Advertisements
Advertisements
Question
Show that the given points form a parallelogram:
A(2.5, 3.5), B(10, – 4), C(2.5, – 2.5) and D(– 5, 5)
Advertisements
Solution
Let A(2.5, 3.5), B(10, – 4), C(2.5, – 2.5) and D(– 5, 5) are the vertices of a parallelogram.
Slope of a line = `(y_2 - y_1)/(x_2 - x_1)`
Slope of AB = `(-4 - 3.5)/(10 - 2.5) = (-7.5)/(7.5)` = – 1
Slope of CD = `(5 + 2.5)/(-5 - 2.5) = (7.5)/(-7.5)` = – 1
Slope of AB = Slope of CD = – 1
∴ AB is Parallel to CD ...(1)
Slope of BC = `(-4 + 2.5)/(10 - 2.5) = (-1.5)/(7.5) = (-15)/75 = -1/5`
Slope of AD = `(5 - 3.5)/(-5 - 2.5) = 1.5/-7.5 = 15/(-75) = -1/5`
Slope of BC = Slope of AD
∴ BC is parallel to AD
From (1) and (2) we get ABCD is a parallelogram.
APPEARS IN
RELATED QUESTIONS
What is the slope of a line whose inclination with positive direction of x-axis is 90°
Find the slope of a line joining the points
`(5, sqrt(5))` with the origin
Find the slope of a line joining the points
(sin θ, – cos θ) and (– sin θ, cos θ)
Show that the given points are collinear: (– 3, – 4), (7, 2) and (12, 5)
If the three points (3, – 1), (a, 3) and (1, – 3) are collinear, find the value of a
The line through the points (– 2, a) and (9, 3) has slope `-1/2` Find the value of a.
Show that the given points form a right angled triangle and check whether they satisfy Pythagoras theorem.
L(0, 5), M(9, 12) and N(3, 14)
The slope of the line which is perpendicular to a line joining the points (0, 0) and (− 8, 8) is
If slope of the line PQ is `1/sqrt(3)` then slope of the perpendicular bisector of PQ is
Without using distance formula, show that the points (−2, −1), (4, 0), (3, 3) and (−3, 2) are vertices of a parallelogram
