/tex]:
[tex]\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\][/tex]
Plugging in the coordinates:
[tex]\[
x_1 = -12, \; y_1 = 0, \; x_2 = \frac{29}{3}, \; y_2 = 0, \; x_3 = 3, \; y_3 = -10
\][/tex]
[tex]\[
\text{Area} = \frac{1}{2} \left| (-12)(0 - (-10)) + \left(\frac{29}{3}\right)(-10 - 0) + 3(0 - 0) \right|
\][/tex]
[tex]\[
= \frac{1}{2} \left| (-12)(10) + \left(\frac{29}{3}\right)(-10) \right|
\][/tex]
[tex]\[
= \frac{1}{2} \left| -120 - \frac{290}{3} \right|
\][/tex]
[tex]\[
= \frac{1}{2} \left| -120 - 96.6667 \right|
\][/tex]
[tex]\[
= \frac{1}{2} \left| -216.6667 \right|
\][/tex]
[tex]\[
= \frac{1}{2} \times 216.6667
\][/tex]
[tex]\[
= 108.3333
\][/tex]
Thus, the area of triangle [tex]\( PQR \)[/tex] is [tex]\( 108.33 \)[/tex] square units.
