High School

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------------------------------------------------ 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].

Answer :

To find the exact area of triangle [tex]\( PQR \)[/tex], we start by calculating the lengths of the sides of the triangle and then apply Heron's formula.

### Step 1: Calculate the Lengths of the Sides

1. Length [tex]\( PQ \)[/tex]:
- Use the distance formula:
[tex]\[
PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\][/tex]
- For [tex]\( P(-1, 2) \)[/tex] and [tex]\( Q(11, 8) \)[/tex]:
[tex]\[
PQ = \sqrt{(11 - (-1))^2 + (8 - 2)^2} = \sqrt{12^2 + 6^2} = \sqrt{144 + 36} = \sqrt{180} = 13.416
\][/tex]

2. Length [tex]\( PR \)[/tex]:
- Use the distance formula:
- For [tex]\( P(-1, 2) \)[/tex] and [tex]\( R(10, 0) \)[/tex]:
[tex]\[
PR = \sqrt{(10 - (-1))^2 + (0 - 2)^2} = \sqrt{11^2 + (-2)^2} = \sqrt{121 + 4} = \sqrt{125} = 11.180
\][/tex]

3. Length [tex]\( RS \)[/tex]:
- Given as [tex]\( 3\sqrt{5} \approx 6.708 \)[/tex].

### Step 2: Apply Heron's Formula

To apply Heron's formula, we first calculate the semi-perimeter [tex]\( s \)[/tex] of triangle [tex]\( PQR \)[/tex]:

[tex]\[
s = \frac{PQ + PR + RS}{2} = \frac{13.416 + 11.180 + 6.708}{2} = 15.652
\][/tex]

Using Heron's formula, the area [tex]\( A \)[/tex] of the triangle is:

[tex]\[
A = \sqrt{s(s - PQ)(s - PR)(s - RS)}
\][/tex]

Calculate each term in Heron's formula:

- [tex]\( s - PQ = 15.652 - 13.416 = 2.236 \)[/tex]
- [tex]\( s - PR = 15.652 - 11.180 = 4.472 \)[/tex]
- [tex]\( s - RS = 15.652 - 6.708 = 8.944 \)[/tex]

Substitute these into Heron's formula:

[tex]\[
A = \sqrt{15.652 \times 2.236 \times 4.472 \times 8.944}
\][/tex]

After calculating, the area [tex]\( A \)[/tex] of the triangle is:

[tex]\[
A = 37.417
\][/tex]

### Conclusion

The exact area of triangle [tex]\( PQR \)[/tex] is approximately [tex]\( 37.417 \)[/tex] square units.