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RD Sharma Class 10 solutions Chapter 14 Coordinate Geometry Ex 14.4

RD Sharma Class 10 solutions Chapter 14 Coordinate Geometry Ex 14.4

RD Sharma Class 10 Solutions Co-Ordinate Geometry Exercise 14.4

Question 1.
Find the centroid of the triangle whose vertices are :
(i) (1, 4), (-1, -1), (3, -2)
(ii) (-2, 3), (2, -1), (4, 0)
Solution:
RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry

Question 2.
Two vertices of a triangle are (1, 2), (3, 5) and its centroid is at the origin. Find the Co-ordinates of the third vertex.
Solution:
Centroid of a triangle is O(0, 0) ….(i)
Co-ordinates of two vertices of a ∆ABC are A (1, 2) and B (3, 5)
Let the third vertex be (x, y)
RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry

Question 3.
Find the third vertex of a triangle, if two of its vertices are at (-3, 1) and (0, -2) and the centroid is at the origin.
Solution:
Let two vertices of a ∆ABC be A (-3, 1) and B (0, -2) and third vertex C be (x, y)
Centroid of the ∆ABC is O (0, 0)
Co-Ordinate Geometry Class 10 RD Sharma

Question 4.
A (3, 2) and B (-2, 1) are two vertices of a triangle ABC whose centroid G has the coordinates (\(\frac { 5 }{ 3 }\) , \(\frac { -1 }{ 3 }\)) . Find the coordinates of the third vertex C of the triangle. [CBSE 2004]
Solution:
A (3, 2) and B (-2, 1) are the two vertices of ∆ABC whose centroid is G (\(\frac { 5 }{ 3 }\) , \(\frac { -1 }{ 3 }\))
Let third vertex C be (x, y)
Co-Ordinate Geometry Class 10 RD Sharma

Question 5.
If (-2, 3), (4, -3) and (4, 5) are the mid-points of the sides of a triangle, find the co-ordinates of its centroid.
Solution:
In ∆ABC, D, E and F are the mid-points of the sides BC, CA and AB respectively.
The co-ordinates of D are (-2, 3), of E are (4,-3) and of F are (4, 5)
Let the co-ordinates of A, B and C be (x1, y1), (x2, y2), (x3, y3) respectively
RD Sharma Class 10 Solutions Co-Ordinate Geometry
RD Sharma Class 10 Solutions Co-Ordinate Geometry Exercise 14.4
RD Sharma Class 10 Solutions Co-Ordinate Geometry
RD Sharma Class 10 Solutions Co-Ordinate Geometry Exercise 14.4

Question 6.
Prove analytically that the line segment joining the middle points of two sides of a triangle is equal to half of the third side.
Solution:
In ∆ABC,
D and E are the mid points of the sides AB and AC respectively
RD Sharma Class 10 Solutions Chapter 14 Co-Ordinate Geometry
RD Sharma Class 10 Pdf Chapter 14 Co-Ordinate Geometry

Question 7.
Prove that the lines joining the middle points of the opposite sides of a quadrilateral and the join of the middle points of its diagonals meet in a point and bisect one another.
Solution:
Let A (x1, y1), B (x2, y2), C (x3, y3) and D (x4, y4) be the vertices of quadrilateral ABCD
E and F are the mid points of side BC and AD respectively and EF is joined G and H are the mid points of diagonal AC and BD.
GH are joined
RD Sharma Class 10 Solutions Chapter 14 Co-Ordinate Geometry
RD Sharma Class 10 Pdf Chapter 14 Co-Ordinate Geometry
RD Sharma Solutions Class 10 Chapter 14 Co-Ordinate Geometry

Question 8.
If G be the centroid of a triangle ABC and P be any other point in the plane, prove that PA² + PB² + PC² = GA² + GB² + GC² + 3GP².
Solution:
In AABC, G is the centroid of it Let P (h, x) is any point in the plane
Let co-ordinates of A are (x1, y1) of B are (x2, y2) and of C are (x3, y3)
Learncbse.In Class 10 Chapter 14 Co-Ordinate Geometry
Learncbse.In Class 10 Chapter 14 Co-Ordinate Geometry
Hence proved.

Question 9.
If G be the centroid of a triangle ABC, prove that AB² + BC² + CA² = 3 (GA² + GB² + GC²)
Solution:
Let the co-ordinates of the vertices of ∆ABC be A (x1, y1), B (x2, y2), C (x3, y3) and let G be the centroid of the triangle
Class 10 RD Sharma Solutions Chapter 14 Co-Ordinate Geometry
RD Sharma Class 10 Pdf Free Download Full Book Chapter 14 Co-Ordinate Geometry
Class 10 RD Sharma Solutions Chapter 14 Co-Ordinate Geometry
Hence proved.

Question 10.
In the figure, 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.
Solution:
RD Sharma Class 10 Solution Chapter 14 Co-Ordinate Geometry
In right ∆OAB, co-ordinates of O are (0, 0) of A are (2a, 0) and of B are (0, 2b)
C is the mid-point of AB
Co-ordinates of C will be
RD Sharma Class 10 Solution Chapter 14 Co-Ordinate Geometry
We see that CO = CA = CB
Hence C is equidistant from the vertices O, A and B.
Hence proved.

Chapter 14 Coordinate Geometry Ex 14.4 Q1
RD-Sharma-class 10-Solutions-Chapter-14-Coordinate Gometry-Ex-14.4-Q1
Coordinate Geometry Ex 14.4 Q2
RD-Sharma-class 10-Solutions-Chapter-14-Coordinate Gometry-Ex-14.4-Q2
Coordinate Geometry Ex 14.4 Q3
RD-Sharma-class 10-Solutions-Chapter-14-Coordinate Gometry-Ex-14.4-Q3
Coordinate Geometry Ex 14.4 Q4
RD-Sharma-class 10-Solutions-Chapter-14-Coordinate Gometry-Ex-14.4-Q4
Q5
RD-Sharma-class 10-Solutions-Chapter-14-Coordinate Gometry-Ex-14.4-Q5
RD-Sharma-class 10-Solutions-Chapter-14-Coordinate Gometry-Ex-14.4-Q5 i
Q6
RD-Sharma-class 10-Solutions-Chapter-14-Coordinate Gometry-Ex-14.4-Q6
Coordinate Geometry Q7
RD-Sharma-class 10-Solutions-Chapter-14-Coordinate Gometry-Ex-14.4-Q7
RD-Sharma-class 10-Solutions-Chapter-14-Coordinate Gometry-Ex-14.4-Q7 i
Ex 14.4 Q8
RD-Sharma-class 10-Solutions-Chapter-14-Coordinate Gometry-Ex-14.4-Q8
Ex 14.4 Q3 Q 9
RD-Sharma-class 10-Solutions-Chapter-14-Coordinate Gometry-Ex-14.4-Q9
RD Sharma Class 10 solutions Chapter 14 Coordinate Geometry Ex 14.4 Q10
RD-Sharma-class 10-Solutions-Chapter-14-Coordinate Gometry-Ex-14.4-Q10

RD Sharma Class 10 Solutions

  • Chapter 14 Co-Ordinate Geometry Ex 14.1
  • Chapter 14 Co-Ordinate Geometry Ex 14.2
  • Chapter 14 Co-Ordinate Geometry Ex 14.3
  • Chapter 14 Co-Ordinate Geometry Ex 14.4
  • Chapter 14 Co-Ordinate Geometry Ex 14.5

 

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