(ii) y-axis. Also, find the coordinates of the point of division in each case. ICBSE 20131
Prove that the points (4,5), (7,6), (6, 3), (3, 2) are the vertices of a parallelogram. Is it a retangular
Answers
Answer:
Let
A(x1,y1) = (4,5)
B(x2,y2) = (7,6)
C(x3,y3) = (6, 3)
D(x4,y4) = (3, 2)
Distance of AB = √{(x2-x1)^2 + (y2-y1)^2}
= √{(7-4)^2 + (6-5)^2}
= √{(3)^2 + (1)^2}
= √{ 9 + 1}
= √10
Distance of BC = √{(x3-x2)^2 + (y3-y2)^2}
= √{(6-7)^2 + (3-6)^2}
= √{(-1)^2 + (-3)^2}
= √{1 + 9}
= √10
Distance of CD = √{(x4-x3)^2 + (y4-y3)^2}
= √{(3-6)^2 + (2-3)^2}
= √{(-3)^2 + (1)^2}
= √{9 + 1}
= √10
Distance of AD = √{(x1-x4)^2 + (y1-y4)^2}
= √{(4-3)^2 + (5-2)^2}
= √{(1)^2 + (3)^2}
= √{1 + 9}
= √10
Since AB=BC=CD=DA
Therefore, ABCD is a parallelogram
Diagonal AC = √{(x3-x1)^2 + (y3-y1)^2}
= √{(6-4)^2 + (3-5)^2}
= √{(2)^2 + (-2)^2}
= √{4 + 4}
= √8
Diagonal BD = √{(x4-x2)^2 + (y4-y2)^2}
= √{(3-7)^2 + (2-6)^2}
= √{(-4)^2 + (-4)^2}
= √{16 + 16}
= √32
Since AC ≠ BD
ABCD is not rectangular