Answer:
[tex]\large \boxed{24.0}[/tex]
Step-by-step explanation:
The formula for the area of triangle is
A = ½bh
In your triangle, b = QR and h = PT.
We can use Pythagoras' Theorem to find the lengths of these line segments.
1. Length of QR
Draw a point U at (-2, -5) and connect line segments QU and RU.
QU = 6 - (-2) = 6 + 2 = 8
RU = -1 - (-5) = -1 + 5 = 4
[tex]\begin{array}{rcl}QR^{2}& = &QU^{2} + RU^{2} \\& = & 8^{2} + 4^{2}\\& = & 64 + 16\\& = &80\\QR & = & \sqrt{80}\\& = & 4\sqrt{5}\\\end{array}[/tex]
2. Length of PT
Draw a point S at (4, -3) and connect line segments PS and ST.
PS = 2 - (-2.8) = 2 + 2.8 = 4.8
ST = 4 - 1.6 = 2.4
[tex]\begin{array}{rcl}PT^{2}& = &PS^{2} + ST^{2} \\& = & 4.8^{2} + 2.4^{2}\\& = & 23.04 + 5.76\\& = &28.80\\PT & = & \sqrt{28.80}\\& \approx & 5.366\\\end{array}[/tex]
3. Area of triangle
[tex]\begin{array}{rcl}A & = & \dfrac{1}{2} \times 4 \sqrt{5} \times 5.366\\& = &24.0\\\end{array}\\\text{The area of the triangle is $\large \boxed{\mathbf{24.0}}$}[/tex]
