Study MaterialsRD Sharma SolutionsRD Sharma Class 10 solutions Chapter 14 Coordinate Geometry Ex 14.4

RD Sharma Class 10 solutions Chapter 14 Coordinate Geometry Ex 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

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    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
    Ex 14.4 Q10
    RD-Sharma-class 10-Solutions-Chapter-14-Coordinate Gometry-Ex-14.4-Q10

     

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