Advertisements
Advertisements
प्रश्न
Show that the given points are collinear: (– 3, – 4), (7, 2) and (12, 5)
उत्तर
The vertices are A(– 3, – 4), B(7, 2) and C(12, 5)
Slope of a line = `(y_2 - y_1)/(x_2 - x_1)`
Slope of AB = `(2 + 4)/(7 + 3) = 6/10 = 3/5`
Slope of BC = `(5 - 2)/(12 - 7) = 3/5`
Slope of AB = Slope of BC = `3/5`
∴ The three points A, B, C are collinear.
APPEARS IN
संबंधित प्रश्न
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)
What is the slope of a line perpendicular to the line joining A(5, 1) and P where P is the mid-point of the segment joining (4, 2) and (–6, 4).
If the three points (3, – 1), (a, 3) and (1, – 3) are collinear, find the value of a
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 parallelogram:
A(2.5, 3.5), B(10, – 4), C(2.5, – 2.5) and D(– 5, 5)
Let A(3, – 4), B(9, – 4), C(5, – 7) and D(7, – 7). Show that ABCD is a trapezium.
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