मराठी
महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

Find the distance between the following pair of points. R(0, -3), S(0, 52) - Geometry Mathematics 2

Advertisements
Advertisements

प्रश्न

Find the distance between the following pair of points.

R(0, -3), S(0, `5/2`)

बेरीज

उत्तर

Suppose co-ordinates of point R are (x1 , y1) and of point S are (x2, y2).

x1 = 0, y1 = -3, x2 = 0, y2 = `5/2`

According to distance formula,

d(R, S) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`

d(R, S) = `sqrt((0 - 0)^2 + [5/2 - (- 3)]^2)`

d(R, S) = `sqrt((0)^2 + [5/2 + 3]^2)`

d(R, S) = `sqrt((0)^2 + (11/2)^2)`

d(R, S) = `sqrt(0 + 121/4)`

d(R, S) = `sqrt(121/4)`

d(R, S) = `11/2`

∴ distance between points R and S is `11/2`.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Co-ordinate Geometry - Practice Set 5.1 [पृष्ठ १०७]

APPEARS IN

बालभारती Geometry (Mathematics 2) [English] 10 Standard SSC Maharashtra State Board
पाठ 5 Co-ordinate Geometry
Practice Set 5.1 | Q 1.3 | पृष्ठ १०७

संबंधित प्रश्‍न

If A(5, 2), B(2, −2) and C(−2, t) are the vertices of a right angled triangle with ∠B = 90°, then find the value of t.


Find the value of x, if the distance between the points (x, – 1) and (3, 2) is 5.


Show that four points (0, – 1), (6, 7), (–2, 3) and (8, 3) are the vertices of a rectangle. Also, find its area


If a≠b≠0, prove that the points (a, a2), (b, b2) (0, 0) will not be collinear.


Find the circumcenter of the triangle whose vertices are (-2, -3), (-1, 0), (7, -6).


Given a triangle ABC in which A = (4, −4), B = (0, 5) and C = (5, 10). A point P lies on BC such that BP : PC = 3 : 2. Find the length of line segment AP.


Using the distance formula, show that the given points are collinear:  

 (1, -1), (5, 2) and (9, 5)


Find x if distance between points L(x, 7) and M(1, 15) is 10. 


Find the distance of the following point from the origin :

(8 , 15)


Find the distance between the following point :

(sec θ , tan θ) and (- tan θ , sec θ)


Find the relation between a and b if the point P(a ,b) is equidistant from A (6,-1) and B (5 , 8).


Prove that the points (7 , 10) , (-2 , 5) and (3 , -4) are vertices of an isosceles right angled triangle.


Prove that the points (a, b), (a + 3, b + 4), (a − 1, b + 7) and (a − 4, b + 3) are the vertices of a parallelogram. 


The distance between the points (3, 1) and (0, x) is 5. Find x.


Find the distance of the following points from origin.
(a+b, a-b) 


KM is a straight line of 13 units If K has the coordinate (2, 5) and M has the coordinates (x, – 7) find the possible value of x.


Find distance between point Q(3, – 7) and point R(3, 3)

Solution: Suppose Q(x1, y1) and point R(x2, y2)

x1 = 3, y1 = – 7 and x2 = 3, y2 = 3

Using distance formula,

d(Q, R) = `sqrt(square)`

∴ d(Q, R) = `sqrt(square - 100)`

∴ d(Q, R) =  `sqrt(square)`

∴ d(Q, R) = `square`


Show that P(– 2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled triangle


Show that the point (0, 9) is equidistant from the points (– 4, 1) and (4, 1)


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×