IV- (6 points) In an orthonormal system of axes( \( \left.\mathrm{x}^{\prime} \mathrm{Ox}, \mathrm{y}^{\prime} \mathrm{Oy}\right) \), consider the points \( \mathrm{A}(2 ; 0), \mathrm{B}(0 ; 4) \) and \( \mathrm{E}(-4 ; 0) \). Let (d) be the line with equation \( y=-2 x+4 \). 1) Plot the points \( A, B \) and \( E \). 2) Verify that \( A \) and \( B \) are two points on (d), then draw (d). 3) Let ( \( \left.d^{\prime}\right) \) be the line passing through \( E \) and perpendicular to (d). Verify that \( y=\frac{1}{2} x+2 \) is the equation of \( \left(d^{\prime}\right) \). 4) The line (d') intersects ( \( \left.\mathrm{y}^{\prime} \mathrm{Oy}\right) \) at \( \mathrm{H}(0 ; 2) \) and intersects (d) at \( \mathrm{F} \). a. Verify that the coordinates of \( \mathrm{F} \) are \( \left(\frac{4}{5} ; \frac{12}{5}\right) \). b. Show that \( \mathrm{H} \) is the orthocenter of triangle \( \mathrm{EAB} \). c. Prove that (AH) is perpendicular to (EB). 5) The line (AH) intersects (EB) at G. a. Prove that the four points E, G, F and \( A \) are on the same circle (C) with diameter to be determined. b. Calculate the radius of (C).
Solution
Step 1 ：1) A(2;0), B(0;4), E(-4;0)
Step 2 ：2) y=-2x+4, A y(-2(2)+4)=0, B y(-2(0)+4)=4, AB on (d)
Step 3 ：3) y=\frac{1}{2}x+2, \frac{1}{2}(-4)+2=0, E on \(\left(d^{\prime}\right)\), \(\left(d^{\prime}\right)\) is perpendicular to (d)
Step 4 ：4a) F coordinates \(\left(\frac{4}{5};\frac{12}{5}\right)\), calculation: x = \(\frac{y-2}{\frac{1}{2}}\), substitute y into (d): x = \(\frac{-2x+4-2}{\frac{1}{2}}\), solve the equation for x: x = \(\frac{4}{5}\)
Step 5 ：4b) H=(0;2), HF (dB\) is perpendicular to EB, AF (dA\) is perpendicular to EA, HA and HE form a right angle
Step 6 ：4c) AH * EB = 2 * (-4) + (0 * 0) = -8
Step 7 ：5a) G is the intersection of AH and EB, four points E, G, F and A are on the same circle with diameter EG
Step 8 ：5b) GE = \(\sqrt{(-4-2)^2+(0-0)^2}\)=6, EF = \(\sqrt{\left(\frac{4}{5}-(-4)\right)^2+\left(\frac{12}{5}-0\right)^2}\)=\(\frac{40}{\sqrt{5}}\), Circle radius = \(\frac{1}{2}\sqrt{40^2+(\frac{40}{\sqrt{5}})^2}\)=\(\frac{40}{\sqrt{5}}\)
