Advertisements
Advertisements
Question
Find distance between points O(0, 0) and B(–5, 12).
Advertisements
Solution
Let O(x1, y1) = O(0, 0) and B(x2, y2) = B(–5, 12)
∴ x1 = 0, y1 = 0, x2 = –5, y2 = 12
By distance formula,
d(O, B) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
= `sqrt((-5 - 0)^2 + (12 - 0)^2`
= `sqrt((-5)^2 + 12^2`
= `sqrt(25 + 144)`
= `sqrt(169)`
∴ d(O, B) = 13 units
∴ The distance between the points O and B is 13 units.
APPEARS IN
RELATED QUESTIONS
If A(4, 3), B(-1, y) and C(3, 4) are the vertices of a right triangle ABC, right-angled at A, then find the value of y.
If two vertices of an equilateral triangle be (0, 0), (3, √3 ), find the third vertex
Find the distance between the following pairs of points:
(2, 3), (4, 1)
Find the values of y for which the distance between the points P (2, -3) and Q (10, y) is 10 units.
The value of 'a' for which of the following points A(a, 3), B (2, 1) and C(5, a) a collinear. Hence find the equation of the line.
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.
Find the distance between the points:
P(a + b, a – b) and Q(a – b, a + b)
Find the distance of the following points from the origin:
C(–4, –6)
Find the distance between the following pairs of point in the coordinate plane :
(4 , 1) and (-4 , 5)
Prove that the points (0 , 0) , (3 , 2) , (7 , 7) and (4 , 5) are the vertices of a parallelogram.
Find the distance between the origin and the point:
(-5, -12)
The centre of a circle is (2x - 1, 3x + 1). Find x if the circle passes through (-3, -1) and the length of its diameter is 20 unit.
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.
Find the distance of the following points from origin.
(a cos θ, a sin θ).
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 point A(7, 5) and B(2, 5).
The points (– 4, 0), (4, 0), (0, 3) are the vertices of a ______.
Find distance between points P(– 5, – 7) and Q(0, 3).
By distance formula,
PQ = `sqrt(square + (y_2 - y_1)^2`
= `sqrt(square + square)`
= `sqrt(square + square)`
= `sqrt(square + square)`
= `sqrt(125)`
= `5sqrt(5)`
|
In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.
|
Based on the above information answer the following questions using the coordinate geometry.
- Find the distance between Lucknow (L) to Bhuj (B).
- If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).
- Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P)
[OR]
Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).
Show that points A(–1, –1), B(0, 1), C(1, 3) are collinear.
