Class 10 Exam  >  Class 10 Notes  >  RD Sharma Solutions for Class 10 Mathematics  >  Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6)

Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 6.29

Ques.16. Prove that (4, 3), (6, 4) (5, 6) and (3, 5)  are the angular points of a square.
Ans.
Let A (4, 3); B (6, 4); C (5, 6) and D (3, 5) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a square. So we should find the lengths of sides of quadrilateral ABCD.
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
All the sides of quadrilateral are equal.
So now we will check the lengths of the diagonals.
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
All the sides as well as the diagonals are equal. Hence ABCD is a square.

Page No 6.29

Ques.17. Prove that the points (−4,−1), (−2, 4), (4, 0) and (2, 3) are the vertices of a rectangle.
Ans.
Let A (−4,−1); B (−2,−4); C (4, 0) and D (2, 3) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a rectangle. So we should find the lengths of opposite sides of quadrilateral ABCD.
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Opposite sides are equal. So now we will check the lengths of the diagonals.
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Opposite sides are equal as well as the diagonals are equal. Hence ABCD is a rectangle.

Page No 6.29

Ques.18. Find the lengths of the medians of a triangle whose vertices are A (−1,3), B(1,−1) and C(5, 1).
Ans.
We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (−1, 3); B (1,−1) and C (5, 1).
So we should find the mid-points of the sides of the triangle.
In general to find the mid-point P(x,y) of two points A(x1,y1) and (x2,y2) we use section formula as,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Therefore mid-point P of side AB can be written as,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Now equate the individual terms to get,
x=0
y=1
So co-ordinates of P is (0, 1)
Similarly mid-point Q of side BC can be written as,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Now equate the individual terms to get,
x=3
y=0
So co-ordinates of Q is (3, 0)
Similarly mid-point R of side AC can be written as,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Now equate the individual terms to get,
x=2
y=2
So co-ordinates of Q is (2, 2)
Therefore length of median from A to the side BC is,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Similarly length of median from B to the side AC is,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Similarly length of median from C to the side AB is
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics

Page No 6.29

Ques.19. Find the ratio in which the line segment joining the points A(3, −3) and B(−2, 7) is divided by the x-axis. Also, find the coordinates of the point of division.    [CBSE 2014]
Ans.
Suppose the x-axis divides the line segment joining the points A(3, −3) and B(−2, 7) in the ratio k : 1.
Using section formula, we get
Coordinates of the point of division =Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Since the point of division lies on the x-axis, so its y-coordinate is 0.
∴ (7k−3)/(k+1)=0
⇒7k−3=0
⇒k=3/7
So, the required ratio is 3/7:1 or 3:7.
Putting k = 3/7, we get
Coordinates of the point of division =Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics=Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
=(15/10,0)
Thus, the coordinates of the point of division are (3/2,0).

Page No 6.29

Ques.20. Find the ratio in which the point P(x, 2) divides the line segment joining the points A(12, 5) and B(4, −3). Also, find the value of x.    [CBSE 2014]
Ans.
Suppose P(x, 2) divides the line segment joining the points A(12, 5) and B(4, −3) in the ratio k : 1.
Using section formula, we get
Coordinates of P = Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
∴((4k+12)/(k+1),(−3k+5)/(k+1))=(x,2)
⇒x=(4k+12)/(k+1) and (−3k+5)/(k+1)=2
Now,
(−3k+5)/(k+1)=2
⇒−3k+5=2k+2
⇒5k=3
⇒k=3/5
So, P divides the line segment AB in the ratio 3 : 5.
Putting k = 3/5 in x=(4k+12)/(k+1),we get
x=Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics=(12+60)/(3+5)=72/8=9
Thus, the value of x is 9.

Page No 6.29

Ques.21. Find the ratio in which the point P(−1, y) lying on the line segment joining A(−3, 10) and B(6 −8) divides it. Also find the value of y.   [CBSE 2013]
Ans.
Suppose P(−1, y) divides the line segment joining A(−3, 10) and B(6 −8) in the ratio k : 1.
Using section formula, we get
Coordinates of P =Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
∴((6k-3)/(k+1), (-8k+10)/(k+1)) = (-1,y)
⇒ (6k-3)/(k+1) = -1 and y= (-8k+10)/(k+1)
Now, (6k-3)/(k+1) = -1
⇒6k-3 = -k-1
⇒7k = 2
⇒k = 2/7
So, P divides the line segment AB in the ratio 2 : 7.
Putting k = 2/7 in y= (-8k+10)/(k+1), we get
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
=(-16+70)/(2+7)
=54/9
=6
Hence, the value of y is 6.

