Advertisements
Advertisements
प्रश्न
Show that points A(–1, –1), B(0, 1), C(1, 3) are collinear.
Advertisements
उत्तर १
A(–1, –1), B(0, 1), C(1, 3)
AB = `sqrt((0 + 1)^2 + (1 + 1)^2`
= `sqrt(1 + 4)`
AB = `sqrt(5)`
BC = `sqrt((1)^2 + (2)^2`
= `sqrt(1 + 4)`
BC = `sqrt(5)`
AC = `sqrt((2)^2 + (3 + 1)^2`
= `sqrt(4 + 16)`
= `sqrt(20)`
= `sqrt(5 xx 4)`
AC = `2sqrt(5)`
AB + BC = AC
`sqrt(5) + sqrt(5) = 2sqrt(5)`
A, B, and C are collinear.
उत्तर २
A`(x_1, y_1) = (-1, -1)`
B`(x_2, y_2) = (0,1)`
C`(x_3, y_3) = (1, 3)`
Slope of line
AB = `(y_2 - y_1)/(x_2 - x_1)`
= `(1 - (-1))/(0 - (-1))`
= `(1+1)/1 = 2`
Slope of line
BC = `(y_3 - y_2)/(x_3 - x_2)`
= `(3 - 1)/(1 - 0)`
= `2/1 = 2.`
As, slope of line AB = slope of line BC
Also AB and BC Hrtes contain common point B
∴ Points A, B, C are collinear. Hence Proved
APPEARS IN
संबंधित प्रश्न
If A(4, 3), B(-1, y) and C(3, 4) are the vertices of a right triangle ABC, right-angled at A, then find the value of y.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes, Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees.
Using distance formula, find which of them is correct.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(- 1, - 2), (1, 0), (- 1, 2), (- 3, 0)
Find the value of a when the distance between the points (3, a) and (4, 1) is `sqrt10`
A(–8, 0), B(0, 16) and C(0, 0) are the vertices of a triangle ABC. Point P lies on AB and Q lies on AC such that AP : PB = 3 : 5 and AQ : QC = 3 : 5. Show that : PQ = `3/8` BC.
`" Find the distance between the points" A ((-8)/5,2) and B (2/5,2)`
Find the distance of the following point from the origin :
(0 , 11)
Prove that the points (6 , -1) , (5 , 8) and (1 , 3) are the vertices of an isosceles triangle.
Find the distance between the following pairs of points:
`(3/5,2) and (-(1)/(5),1(2)/(5))`
Point P (2, -7) is the center of a circle with radius 13 unit, PT is perpendicular to chord AB and T = (-2, -4); calculate the length of: AT
Point P (2, -7) is the centre of a circle with radius 13 unit, PT is perpendicular to chord AB and T = (-2, -4); calculate the length of AB.
Find the distance of the following points from origin.
(5, 6)
Find distance of point A(6, 8) from origin
Find distance between point Q(3, – 7) and point R(3, 3)
Solution: Suppose Q(x1, y1) and point R(x2, y2)
x1 = 3, y1 = – 7 and x2 = 3, y2 = 3
Using distance formula,
d(Q, R) = `sqrt(square)`
∴ d(Q, R) = `sqrt(square - 100)`
∴ d(Q, R) = `sqrt(square)`
∴ d(Q, R) = `square`
The distance between the points (0, 5) and (–5, 0) is ______.
If the distance between the points (4, P) and (1, 0) is 5, then the value of p is ______.
The distance between the point P(1, 4) and Q(4, 0) is ______.
The points (– 4, 0), (4, 0), (0, 3) are the vertices of a ______.
Find the distance between the points O(0, 0) and P(3, 4).
|
Tharunya was thrilled to know that the football tournament is fixed with a monthly timeframe from 20th July to 20th August 2023 and for the first time in the FIFA Women’s World Cup’s history, two nations host in 10 venues. Her father felt that the game can be better understood if the position of players is represented as points on a coordinate plane. |
- At an instance, the midfielders and forward formed a parallelogram. Find the position of the central midfielder (D) if the position of other players who formed the parallelogram are :- A(1, 2), B(4, 3) and C(6, 6)
- Check if the Goal keeper G(–3, 5), Sweeper H(3, 1) and Wing-back K(0, 3) fall on a same straight line.
[or]
Check if the Full-back J(5, –3) and centre-back I(–4, 6) are equidistant from forward C(0, 1) and if C is the mid-point of IJ. - If Defensive midfielder A(1, 4), Attacking midfielder B(2, –3) and Striker E(a, b) lie on the same straight line and B is equidistant from A and E, find the position of E.
