Advertisements
Advertisements
Question
Find the values of x, y if the distances of the point (x, y) from (-3, 0) as well as from (3, 0) are 4.
Advertisements
Solution
We have P(x, y), Q(-3, 0) and R(3, 0)
`PQ = sqrt((x + 3)^2 + (y - 0)^2)`
`=> 4 = sqrt(x^2 + 9 + 6x + y^2)`
Squaring both sides
`=> (4)^2 = (sqrt(x^2 + 9 + 6x + y^2))`
`=> 16 = x^2 + 9 + 6x + y^2`
`=> x^2 + y^2 = 16 - 9 - 6x`
`=> x^2 + y^2 = 7 - 6x` ......(1)
`PR = (sqrt((x - 3)^2 + (y - 0)^2)`
`=> 4 = sqrt(x^2 + 9 - 6x + y^2)`
Squaring both sides
`(4)^2 = (sqrt(x^2 + 9 - 6x + y^2))`
`=> 16 = x^2 + 9 - 6x + y^2`
`=> x^2 + y^2 = 16 - 9 + 6x`
`=> x^2 + y^2 = 7 + 6x` .....(2)
Equating (1) and (2)
7 - 6x = 7 + 6x
⇒ 7 - 7 = 6x + 6x
⇒ 0 = 12x
⇒ x = 0
Substituting the value of x = 0 in (2)
`x^2 + y^2 = 7 + 6x`
`0 + y^2 = 7 + 6 xx 0`
`y^2 = 7`
`y = +- sqrt7`
APPEARS IN
RELATED QUESTIONS
If A(5, 2), B(2, −2) and C(−2, t) are the vertices of a right angled triangle with ∠B = 90°, then find the value of t.
Find the coordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, – 2) and (2, – 2). Also, find its circum radius
If a≠b≠0, prove that the points (a, a2), (b, b2) (0, 0) will not be collinear.
The length of a line segment is of 10 units and the coordinates of one end-point are (2, -3). If the abscissa of the other end is 10, find the ordinate of the other end.
Using the distance formula, show that the given points are collinear:
(1, -1), (5, 2) and (9, 5)
`" Find the distance between the points" A ((-8)/5,2) and B (2/5,2)`
Find the distance between the following pair of point.
T(–3, 6), R(9, –10)
Determine whether the points are collinear.
A(1, −3), B(2, −5), C(−4, 7)
Find the distance between the following pairs of point in the coordinate plane :
(4 , 1) and (-4 , 5)
Find the distance between the following point :
(p+q,p-q) and (p-q, p-q)
Show that the points (2, 0), (–2, 0), and (0, 2) are the vertices of a triangle. Also, a state with the reason for the type of triangle.
Find the distance of the following points from origin.
(a+b, a-b)
Use distance formula to show that the points A(-1, 2), B(2, 5) and C(-5, -2) are collinear.
The distance between points P(–1, 1) and Q(5, –7) is ______.
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`
If the distance between the points (4, P) and (1, 0) is 5, then the value of p is ______.
The equation of the perpendicular bisector of line segment joining points A(4,5) and B(-2,3) 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:

The coordinates of the centroid of ΔEHJ are ______.
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?
What is the distance of the point (– 5, 4) from the origin?
