Advertisements
Advertisements
Question
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`
Advertisements
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) = \[\sqrt{\boxed{(x_2 - x_1)^2 + (y_2 - y_1)^2}}\]
= `sqrt((3 - 3)^2 - [3 - (- 7)]^2`
= `sqrt(0^2 + (10)^2)`
∴ d(Q, R) = \[\sqrt{\boxed{0} - 100}\]
∴ d(Q, R) = \[\sqrt{\boxed{100}}\]
∴ d(Q, R) = \[\boxed{10}\]
APPEARS IN
RELATED QUESTIONS
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.
If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find p. Also, find the length of AB.
Find the distance between the following pairs of points:
(−5, 7), (−1, 3)
ABC is a triangle and G(4, 3) is the centroid of the triangle. If A = (1, 3), B = (4, b) and C = (a, 1), find ‘a’ and ‘b’. Find the length of side BC.
If the points (2, 1) and (1, -2) are equidistant from the point (x, y), show that x + 3y = 0.
Find the distance of the following points from the origin:
B(–5, 5)
Find all possible values of x for which the distance between the points A(x, –1) and B(5, 3) is 5 units.
Find the values of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units.
Determine whether the points are collinear.
P(–2, 3), Q(1, 2), R(4, 1)
Find the distance between the following point :
(sin θ , cos θ) and (cos θ , - sin θ)
Find the value of m if the distance between the points (m , -4) and (3 , 2) is 3`sqrt 5` units.
Find the coordinate of O , the centre of a circle passing through P (3 , 0), Q (2 , `sqrt 5`) and R (`-2 sqrt 2` , -1). Also find its radius.
Find the distance between the origin and the point:
(-5, -12)
A point P lies on the x-axis and another point Q lies on the y-axis.
If the abscissa of point P is -12 and the ordinate of point Q is -16; calculate the length of line segment PQ.
Prove that the points P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1) are the vertices of a rectangle PQRS.
The points A (3, 0), B (a, -2) and C (4, -1) are the vertices of triangle ABC right angled at vertex A. Find the value of a.
If the distance between the points (x, -1) and (3, 2) is 5, then the value of x is ______.
Case Study -2
A hockey field is the playing surface for the game of hockey. Historically, the game was played on natural turf (grass) but nowadays it is predominantly played on an artificial turf.
It is rectangular in shape - 100 yards by 60 yards. Goals consist of two upright posts placed equidistant from the centre of the backline, joined at the top by a horizontal crossbar. The inner edges of the posts must be 3.66 metres (4 yards) apart, and the lower edge of the crossbar must be 2.14 metres (7 feet) above the ground.
Each team plays with 11 players on the field during the game including the goalie. Positions you might play include -
- Forward: As shown by players A, B, C and D.
- Midfielders: As shown by players E, F and G.
- Fullbacks: As shown by players H, I and J.
- Goalie: As shown by player K.
Using the picture of a hockey field below, answer the questions that follow:
What are the coordinates of the position of a player Q such that his distance from K is twice his distance from E and K, Q and E are collinear?
Name the type of triangle formed by the points A(–5, 6), B(–4, –2) and C(7, 5).
Find distance between points P(– 5, – 7) and Q(0, 3).
By distance formula,
PQ = `sqrt(square + (y_2 - y_1)^2`
= `sqrt(square + square)`
= `sqrt(square + square)`
= `sqrt(square + square)`
= `sqrt(125)`
= `5sqrt(5)`
