Masterscorp wrote:
Why can't we use the formula for the areas of parallelograms here?
I calculated that Point A must be at (-4;3) so I had the coordinates of all four points. I took AC = \(\sqrt{2}\) as the height and DC = 5 as the base. As \(\sqrt{2} =~ 1.4\) I calculated 1.4*5 which is 7. Of course 1.4 is only an approximation so the result is not exact.
I am aware that this solution doesn't work here but I want to understand why.
Masterscorpto use the area of parallelogram formula here.
First we have to find the equation of CD,
\(y= \frac{{4-0}}{{-3-0}}x+c\)
As CD passes through (0,0); c would be zero
Equation of CD : \(y = - \frac{4}{3}x\)
\(y+\frac{4}{3}x=0\)
As you have already found out the coordinates of point A(-4,3)
We can find perpendicular distance between CD and Point A using
\(D =\frac{{|ax_1 +by_1+c|}}{{\sqrt{a^2+b^2}}}\)
In this case , \(a=\frac{4}{3},b=1,c=0,x_1=-4,y_1=3\)
Solving for D, we will get \(D = \frac{7}{5}=1.4\)
As length of \(CD =5\)
Area of Parallelogram ABCD = \(5*\frac{7}{5}= 7\)
AC is not perpendicular to CD as you assumed
Line equation of AC will \(y = \frac{{4-3}}{{-3+4}}x+c\)
\(y = x+c\)
putting (-3,4) in above equation, we will get \(c=7\)
So equation of line AC would be \(y = x+7\)
As product of slopes of two non vertical perpendicular line should be -1.
Let us check the product of slope of AC and CD, it is coming out be \(1*\frac{-4}{3}\) = \(\frac{-4}{3}\)
So line AC and CD are not perpendicular