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Question: A rectangle has two opposite vertices at the points (1, 2) and (5, 5). If the other vertices lie on ...

A rectangle has two opposite vertices at the points (1, 2) and (5, 5). If the other vertices lie on the line x=3x=3, then the coordinates of the other vertices are
(a) (3, -1), (3, -6)
(b) (3, 1), (3, 5)
(c) (3, 2), (3, 6)
(d) (3, 1), (3, 6)

Explanation

Solution

Assume that the coordinates of the other two vertices are of the form (3, a) and (3, b). Write the slope of lines of sides of the rectangle and use the fact that the product of slopes of two perpendicular lines is -1 to calculate the value of variables ‘a’ and ‘b’.

Complete Step-by-step answer:
We know that two opposite vertices of a rectangle are at points (1, 2) and (5, 5) and the other two vertices lie on the line x=3x=3. We have to find the coordinates of the other two vertices.
Let’s assume that we have a rectangle ABCD such that A=(1,2),C=(5,5)A=\left( 1,2 \right),C=\left( 5,5 \right). Let’s assume that the other two vertices are B=(3,a),D=(3,b)B=\left( 3,a \right),D=\left( 3,b \right), as shown in the figure.

We will now write the slopes of sides of rectangle ABCD.
We know that the slope of the line joining any two points (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) and (x2,y2)\left( {{x}_{2}},{{y}_{2}} \right) is y2−y1x2−x1\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}.
Substituting x1=1,y1=2,x2=3,y2=a{{x}_{1}}=1,{{y}_{1}}=2,{{x}_{2}}=3,{{y}_{2}}=a in the above expression, the slope of line AB is =a−23−1=a−22=\dfrac{a-2}{3-1}=\dfrac{a-2}{2}.
Substituting x1=5,y1=5,x2=3,y2=a{{x}_{1}}=5,{{y}_{1}}=5,{{x}_{2}}=3,{{y}_{2}}=a in the above expression, the slope of line BC is =a−53−5=a−5−2=\dfrac{a-5}{3-5}=\dfrac{a-5}{-2}.
Substituting x1=5,y1=5,x2=3,y2=b{{x}_{1}}=5,{{y}_{1}}=5,{{x}_{2}}=3,{{y}_{2}}=b in the above expression, the slope of the line CD is =b−53−5=b−5−2=\dfrac{b-5}{3-5}=\dfrac{b-5}{-2}.
Substituting x1=1,y1=2,x2=3,y2=b{{x}_{1}}=1,{{y}_{1}}=2,{{x}_{2}}=3,{{y}_{2}}=b in the above expression, the slope of line AD is =b−23−1=b−22=\dfrac{b-2}{3-1}=\dfrac{b-2}{2}.
We know that any two adjoining lines of a rectangle are perpendicular to each other. Thus, we have AB⊥BCAB\bot BC and CD⊥ADCD\bot AD.
We know that the product of slopes of two perpendicular lines is -1.
Thus, considering the lines AB and BC, we have (a−22)(a−5−2)=−1\left( \dfrac{a-2}{2} \right)\left( \dfrac{a-5}{-2} \right)=-1.
Simplifying the above equation, we have a2−2a−5a+10=4{{a}^{2}}-2a-5a+10=4. Thus, we have a2−7a+6=0{{a}^{2}}-7a+6=0.
Factorizing the above equation by splitting the middle term, we have a2−a−6a+6=0{{a}^{2}}-a-6a+6=0. Taking out the common terms, we have a(a−1)−6(a−1)=0a\left( a-1 \right)-6\left( a-1 \right)=0.
Thus, we have (a−1)(a−6)=0\left( a-1 \right)\left( a-6 \right)=0. So, we have a=1,6a=1,6.
Similarly, we will now consider the lines CD and AD. Thus, we have (b−22)(b−5−2)=−1\left( \dfrac{b-2}{2} \right)\left( \dfrac{b-5}{-2} \right)=-1.
Simplifying the above equation, we have b2−2b−5b+10=4{{b}^{2}}-2b-5b+10=4. Thus, we have b2−7b+6=0{{b}^{2}}-7b+6=0.
Factorizing the above equation by splitting the middle term, we have b2−b−6b+6=0{{b}^{2}}-b-6b+6=0. Taking out the common terms, we have b(b−1)−6(b−1)=0b\left( b-1 \right)-6\left( b-1 \right)=0.
Thus, we have (b−1)(b−6)=0\left( b-1 \right)\left( b-6 \right)=0. So, we have b=1,6b=1,6.
So, we have b=1,a=6b=1,a=6. Thus, the coordinates of B and D are B=(3,a)=(3,6)B=\left( 3,a \right)=\left( 3,6 \right) and D=(3,b)=(3,1)D=\left( 3,b \right)=\left( 3,1 \right).
Hence, the coordinates of the other two vertices of the rectangle are B=(3,6)B=\left( 3,6 \right) and D=(3,1)D=\left( 3,1 \right), which is option (d).

Note: We can also solve this question by writing the equations of the slope of diagonals and then use the fact that the diagonals of a rectangle bisect each other. Solve the equations to calculate the other two vertices of the rectangle.