Show that the configuration below exists in the Cartesian plane IIg for some field F with char(F)=0. The figure is to scale, so you may assume anything that looks parallel, perpendicular, or the same length is. Reminder: you can not use similar triangles or properties of similar triangles. a) Using the given points: D=(0,0), 1=(0,1), H=(1,0), & C=(a,0) as well as the vertical and horizontal lines, we can find the coordinates of many of the points. N b) Looking at the lines with positive slope: DF || HG || IB || JE DF: its slope is m = then we find that it's equation is y = So, we get that point F= HG : its slope is on = then we find that it's equation is y = So, we get that points M = and G = IB: its slope is m = then we find that it's equation is y So, we get that points B = and L= Asside: before we look at line JE.we instead consider the horizontal line AB. It's equation is y = so we get that points A= and E = JE: its slope is m = : then we find that it's equation is y= So, we get that points J = and K = So, we get that points J = and Ka c) Can the coordinates of all the points and the equations of all lines in this configuration be written in terms of a? (yeso) d) Now, we need to solve for a. Consider the line AC. We will find its slope in 2 ways: Find the slope of AC using the points and M: m = 1/ Find the slope of AC using the points A and M: m= Use this information to find a quadratic equation involving a a+ 0 Solve for a to get that a = $ $ Vt, where 8 = and t = (simplified) e) What is the necessary and sufficient condition for this configuration to exist in the Cartesian plane II, for some field F with char(F)=0? EF f) This means that the smallest field that contains this configuration in its Cartesian plane is QC directions: don't put spaces in your answers be sure to include() for your points type square roots as V2 =sqrt(2) Show that the configuration below exists in the Cartesian plane IIg for some field F with char(F)=0. The figure is to scale, so you may assume anything that looks parallel, perpendicular, or the same length is. Reminder: you can not use similar triangles or properties of similar triangles. a) Using the given points: D=(0,0), 1=(0,1), H=(1,0), & C=(a,0) as well as the vertical and horizontal lines, we can find the coordinates of many of the points. N b) Looking at the lines with positive slope: DF || HG || IB || JE DF: its slope is m = then we find that it's equation is y = So, we get that point F= HG : its slope is on = then we find that it's equation is y = So, we get that points M = and G = IB: its slope is m = then we find that it's equation is y So, we get that points B = and L= Asside: before we look at line JE.we instead consider the horizontal line AB. It's equation is y = so we get that points A= and E = JE: its slope is m = : then we find that it's equation is y= So, we get that points J = and K = So, we get that points J = and Ka c) Can the coordinates of all the points and the equations of all lines in this configuration be written in terms of a? (yeso) d) Now, we need to solve for a. Consider the line AC. We will find its slope in 2 ways: Find the slope of AC using the points and M: m = 1/ Find the slope of AC using the points A and M: m= Use this information to find a quadratic equation involving a a+ 0 Solve for a to get that a = $ $ Vt, where 8 = and t = (simplified) e) What is the necessary and sufficient condition for this configuration to exist in the Cartesian plane II, for some field F with char(F)=0? EF f) This means that the smallest field that contains this configuration in its Cartesian plane is QC directions: don't put spaces in your answers be sure to include() for your points type square roots as V2 =sqrt(2)