x+10y−10=0→y=1010−x→y=−10x+1
x+10y−5=0→y=105−x→y=−10x+21
The picture is presented below
ABCD is the square, then CD is perpendicular to both BC and AD, then it's a distance between them
The area of the ABCD is equal to CD2
So, the goal is to find the distance between two given lines (CD).
Since two of the given lines has equal slopes, then they are parallel, and DK equal to ∣b1−b2∣ , where b1,b2 - free terms
In the given case DK=∣b1−b2∣=∣1−21∣=0.5
∠СDK=∠MDA
MD=DC
From two statements above we can conclude that triangles MDA and CDK is equals, so AD = 0.5
From triangle MDA(∠A=2π): AM=AD∗tan(∠MDA)
∠MDA=π−∠MDF→tan(∠MDA)=tan(π−∠MDF)=−tan(∠MDF)=−k=
=−(−0.1)=0.1 (k is the slope of the given lines)
AM=AD∗0.1=0.05
Now from the triangle MDA using the Pythagorian theorem we receive:
AD2=MA2+AD2=0.2525=CD2 = area of ABCD
The area of the square is 0.2525
Comments