Advertisements
Advertisements
प्रश्न
Find the distance between the following pair of point.
T(–3, 6), R(9, –10)
Advertisements
उत्तर
T(–3, 6), R(9, –10)
Let T (x1, y1) and R (x2, y2) be the given points.
∴ x1 = −3, y1 = 6, x2 = 9, y2 = −10
\[\mathrm{d(T,R)}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\]
= \[\sqrt{\left[9-(-3)\right]^{2}+\left(-10-6\right)^{2}}\]
= \[\sqrt{\left(9+3\right)^{2}+\left(-10-6\right)^{2}}\]
= \[\sqrt{12^{2}+\left(-16\right)^{2}}\]
= \[\sqrt{144 + 256}\]
= \[\sqrt{400}\]
= 20
∴ d(T, R) = 20 units
∴ The distance between the points T and R 20 units.
APPEARS IN
संबंधित प्रश्न
Find the distance between two points
(i) P(–6, 7) and Q(–1, –5)
(ii) R(a + b, a – b) and S(a – b, –a – b)
(iii) `A(at_1^2,2at_1)" and " B(at_2^2,2at_2)`
If the opposite vertices of a square are (1, – 1) and (3, 4), find the coordinates of the remaining angular points.
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.
If the distances of P(x, y) from A(5, 1) and B(–1, 5) are equal, then prove that 3x = 2y
Find the value of a when the distance between the points (3, a) and (4, 1) is `sqrt10`
Find the distance between the points:
A(9, 3) and B(15, 11)
Find the distance between the points:
A(–6, –4) and B(9, –12)
Find all possible values of x for which the distance between the points A(x, –1) and B(5, 3) is 5 units.
Find the value of y for which the distance between the points A (3, −1) and B (11, y) is 10 units.
Find the value of m if the distance between the points (m , -4) and (3 , 2) is 3`sqrt 5` units.
Prove that the points (4 , 6) , (- 1 , 5) , (- 2, 0) and (3 , 1) are the vertices of a rhombus.
The distance between the points (3, 1) and (0, x) is 5. Find x.
Calculate the distance between A (7, 3) and B on the x-axis whose abscissa is 11.
Find the distance of the following points from origin.
(5, 6)
Give the relation that must exist between x and y so that (x, y) is equidistant from (6, -1) and (2, 3).
The distance between point P(2, 2) and Q(5, x) is 5 cm, then the value of x = ______.
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`
Find distance CD where C(–3a, a), D(a, –2a).
The distance of the point P(–6, 8) from the origin is ______.
The distance of the point (5, 0) from the origin is ______.
