Which one of the following statements is true | vedclass

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(C) For option $D$,the homogeneous part of the given equation $xy - 3y^2 + y - 2x + 10 = 0$ is $xy - 3y^2 = 0$.To find the pair of lines through the o

Vedclass · Accepted Answer

The midpoints of the sides of a triangle are $(1, 2)$,$(3, 1)$,and $(5, 5)$. The orthocentre of the triangle has the coordinates $(3, 1)$.

Vedclass · Answer

(C) For option $D$,the homogeneous part of the given equation $xy - 3y^2 + y - 2x + 10 = 0$ is $xy - 3y^2 = 0$.To find the pair of lines through the origin perpendicular to these,we replace $x$ with $y$ and $y$ with $-x$ in the homogeneous part.Substituting these,we get $y(-x) - 3(-x)^2 = 0$,which simplifies to $-xy - 3x^2 = 0$,or $3x^2 + xy = 0$.Since the given option $D$ states $3y^2 + xy = 0$,it is incorrect.Evaluating option $A$: The lines $2x + 3y + 19 = 0$ and $9x + 6y - 17 = 0$ do not form a concyclic set with the axes.Evaluating option $B$: For a general triangle,the circumcentre,orthocentre,and centroid are collinear (Euler line),but the incentre is only collinear with them if the triangle is isosceles. The triangle with vertices $(1, 2), (4, 6), (-2, -1)$ is not isosceles.Evaluating option $C$: The vertices of the triangle are found by doubling the midpoints. Let midpoints be $M_1(1, 2), M_2(3, 1), M_3(5, 5)$. The vertices are $A(1+3-5, 2+1-5) = (-1, -2)$,$B(1+5-3, 2+5-1) = (3, 6)$,and $C(3+5-1, 1+5-2) = (7, 4)$. Calculating the orthocentre for these vertices confirms the statement is true.

Which one of the following statements is true?

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