मराठी

The Coordinates of the Point P Are (−3, 2). Find the Coordinates of the Point Q Which Lies on the Line Joining P and Origin Such that Op = Oq.

Advertisements
Advertisements

प्रश्न

The coordinates of the point P are (−3, 2). Find the coordinates of the point Q which lies on the line joining P and origin such that OP = OQ.

Advertisements

उत्तर

If `(x_1,y_1)` and `(x_2, y_2)` are given as two points, then the co-ordinates of the midpoint of the line joining these two points is given as

`(x_m,y_m) = ((x_1 + x_2)/2, (y_1 + y_2)/2)`

It is given that the point ‘P’ has co-ordinates (32)

Here we are asked to find out the co-ordinates of point ‘Q’ which lies along the line joining the origin and point ‘P’. Thus we can see that the points ‘P’, ‘Q’ and the origin are collinear.

Let the point ‘Q’ be represented by the point (x, y)

Further it is given that the OP = OQ

This implies that the origin is the midpoint of the line joining the points ‘P’ and ‘Q’.

So we have that `(x_m,y_m) = (0,0)`

Substituting the values in the earlier mentioned formula we get,

`(x_m,y_m) = ((-3 + x)/2, (2 + y)/2)`

`(0,0) = ((-3 + x)/2, (2 + x)/2)`

Equating individually we have, x = 3 and y = -2

Thus the co−ordinates of the point ‘Q’ is (3, -2)

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 6: Co-ordinate Geometry - Exercise 6.2 [पृष्ठ १६]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 10
पाठ 6 Co-ordinate Geometry
Exercise 6.2 | Q 18 | पृष्ठ १६

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

In Fig. 14.36, a right triangle BOA is given C is the mid-point of the hypotenuse AB. Show that it is equidistant from the vertices O, A  and B. 

    

We have a right angled triangle,`triangle BOA`  right angled at O. Co-ordinates are B (0,2b); A (2a0) and C (0, 0).

 

 

 


Name the quadrilateral formed, if any, by the following points, and given reasons for your answers:

A(-1,-2) B(1, 0), C (-1, 2), D(-3, 0)


Find the co-ordinates of the point equidistant from three given points A(5,3), B(5, -5) and C(1,- 5).


Show that the following points are the vertices of a square:

(i) A (3,2), B(0,5), C(-3,2) and D(0,-1)


Find the ratio which the line segment joining the pints A(3, -3) and B(-2,7) is divided by x -axis Also, find the point of division.


The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are (0, −3). The origin is the midpoint of the base. Find the coordinates of the points A and B. Also, find the coordinates of another point D such that ABCD is a rhombus.


ABCD is rectangle formed by the points A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1). If P,Q,R and S be the midpoints of AB, BC, CD and DA respectively, Show that PQRS is a rhombus.


If the vertices of ΔABC  be A(1, -3) B(4, p) and C(-9, 7) and its area is 15 square units, find the values of p


If the point C(k,4) divides the join of A(2,6) and B(5,1) in the ratio 2:3 then find the value of k. 


Find the centroid of ΔABC  whose vertices are A(2,2) , B (-4,-4) and C (5,-8).


If the point P (m, 3) lies on the line segment joining the points \[A\left( - \frac{2}{5}, 6 \right)\] and B (2, 8), find the value of m.

 
 

Write the ratio in which the line segment doining the points A (3, −6), and B (5, 3) is divided by X-axis.


If P (x, 6) is the mid-point of the line segment joining A (6, 5) and B (4, y), find y.

 

If x is a positive integer such that the distance between points P (x, 2) and Q (3, −6) is 10 units, then x =


If the area of the triangle formed by the points (x, 2x), (−2, 6)  and (3, 1) is 5 square units , then x =


If the points P (xy) is equidistant from A (5, 1) and B (−1, 5), then


Write the equations of the x-axis and y-axis. 


In the above figure, seg PA, seg QB and RC are perpendicular to seg AC. From the information given in the figure, prove that: `1/x + 1/y = 1/z`


If point P is midpoint of segment joining point A(–4, 2) and point B(6, 2), then the coordinates of P are ______.


Find the coordinates of the point which lies on x and y axes both.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×