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10th Class Mathematics Coordinate Geometry Question Bank Co-ordinate Geometry

  • question_answer
    Find the coordinates of in centre of a triangle PQR whose vertices are P (15, 15); Q (47, 40); R (65, 20) and whose sides are QR = 26.9; RP = 50.2 and PQ = 40.6.

    A)  (46, 20)                       

    B)  (46, 27)     

    C)  (27, 27)                       

    D)  (44, 25)

    Correct Answer: B

    Solution :

    (b): \[Ix=\frac{pPx+qQx+rRx}{p+q+r}\] \[Iy=\frac{pPy+qQy+rRy}{p+q+r}\] \[\therefore Ix=\frac{26.9\times 15+50.2\times 47+40.6\times 65}{117.7}=46\] \[Iy=\frac{26.9\times 15+50.2\times 40+40.6\times 65}{117.7}=27\]     


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