How i can solve the problem of ncert class 10 coordinate geometry question easily . What is the best way to solve the exercise 7.4 question no.8 easily , please give me easy solution for this question. ABCD is a rectangle formed by the points A (-1, – 1), B (-1, 4), C (5, 4) and D (5, -1). P, Q, R and S are the midpoints of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.
P id the mid-point of side AB,
Coordinate of P = ( (-1 – 1)/2, (-1 + 4)/2 ) = (-1, 3/2)
Similarly, Q, R and S are (As Q is mid-point of BC, R is midpoint of CD and S is midpoint of AD)
Coordinate of Q = (2, 4)
Coordinate of R = (5, 3/2)
Coordinate of S = (2, -1)
Now,
Length of PQ = √[(-1 – 2)2 + (3/2 – 4)2] = √(61/4) = √61/2
Length of SP = √[(2 + 1)2 + (-1 – 3/2)2] = √(61/4) = √61/2
Length of QR = √[(2 – 5)2 + (4 – 3/2)2] = √(61/4) = √61/2
Length of RS = √[(5 – 2)2 + (3/2 + 1)2] = √(61/4) = √61/2
Length of PR (diagonal) = √[(-1 – 5)2 + (3/2 – 3/2)2] = 6
Length of QS (diagonal) = √[(2 – 2)2 + (4 + 1)2] = 5
The above values show that, PQ = SP = QR = RS = √61/2, i.e. all sides are equal.
But PR ≠ QS i.e. diagonals are not of equal measure.
Hence, the given figure is a rhombus