The line [tex]l_1[/tex] passes through the points [tex]P(-1, 2)[/tex] and [tex]Q(11, 8)[/tex]. The line [tex]l_2[/tex] passes through the point [tex]R(10, 0)[/tex] and is perpendicular to [tex]l_1[/tex]. The lines [tex]l_1[/tex] and [tex]l_2[/tex] intersect at the point [tex]S(7, 6)[/tex]. The length [tex]RS[/tex] is [tex]3 \sqrt{5}[/tex].

Hence, or otherwise, find the exact area of triangle [tex]PQR[/tex].

Certainly! Let's find the exact area of triangle [tex]\( PQR \)[/tex] by using the coordinates of the points. We have:

- Point [tex]\( P \)[/tex] at [tex]\((-1, 2)\)[/tex]
- Point [tex]\( Q \)[/tex] at [tex]\((11, 8)\)[/tex]
- Point [tex]\( R \)[/tex] at [tex]\((10, 0)\)[/tex]

To calculate the area of triangle [tex]\( PQR \)[/tex], we can use the formula for the area of a triangle given its vertices [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]

Substitute the coordinates of [tex]\( P \)[/tex], [tex]\( Q \)[/tex], and [tex]\( R \)[/tex] into the formula:

- [tex]\( x_1 = -1, y_1 = 2 \)[/tex]
- [tex]\( x_2 = 11, y_2 = 8 \)[/tex]
- [tex]\( x_3 = 10, y_3 = 0 \)[/tex]

The formula becomes:

[tex]\[
\text{Area} = \frac{1}{2} \left| -1(8 - 0) + 11(0 - 2) + 10(2 - 8) \right|
\][/tex]

Calculate each term:

1. [tex]\( -1(8 - 0) = -1 \times 8 = -8 \)[/tex]
2. [tex]\( 11(0 - 2) = 11 \times (-2) = -22 \)[/tex]
3. [tex]\( 10(2 - 8) = 10 \times (-6) = -60 \)[/tex]

Add these values together to find the determinant:

[tex]\[
-8 - 22 - 60 = -90
\][/tex]

Now, take the absolute value of the determinant:

[tex]\[
|-90| = 90
\][/tex]

Finally, calculate the area by dividing by 2:

[tex]\[
\text{Area} = \frac{1}{2} \times 90 = 45
\][/tex]

Therefore, the exact area of triangle [tex]\( PQR \)[/tex] is 45 square units.