Advertisements
Advertisements
प्रश्न
Find distance CD where C(–3a, a), D(a, –2a).
Advertisements
उत्तर
Let C(x1, y1) and D(x2, y2) be the given points.
∴ x1 = –3a, y1 = a, x2 = a, y2 = –2a
By distance formula,
d(C, D) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
= `sqrt([a - (-3a)]^2 + (-2a - a)^2`
= `sqrt((a + 3a)^2 + (-2a - a)^2`
= `sqrt((4a)^2 + (-3a)^2`
= `sqrt(16a^2 + 9a^2)`
= `sqrt(25a^2)`
∴ d(C, D) = 5a units
संबंधित प्रश्न
If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b). Prove that bx = ay.
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.
L(5, –8), M(–7, –3)
Show that the ▢PQRS formed by P(2, 1), Q(–1, 3), R(–5, –3) and S(–2, –5) is a rectangle.
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 pairs of point in the coordinate plane :
(13 , 7) and (4 , -5)
Find the distance between the following point :
(Sin θ - cosec θ , cos θ - cot θ) and (cos θ - cosec θ , -sin θ - cot θ)
Prove that the points (7 , 10) , (-2 , 5) and (3 , -4) are vertices of an isosceles right angled triangle.
Prove that the points (0 , 0) , (3 , 2) , (7 , 7) and (4 , 5) are the vertices of a parallelogram.
Find the distance between the points (a, b) and (−a, −b).
Find the coordinates of the points on the y-axis, which are at a distance of 10 units from the point (-8, 4).
Prove that the points A (1, -3), B (-3, 0) and C (4, 1) are the vertices of an isosceles right-angled triangle. Find the area of the triangle.
Calculate the distance between A (5, -3) and B on the y-axis whose ordinate is 9.
Find the distance of the following points from origin.
(a+b, a-b)
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).
Find distance between points O(0, 0) and B(–5, 12).
Find the value of a, if the distance between the points A(–3, –14) and B(a, –5) is 9 units.
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).
Find the distance between the points O(0, 0) and P(3, 4).
