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प्रश्न
Find the values of y for which the distance between the points P (2, -3) and Q (10, y) is 10 units.
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उत्तर १
PQ = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`
⇒ `sqrt((10 - 2)^2 + (y + 3)^2) = 10`
⇒ (8)2 + (y + 3)2 = 100
⇒ 64 + y2 + 6y + 9 = 100
⇒ y2 + 6y + 73 - 100 = 0
⇒ y2 + 6y - 27 = 0
⇒ y2 + 9y - 3y - 27 = 0
⇒ y(y + 9) - 3(y + 9) = 0
⇒ (y + 9) (y - 3) = 0
⇒ y + 9 = 0
⇒ y = -9
and y - 3 = 0
⇒ y = 3
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)`
The distance between two points P(2,−3) and Q(10,y) is given as 10 units. Substituting these values in the formula for distance between two points, we have,
10 = `sqrt((2 - 10)^2 + (-3 - y)^2)`
Now, squaring the above equation on both sides of the equals sign
100 = (-8)2 + (-3 - y)2
100 = 64 + 9 + y2 + 6y
27 = y2 + 6y
Thus, we arrive at a quadratic equation. Let us solve this now.
y2 + 6y - 27 = 0
y2 + 9y - 3y - 27 = 0
y(y + 9) - 3(y + 9) = 0
(y - 3) (y + 9) = 0
The roots of the above quadratic equation are thus 3 and −9.
Thus, the value of ‘y’ could either be 3 or -9.
संबंधित प्रश्न
If Q (0, 1) is equidistant from P (5, − 3) and R (x, 6), find the values of x. Also find the distance QR and PR.
Find all possible values of y for which distance between the points is 10 units.
Using the distance formula, show that the given points are collinear:
(1, -1), (5, 2) and (9, 5)
Determine whether the points are collinear.
L(–2, 3), M(1, –3), N(5, 4)
Determine whether the points are collinear.
P(–2, 3), Q(1, 2), R(4, 1)
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.
x (1,2),Y (3, -4) and z (5,-6) are the vertices of a triangle . Find the circumcentre and the circumradius of the triangle.
ABCD is a square . If the coordinates of A and C are (5 , 4) and (-1 , 6) ; find the coordinates of B and D.
A(2, 5), B(-2, 4) and C(-2, 6) are the vertices of a triangle ABC. Prove that ABC is an isosceles triangle.
Find the co-ordinates of points on the x-axis which are at a distance of 17 units from the point (11, -8).
The length of line PQ is 10 units and the co-ordinates of P are (2, -3); calculate the co-ordinates of point Q, if its abscissa is 10.
The distances of point P (x, y) from the points A (1, - 3) and B (- 2, 2) are in the ratio 2: 3.
Show that: 5x2 + 5y2 - 34x + 70y + 58 = 0.
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).
If the point (x, y) is at equidistant from the point (a + b, b – a) and (a-b, a + b). Prove that ay = bx.
Find distance between points O(0, 0) and B(–5, 12).
The distance of the point (α, β) from the origin 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 point on x axis equidistant from I and E is ______.
If the point A(2, – 4) is equidistant from P(3, 8) and Q(–10, y), find the values of y. Also find distance PQ.
What is the distance of the point (– 5, 4) from the origin?
