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Centroid, circumcentre or orthocentre of the triangle formed by three given lines

This type of question, asked 3 times: 8 Apr 2023, 13 Apr 2023, 4 Apr 2025. All 3 questions below, oldest first.

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  1. Question 1

    8 Apr 2023

    64
    Let C(α,β)C(\alpha,\beta) be the circumcenter of the triangle formed by the lines
    4x+3y=694x + 3y = 69
    4y−3x=174y - 3x = 17 and
    x+7y=61x + 7y = 61
    Then (α−β)2+α+β(\alpha - \beta)^{2}+ \alpha + \beta is equal to

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  2. Question 2

    13 Apr 2023

    156
    Let (α,β)(\alpha,\beta) be the centroid of the triangle formed by the lines 15x−y=82,6x−5y=−415x - y = 82,6x - 5y = - 4 and 9x+4y=179x + 4y = 17. Then α+2β\alpha + 2\beta and 2α−β2\alpha - \beta are the roots of the equation

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  3. Question 3

    4 Apr 2025

    65
    Let the three sides of a triangle are on the lines 4x−7y+10=0,x+y=54x- 7y+ 10 = 0,x+y=5 and 7x+4y=157x+ 4y= 15. Then the distance of its orthocenter from the orthocenter of the triangle formed by the lines x=0,y=0x= 0,y= 0 and x+y=1x+y= 1 is.

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