+295 In the figure, what is the area of triangle \(ABD\)? Express your answer as a common fraction.
+118687 The only way I can think to do this is with co-ordinate geometry.
Equation of AF is y=(-4/3)x+4
Equation of EB is y=(-2/7)x+2
Simultaneous solution of these gives the point D which is \((\frac{21}{11},\frac{16}{11})\)
Distance AB can be found using Pythagoras's theoem. It is \(\sqrt{65}\)
Equation of AB is
\(y=\frac{-4x}{7}+4\\ 7y=-4x+28\\ 4x+7y-28=0 \)
Now I can use the perpendicular distance formula of a point to a line to find the perpendicular height between AB and D
\(d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\\ d=\frac{|4*\frac{21}{11}+7*\frac{16}{11}-28|}{\sqrt{16+49}}\\ d=\frac{|4*\frac{21}{11}+7*\frac{16}{11}-28|}{\sqrt{16+49}}\\ d=\frac{112}{11}*\frac{1}{\sqrt{65}}\\ d=\frac{112\sqrt{65}}{11*65}\\ d=\frac{112\sqrt{65}}{715}\\ \)
Area of ABD =
\(\frac{1}{2}*\sqrt{65}*\frac{112\sqrt{65}}{715}\\ =\frac{56*65}{715}\\ =\frac{56*13}{143}\\ =\frac{728}{143}\\ =5\frac{13}{143}u^2\)
That is assuming I have made not stupid errors.
You could check this using heron's formula but that would be a lot of work too.
Since the area of ABC is 14 my answer does sound reasonable.
