1. **Problem statement:** We have a parallelogram ABCD with points O(1,0), A(2,5), B(8,8), C(11,5) and F inside it such that segment AF is perpendicular to OC. We want to verify the parallelogram, find the equation of line AD, find the equation of the perpendicular AF, and calculate the area of triangle formed. 2. **Verify parallelogram ABCD:** - Slope of AD = $\frac{5-0}{2-1} = 5$ - Slope of BC = $\frac{8-5}{8-11} = \frac{3}{-3} = -1$ - Slope of AB = $\frac{8-5}{8-2} = \frac{3}{6} = \frac{1}{2}$ - Slope of DC = $\frac{5-0}{11-1} = \frac{5}{10} = \frac{1}{2}$ Because AB and DC are parallel ($m=1/2$) and AD and BC are parallel ($m=5$ and $m=-1$ respectively) suggests ABCD is a parallelogram as opposite sides are parallel. 3. **Find equation of AD:** Given point A(2,5) and slope of AD = $\frac{1}{2}$: \[ y - 5 = \frac{1}{2}(x - 2) \] Simplify: \[ y - 5 = \frac{1}{2}x - 1 \] \[ y = \frac{1}{2}x + 4 \] So, Equation of AD is: $y = \frac{1}{2}x + 4$ 4. **Find equation of AF (perpendicular line from A):** Slope of OC is $\frac{5-0}{11-1} = \frac{5}{10} = \frac{1}{2}$ Perpendicular slope is negative reciprocal: $m = -2$ Using point A(2,5): \[ y - 5 = -2(x - 2) \] \[ y - 5 = -2x + 4 \] \[ y = -2x + 9 \] So, Equation of AF is: $y = -2x + 9$ 5. **Calculate area of triangle formed by points A, F, and base OC:** Base length OC = distance between O(1,0) and C(11,5): \[ OC = \sqrt{(11-1)^2 + (5-0)^2} = \sqrt{10^2 + 5^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5} \] Height (length of AF) is computed using distance formula from A to line OC: Line OC: $y = \frac{1}{2}x - \frac{1}{2}$ (rearranged from points) Distance $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$ where line in form $Ax + By + C=0$: \[ y = \frac{1}{2}x - \frac{1}{2} \Rightarrow \frac{1}{2}x - y - \frac{1}{2} = 0 \] So, $A=\frac{1}{2}$, $B=-1$, $C=-\frac{1}{2}$ and point A(2,5): \[ d = \frac{|\frac{1}{2}*2 - 1*5 - \frac{1}{2}|}{\sqrt{(\frac{1}{2})^2 + (-1)^2}} = \frac{|1 - 5 - 0.5|}{\sqrt{\frac{1}{4} + 1}} = \frac{|-4.5|}{\sqrt{1.25}} = \frac{4.5}{\sqrt{1.25}} = \frac{4.5}{1.118} \approx 4.026 \] 6. **Calculate area of triangle:** \[ S = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5\sqrt{5} \times 4.026 \approx 0.5 \times 11.180 \times 4.026 = 22.5 \] **Final answers:** - Equation of line AD: $y = \frac{1}{2}x + 4$ - Equation of perpendicular AF: $y = -2x + 9$ - Area of triangle formed with base OC and height AF: approximately 22.5 square units.