Let’s assume that we have a rectangle ABCD such that
A = (1, 2), C = (5, 5)
B = (3, a), D = (3, b)
We know that the slope of the line joining any two points (x1, y1) and (x2, y2) is
\(\frac{y_2 - y_1}{x_2-x_1}\)
Substituting x1 = 1, y1 = 2, x2 = 3, y2 = a in the above expression, the slope of line AB is
\(\frac{a-2}{3-1} = \frac{a-2}{2}\)
Substituting x1 = 5, y1 = 5, x2 = 3, y2 = a in the above expression, the slope of line BC is
\(\frac{a-5}{3-5} = \frac{a-5}{-2}\)
Substituting x1 = 5, y1 = 5, x2 = 3, y2 = b in the above expression, the slope of line CD is
\(\frac{b-5}{3-5} = \frac{b-5}{-2}\)
Substituting x1 = 1, y1 = 2, x2 = 3, y2 = b in the above expression, the slope of line AD is
\(\frac{b-2}{3-1} = \frac{b-2}{2}\)
We know that any two adjoining lines of a rectangle are perpendicular to each other. Thus, we have AB ⊥ BC and CD ⊥ AD.
We know that the product of slopes of two perpendicular lines is -1.
Thus, considering the lines AB and BC, we have \((\frac{a-2}{2}) (\frac{a-5}{-2}) = -1\)
Simplifying the above equation, we have a2 − 2a − 5a + 10 = 4. Thus, we have a2 − 7a + 6 = 0.
Factorizing the above equation by splitting the middle term, we have a2 − a − 6a + 6 = 0.
Taking out the common terms, we have a(a − 1) − 6(a − 1) = 0.
Thus, we have (a − 1)(a − 6) = 0. So, we have a = 1, 6.
Similarly, we will now consider the lines CD and AD. Thus, we have
\((\frac{b-2}{2}) (\frac{b-5}{-2}) = -1\)
Simplifying the above equation, we have b2 − 2b − 5b + 10 = 4. Thus, we have b2 − 7b + 6 = 0
Factorizing the above equation by splitting the middle term, we have b2 − b − 6b + 6 = 0. Taking out the common terms, we have b(b − 1) −6(b − 1) = 0
Thus, we have (b − 1)(b − 6) = 0. So, we have b = 1,6.
So, we have b = 1, a = 6. Thus, the coordinates of B and D are
B = (3, a) = (3, 6) and D = (3, b) = (3,1)
Hence, the coordinates of the other two vertices of the rectangle are B =(3, 6) and D = (3, 1).
