If $R\ ( x,\ y)$ is a point on the line segment joining the points $P\ ( a,\ b)$ and $Q\ ( b,\ a)$, then prove that $a+b=x+y$
Given: There is a point $R\ ( x,\ y)$ \ on the line segment joining the points $P\ ( a,\ b)$ and $Q\ ( b,\ a)$.
To do: To prove that $a+b=x+y$.
Solution:
$\because \ R( x,\ y)$ lies on the line segment joining the points $P( a,\ b)$ and $Q( b,\ a)$
$R( x,\ y) ,\ P( a,\ b) \ and\ Q( b,\ a)$ are collinear .
The area A of $\vartriangle PQR\ $ should be 0.
And we know that area of a triangle with vertices $( x_{1} ,\ y_{1}) ,\ ( x_{2} ,\ y) \ and\ ( x_{3} ,\ y_{3} )$
$=\frac{1}{2}[ x_{1}( y_{2} -y_{3}) +x_{2}( y_{3} -y_{1}) +x_{3}( y_{1} -y_{2})]$
Here we find the vertices $R( x,\ y) ,\ P( a,\ b)$ and $Q( b,\ a)$
$\therefore$ Area of $\vartriangle PQR=\frac{1}{2}[ x( b-a) +a( a-y) +b( y-b)]$
$0=\frac{1}{2}\left( bx-ax+a^{2} -ay+by-b^{2}\right)$
$\Rightarrow \ bx-ax+a^{2} -ay+by-b^{2} =0$
$\Rightarrow \ a^{2} -b^{2} -x( a-b) -y( a-b) =0$
$\Rightarrow ( a-b)( a+b) -x( a-b) -y( a-b) =0$
$\Rightarrow ( a-b)[( a+b) -( x+y)] =0$
$\Rightarrow ( a+b) -( x+y) =0$
$\Rightarrow \ a+b=x+y$
Hence proved.
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