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प्रश्न
Show that the point (0, 9) is equidistant from the points (–4, 1) and (4, 1).
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उत्तर
Let P(x1, y1) = P(0, 9), Q(x2, y2) = Q(–4, 1), R(x3, y3) = R(4, 1)
By distance formula,
d(P, Q) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
= `sqrt([(-4) - 0]^2 + (1 - 9)^2`
= `sqrt((-4)^2 + (-8)^2`
= `sqrt(16 + 64)`
= `sqrt(80)`
= `4sqrt(5)`
And
d(P, R) = `sqrt((x_3 - x_1)^2 + (y_3 - y_1)^2`
= `sqrt((4 - 0)^2 + (1 - 9)^2`
= `sqrt(4^2 + (-8)^2`
= `sqrt(16 + 64)`
= `sqrt(80)`
= `4sqrt(5)`
Here, d(P, Q) = d(P, R)
∴ The point (0, 9) is equidistant from (–4, 1) and (4, 1).
संबंधित प्रश्न
If P and Q are two points whose coordinates are (at2 ,2at) and (a/t2 , 2a/t) respectively and S is the point (a, 0). Show that `\frac{1}{SP}+\frac{1}{SQ}` is independent of t.
Name the type of quadrilateral formed, if any, by the following point, and give reasons for your answer:
(−3, 5), (3, 1), (0, 3), (−1, −4)
Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
If a ≠ b ≠ 0, prove that the points (a, a2), (b, b2) (0, 0) will not be collinear.
Find the value of a when the distance between the points (3, a) and (4, 1) is `sqrt10`
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.
Find the distance between the points:
A(–6, –4) and B(9, –12)
For what values of k are the points (8, 1), (3, –2k) and (k, –5) collinear ?
Find the distance between the following pair of point.
T(–3, 6), R(9, –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 of the following point from the origin :
(0 , 11)
Find the distance of the following point from the origin :
(13 , 0)
Find the coordinates of O, the centre passing through A( -2, -3), B(-1, 0) and C(7, 6). Also, find its radius.
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.
Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square ABCD.
Show that the points (a, a), (-a, -a) and `(-asqrt(3), asqrt(3))` are the vertices of an equilateral triangle.
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 y axis equidistant from B and C is ______.
The distance of the point P(–6, 8) from the origin is ______.
∆ABC with vertices A(–2, 0), B(2, 0) and C(0, 2) is similar to ∆DEF with vertices D(–4, 0), E(4, 0) and F(0, 4).
The distance between the points (0, 5) and (–3, 1) is ______.
