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( a ) x + y = 1

( b ) x + y = 2

( c ) x + y = 2xy

( d ) 2x + 2y = 1

Answer
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In figure we can see that P is mid point of line joining points $B({{x}_{1}},0)$ and $C(0,{{y}_{1}})$.

So using mid point formula $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ , we can find out coordinates of point P which is,

$P(h,k)=P\left( \dfrac{{{x}_{1}}+0}{2},\dfrac{0+{{y}_{1}}}{2} \right)$

$P(h,k)=P\left( \dfrac{{{x}_{1}}}{2},\dfrac{{{y}_{1}}}{2} \right)$

On comparing, we get

$h=\dfrac{{{x}_{1}}}{2},k=\dfrac{{{y}_{1}}}{2}$

Or, ${{x}_{1}}=2h,{{y}_{1}}=2k$

So, coordinate of point B will be ( 2h, 0 ) and coordinate of point C will be ( 0, 2k ).

Now, if we have two points say, $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ lying on same line, then slope of line is equals to $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{2}}}$ .

So slope of line AB will be equals to whose two pints are A ( 1, 1 ) and B ( 2h, 0 ),

${{m}_{AB}}=\dfrac{1-0}{1-2h}$

Or, ${{m}_{AB}}=\dfrac{1}{1-2h}$

And, slope of line AC will be equals to whose two pints are A ( 1, 1 ) and C ( 0, 2k ),

${{m}_{AC}}=\dfrac{1-2k}{1-0}$

Or, ${{m}_{AC}}=\dfrac{1-2k}{1}$

We know that when two lines are perpendicular and slope of those two lines are say ${{m}_{1}}$ and \[{{m}_{2}}\]then, ${{m}_{1}}\times {{m}_{2}}=-1$

We are given that AC and AB are perpendicular and slope of line AB is ${{m}_{AB}}=\dfrac{1}{1-2h}$ and slope of AC is ${{m}_{AC}}=\dfrac{1-2k}{1}$ so,

${{m}_{AC}}\times {{m}_{AB}}=-1$

Putting values of ${{m}_{AB}}=\dfrac{1}{1-2h}$ and ${{m}_{AC}}=\dfrac{1-2k}{1}$in ${{m}_{AC}}\times {{m}_{AB}}=-1$, we get

$\dfrac{1-2k}{1}\times \dfrac{1}{1-2h}=-1$

On simplifying, we get

$\dfrac{2k-1}{2h-1}=-1$

Taking 2h – 1 from denominator of left hand side to numerator of right hand side, we get

$2k-1=-(2h-1)$

On simplifying, we get

$2k-1=-2h+1$

On solving we get

$h+k=1$….( i )

In general we can write ( h, k ) in terms of ( x, y ) as

$x+y=1$