Answer to Question #259023 in Analytic Geometry for Aliyah

Question

Two sides of a square are on the line x + 10y – 10 = 0 and x + 10y – 5 = 0. Find the area of the square.

Answer 1

$x+10y-10=0\to y={\frac {10-x} {10}}\to y=-{\frac x {10}}+1$

$x+10y-5=0\to y={\frac {5-x} {10}}\to y=-{\frac x {10}}+{\frac 1 2}$

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 $CD^{2}$

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 $|b{\scriptscriptstyle 1}-b{\scriptscriptstyle 2}|$ , where $b{\scriptscriptstyle 1},b{\scriptscriptstyle 2}$ - free terms

In the given case $DK=|b{\scriptscriptstyle 1}-b{\scriptscriptstyle 2}|=|1-{\frac 1 2}|=0.5$

$∠СDK = ∠MDA$

$MD=DC$

From two statements above we can conclude that triangles MDA and CDK is equals, so AD = 0.5

From triangle MDA($∠A={\frac \pi 2}$): $AM=AD*tan(∠MDA)$

$∠MDA=\pi - ∠MDF \to tan(∠MDA)=tan(\pi -∠MDF)=-tan(∠MDF) = -k =$

$= -(-0.1)=0.1$ (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:

$AD^{2}=MA^{2}+AD^{2}=0.2525=CD^{2}$ = area of ABCD

The area of the square is 0.2525