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Question: Three vertices of a parallelogram are (a+b,a-b), (2a+b,2a-b), (a-b,a+b). The coordinates of fourth v...

Three vertices of a parallelogram are (a+b,a-b), (2a+b,2a-b), (a-b,a+b). The coordinates of fourth vertex are:
[a] (-b,b)
[b] (b,b)
[c] (-b,-b)
[d] None of these.

Explanation

Solution

Hint: The question can be solved using the properties of a parallelogram. Since diagonals of a parallelogram bisect each other, the midpoint of one diagonal should coincide with the midpoint of the other diagonal. So, assume the coordinates of the last point be (x,y). Then find the midpoints of both the diagonals. Constraints these midpoints to be equal to each other. Alternatively, let ABCD be the parallelogram and let us assume we have to find the coordinates of D. Then find the equation of lines AD and CD. Solve the two equations to get the coordinates of D as D is the point of intersection of AD and CD. Alternatively, you can use the following result:
If A(x1,y1),B(x2,y2),C(x3,y3),D(x,y)\text{A}\left( {{x}_{1}},{{y}_{1}} \right),\text{B}\left( {{x}_{2}},{{y}_{2}} \right),\text{C}\left( {{x}_{3}},{{y}_{3}} \right),\text{D}\left( x,y \right)be the sides of a parallelogram ABCD, then
x=x1−x2+x3x={{x}_{1}}-{{x}_{2}}+{{x}_{3}} and y=y1−y2+y3y={{y}_{1}}-{{y}_{2}}+{{y}_{3}}.

Complete step-by-step answer:
Let ABCD be the given parallelogram with
A≡(a+b,a−b),B≡(2a+b,2a−b),C≡(a−b,a+b)\text{A}\equiv \left( a+b,a-b \right),\text{B}\equiv \left( 2a+b,2a-b \right),\text{C}\equiv \left( a-b,a+b \right) and D≡(x,y)\text{D}\equiv \left( x,y \right).
Let O be the midpoint of diagonals AC and BD {Because diagonals of a parallelogram bisect each other.}
Finding midpoint of diagonal AC:
We know that if A≡(x1,y1)\text{A}\equiv \left( {{x}_{1}},{{y}_{1}} \right)and B≡(x2,y2)\text{B}\equiv \left( {{x}_{2}},{{y}_{2}} \right), then the midpoint C of AB is given by C≡(x1+x22,y1+y22)\text{C}\equiv \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right).
Here x1=a+b,x2=a−b,y1=a−b{{x}_{1}}=a+b,{{x}_{2}}=a-b,{{y}_{1}}=a-b and y2=a+b{{y}_{2}}=a+b
Using the above result, we get
O≡(a+b+a−b2,a−b+a+b2)=(2a2,2a2)=(a,a)\text{O}\equiv \left( \dfrac{a+b+a-b}{2},\dfrac{a-b+a+b}{2} \right)=\left( \dfrac{2a}{2},\dfrac{2a}{2} \right)=\left( a,a \right)
Finding midpoint of diagonal BD:
We know that if A≡(x1,y1)\text{A}\equiv \left( {{x}_{1}},{{y}_{1}} \right)and B≡(x2,y2)\text{B}\equiv \left( {{x}_{2}},{{y}_{2}} \right), then the midpoint C of AB is given by C≡(x1+x22,y1+y22)\text{C}\equiv \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right).
Here x1=2a+b,x2=x,y1=2a−b{{x}_{1}}=2a+b,{{x}_{2}}=x,{{y}_{1}}=2a-b and y2=y{{y}_{2}}=y
O≡(2a+b+x2,2a−b+y2)\text{O}\equiv \left( \dfrac{2a+b+x}{2},\dfrac{2a-b+y}{2} \right)
Hence we have
(2a+b+x2,2a−b+y2)=(a,a)\left( \dfrac{2a+b+x}{2},\dfrac{2a-b+y}{2} \right)=\left( a,a \right)
Comparing x-coordinates, we get
2a+b+x2=a ⇒2a+b+x=2a ⇒x=−b \begin{aligned} & \dfrac{2a+b+x}{2}=a \\\ & \Rightarrow 2a+b+x=2a \\\ & \Rightarrow x=-b \\\ \end{aligned}
Comparing y -coordinates, we get
2a−b+y2=a ⇒2a−b+y=2a ⇒y=b \begin{aligned} & \dfrac{2a-b+y}{2}=a \\\ & \Rightarrow 2a-b+y=2a \\\ & \Rightarrow y=b \\\ \end{aligned}
Hence we have coordinates of D are (-b,b).

Note: :Alternative solution:
We know that if A(x1,y1),B(x2,y2),C(x3,y3),D(x,y)\text{A}\left( {{x}_{1}},{{y}_{1}} \right),\text{B}\left( {{x}_{2}},{{y}_{2}} \right),\text{C}\left( {{x}_{3}},{{y}_{3}} \right),\text{D}\left( x,y \right)be the sides of a parallelogram ABCD, then
x=x1−x2+x3x={{x}_{1}}-{{x}_{2}}+{{x}_{3}} and y=y1−y2+y3y={{y}_{1}}-{{y}_{2}}+{{y}_{3}}.
Here x1=a+b,x2=2a+b,x3=a−b,y1=a−b,y2=2a−b{{x}_{1}}=a+b,{{x}_{2}}=2a+b,{{x}_{3}}=a-b,{{y}_{1}}=a-b,{{y}_{2}}=2a-b and y3=a+b{{y}_{3}}=a+b
Hence
x=a+b−2a−b+a−b=−b y=a−b−2a+b+a+b=b \begin{aligned} & x=a+b-2a-b+a-b=-b \\\ & y=a-b-2a+b+a+b=b \\\ \end{aligned}
Hence the coordinates of D are (-b,b).