Advertisements
Advertisements
प्रश्न
Show that the point (11, –2) is equidistant from (4, –3) and (6, 3).
Advertisements
उत्तर
Let P(x1, y1) = P(11, –2), Q(x2, y2) = Q(4, –3), R(x3, y3) = R(6, 3)
By distance formula,
d(P, Q) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
= `sqrt((4 - 11)^2 + [-3 - (-2)]^2`
= `sqrt((-7)^2 + (-1)^2`
= `sqrt(49 + 1)`
= `sqrt(50)`
= `5sqrt(2)`
And
d(P, R) = `sqrt((x_3 - x_1)^2 + (y_3 - y_1)^2`
= `sqrt((6 - 11)^2 + [3 - (-2)]^2`
= `sqrt((-5)^2 + (5)^2`
= `sqrt(25 + 25)`
= `sqrt(50)`
= `5sqrt(2)`
Here, d(P, Q) = d(P, R)
∴ Point (11, –2) is equidistant from (4, –3) and (6, 3).
APPEARS IN
संबंधित प्रश्न
If the point P(2, 2) is equidistant from the points A(–2, k) and B(–2k, –3), find k. Also, find the length of AP.
If the point (x, y) is equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.
Find the value of a when the distance between the points (3, a) and (4, 1) is `sqrt10`.
Find x if distance between points L(x, 7) and M(1, 15) is 10.
If the point P(2, 1) lies on the line segment joining points A(4, 2) and B(8, 4), then ______.
Find the distance between the following pairs of point in the coordinate plane :
(13 , 7) and (4 , -5)
Find the distance between the following point :
(p+q,p-q) and (p-q, p-q)
Prove that the following set of point is collinear :
(5 , 5),(3 , 4),(-7 , -1)
Prove that the points (1 , 1) , (-1 , -1) and (`- sqrt 3 , sqrt 3`) are the vertices of an equilateral triangle.
Find the distance between the following pair of points:
`(sqrt(3)+1,1)` and `(0, sqrt(3))`
Find the distance between the origin and the point:
(-5, -12)
Points A (-3, -2), B (-6, a), C (-3, -4) and D (0, -1) are the vertices of quadrilateral ABCD; find a if 'a' is negative and AB = CD.
By using the distance formula prove that each of the following sets of points are the vertices of a right angled triangle.
(i) (6, 2), (3, -1) and (- 2, 4)
(ii) (-2, 2), (8, -2) and (-4, -3).
Show that the points (a, a), (-a, -a) and `(-asqrt(3), asqrt(3))` are the vertices of an equilateral triangle.
Find distance between point A(–1, 1) and point B(5, –7):
Solution: Suppose A(x1, y1) and B(x2, y2)
x1 = –1, y1 = 1 and x2 = 5, y2 = –7
Using distance formula,
d(A, B) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
∴ d(A, B) = `sqrt(square +[(-7) + square]^2`
∴ d(A, B) = `sqrt(square)`
∴ d(A, B) = `square`
The distance between the points A(0, 6) and B(0, -2) is ______.
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:

If a player P needs to be at equal distances from A and G, such that A, P and G are in straight line, then position of P will be given by ______.
The point A(2, 7) lies on the perpendicular bisector of line segment joining the points P(6, 5) and Q(0, – 4).
Find the points on the x-axis which are at a distance of `2sqrt(5)` from the point (7, – 4). How many such points are there?