Page No 6.29

Ques.22. Find the coordinates of a point A, where AB is a diameter of the circle whose centre is (2, −3) and B is (1, 4).
Ans. 
Let the co-ordinates of point A be (x,y).
Centre lies on the mid-point of the diameter. So applying the mid-point formula we get,
(x+1)/2=2
x=3
Similarly,
(y+4)/2=-3
y=-10
So the co-ordinates of A are (3,−10)

Page No 6.29

Ques.23. If the points (−2, −1), (1, 0), (x, 3) and  (1, y) form a parallelogram, find the values of x and y.
Ans.
Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (−2,−1); B (1, 0); C (x, 3) and D (1, y).
Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide.
In general to find the mid-point P(x,y) of two points A(x1,y1) and B(x2,y2) we use section formula as,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
The mid-point of the diagonals of the parallelogram will coincide.
So,
Co-ordinate of mid-point of AC = Co-ordinate of mid-point of BD
Therefore,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Now equate the individual terms to get the unknown value. So,
(x-2)/2 = 1
x = 4
Similarly,
(y-0)/2 = 1
y = 2
Therefore, x=4, y=2

Page No 6.29

Ques. 24. The points A(2, 0), B(9, 1) C(11, 6) and D(4, 4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not.
Ans. 
Let A (2, 0); B (9, 1); C (11, 6) and D (4, 4) be the vertices of a quadrilateral. We have to check if the quadrilateral ABCD is a rhombus or not.
So we should find the lengths of sides of quadrilateral ABCD.Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
All the sides of quadrilateral are unequal. Hence ABCD is not a rhombus.

Page No 6.29

Ques.25. In what ratio does the point (−4, 6) divide the line segment joining the points A(−6, 10) and B(3,−8)?
Ans. 
The co-ordinates of a point which divided two points  and  internally in the ratio  is given by the formula,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Here it is said that the point (−4,6) divides the points A(−6,10) and B(3,−8). Substituting these values in the above formula we have,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Equating the individual components we have,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
-4m-4n = 3m-6n
7m = 2n
m/n = 2/7
Therefore the ratio in which the line is divided is 2:7

Page No 6.29

Ques. 26. Find the ratio in which the y-axis divides the line segment joining the points (5, −6)  and (−1,−4). Also, find the coordinates of the point of division.
Ans. 
The ratio in which the y-axis divides two points (x1,y1) and (x2,y2) is λ:1.
The co-ordinates of the point dividing two points (x1,y1) and (x2,y2) in the ratio m:n is given as,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematicswhere, λ = m/n
Here the two given points are A(5,−6) and B(−1,−4).
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Since, the y-axis divided the given line, so the x coordinate will be 0.
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Thus the given points are divided by the y-axis in the ratio 5:1.
The co-ordinates of this point (x, y) can be found by using the earlier mentioned formula.
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Thus the co-ordinates of the point which divides the given points in the required ratio are Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics.

Page No 6.29

Ques.27. Show that A (−3, 2), B (−5, −5), C (2,−3), and D (4, 4) are the vertices of a rhombus.
Ans.
Let A (−3, 2); B (−5,−5); C (2,−3) and D (4, 4) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a rhombus. So we should find the lengths of sides of quadrilateral ABCD.
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
All the sides of quadrilateral are equal. Hence ABCD is a rhombus.

Page No 6.29

Ques.28. Find the length of the medians of a ΔABC having vertices at A(0, −1), B(2, 1) and C(0, 3).
Ans. 
We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (0,−1); B (2, 1) and C (0, 3).
So we should find the mid-points of the sides of the triangle.
In general to find the mid-point P(x,y) of two points A(x1,y1) and B(x2,y2) we use section formula as,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Therefore mid-point P of side AB can be written as,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Now equate the individual terms to get,
x=1
y=0
So co-ordinates of P is (1, 0)
Similarly mid-point Q of side BC can be written as,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Now equate the individual terms to get,
x=1
y=2
So co-ordinates of Q is (1, 2)
Similarly mid-point R of side AC can be written as,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Now equate the individual terms to get,
x=0
y=1
So co-ordinates of R is (0, 1)
Therefore length of median from A to the side BC is,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Similarly length of median from B to the side AC is,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Similarly length of median from C to the side AB is
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics

Page No 6.29

Ques.29. Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, –3). Hence, find m.
Ans.
Let P divides AB in a ratio of λ : 1
Therefore, coordinates of the point P areChapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Given that coordinates of the point P are (4, m).
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics=4
⇒6λ+2=4λ+4
⇒λ=1
Hence, the point P divides AB in a ratio of 1 : 1.
Replacing the value of λ = 1 in y-coordinate of P, we get
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics= m
⇒ m=0
Thus, y-coordinate of P is equal to 0.

Page No 6.30

Ques.30. Find the coordinates of the points which divide the line segment joining the points (−4, 0) and (0, 6) in four equal parts.
Ans.
The co-ordinates of the midpoint (xm,ym)  between two points (x1,y1) and (x2,y2) is given by,
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Here we are supposed to find the points which divide the line joining A(−4,0) and B(0,6) into 4 equal parts.
We shall first find the midpoint M(x, y) of these two points since this point will divide the line into two equal parts
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
So the point M(−2,3) splits this line into two equal parts.
Now, we need to find the midpoint of A(−4,0) and M(−2,3) separately and the midpoint of B(0,6) and M(−2,3). These two points along with M(−2,3) split the line joining the original two points into four equal parts.
Let M1(e,d) be the midpoint of A(−4,0) and M(−2,3).
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Now let M2(g,h) bet the midpoint of B(0,6) and M(−2,3).
Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics
Hence the co-ordinates of the points which divide the line joining the two given points are Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics.

The document Chapter 7 - Coordinate Geometry, RD Sharma Solutions - (Part-6) | RD Sharma Solutions for Class 10 Mathematics is a part of the Class 10 Course RD Sharma Solutions for Class 10 Mathematics.
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