Advertisements
Advertisements
प्रश्न
Without using distance formula, show that the points (−2, −1), (4, 0), (3, 3) and (−3, 2) are vertices of a parallelogram
Advertisements
उत्तर
The vertices A(−2, −1), B(4, 0), C(3, 3) and D(−3, 2)
Slope of a line = `(y_2 - y_1)/(x_2 - x_1)`
Slope of AB = `(0 + 1)/(4 + 2) = 1/6`
Slope of BC = `(3 - 0)/(3 - 4) = 3/(-1)` = −3
Slope of CD = `(2 - 3)/(-3 - 3) = (-1)/(-6) = 1/6`
Slope of AD = `(2 + 1)/(-3 + 2) = 3/(-1)` = −3
Slope of AB = Slope of CD = `1/6`
∴ AB || CD ...(1)
Slope of BC = Slope of AD = −3
∴ BC || AD ...(2)
From (1) and (2) we get ABCD is a parallelogram.
APPEARS IN
संबंधित प्रश्न
What is the slope of a line whose inclination with positive direction of x-axis is 0°
What is the inclination of a line whose slope is 0
What is the inclination of a line whose slope is 1
Find the slope of a line joining the points
`(5, sqrt(5))` with the origin
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.
The line through the points (– 2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x.
Show that the given points form a right angled triangle and check whether they satisfy Pythagoras theorem
A(1, – 4), B(2, – 3) and C(4, – 7)
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)
Let A(3, – 4), B(9, – 4), C(5, – 7) and D(7, – 7). Show that ABCD is a trapezium.
