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प्रश्न
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (−3, 4).
उत्तर १
Point (x, y) is equidistant from (3, 6) and (−3, 4).
∴ `sqrt((x-3)^2+(y-6)^2)= sqrt((x-(-3))^2 + (y -4)^2)`
⇒ `sqrt((x-3)^2+(y-6)^2)= sqrt((x+3)^2+(y-4)^2)`
⇒ `(x-3)^2 + (y -6)^2 = (x+3)^2 + (y-4)^2`
⇒ x2 + 9 - 6x + y2 + 36 - 12y = x2 + 9 + 6x + y2 + 16 - 8y
⇒ 36 - 16 = 6x + 6x + 12y - 8y
⇒ 20 = 12x + 4y
⇒ 3x + y = 5
⇒ 3x + y - 5 = 0
उत्तर २
The distance d between two points `(x_1, y_1)` and `(x_2, y_2)` is given by the formula
d = `sqrt((x_1-x_2)^2 + (y_1 - y_2)^2)`
Let the three given points be P(x, y), A(3, 6) and B(−3, 4).
Now let us find the distance between ‘P’ and ‘A’.
PA = `sqrt((x - 3)^2 + (y - 3))`
Now, let us find the distance between ‘P’ and ‘B’.
PB = `sqrt((x + 3)^2 + (y - 6)^2)`
It is given that both these distances are equal. So, let us equate both the above equations,
PA = PB
`sqrt((x - 3)^2 + (y - 6)^2 )`
Squaring on both sides of the equation, we get
(x - 3)2 + (y - 6)2
= (x + 3)2 + (y - 4)2
= x2 + 9 - 6x + y2 + 36 - 12y
= x2 + 9 + 6x + y2 + 16 - 8y
= 12x + 4y = 20
= 3x + y = 5
Hence, the relationship between ‘x’ and ‘y’ based on the given condition is 3x + y = 5
संबंधित प्रश्न
Prove that the points (–3, 0), (1, –3) and (4, 1) are the vertices of an isosceles right angled triangle. Find the area of this triangle
Find the values of y for which the distance between the points P (2, -3) and Q (10, y) is 10 units.
Find the distance between the following pair of points:
(asinα, −bcosα) and (−acos α, bsin α)
Show that the quadrilateral whose vertices are (2, −1), (3, 4) (−2, 3) and (−3,−2) is a rhombus.
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.
If the point A(x,2) is equidistant form the points B(8,-2) and C(2,-2) , find the value of x. Also, find the value of x . Also, find the length of AB.
`" Find the distance between the points" A ((-8)/5,2) and B (2/5,2)`
Find the distance between the following pair of points.
R(0, -3), S(0, `5/2`)
AB and AC are the two chords of a circle whose radius is r. If p and q are
the distance of chord AB and CD, from the centre respectively and if
AB = 2AC then proove that 4q2 = p2 + 3r2.
Find the distance between the following point :
(p+q,p-q) and (p-q, p-q)
Prove that the points (7 , 10) , (-2 , 5) and (3 , -4) are vertices of an isosceles right angled triangle.
A point P (2, -1) is equidistant from the points (a, 7) and (-3, a). Find a.
Given A = (x + 2, -2) and B (11, 6). Find x if AB = 17.
Calculate the distance between A (5, -3) and B on the y-axis whose ordinate is 9.
Calculate the distance between A (7, 3) and B on the x-axis, whose abscissa is 11.
Show that the points (2, 0), (– 2, 0) and (0, 2) are vertices of a triangle. State the type of triangle with reason
The distance between the points A(0, 6) and B(0, -2) 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 distance of the point (5, 0) from the origin is ______.
