Let A equiv (4,4), B equiv (8,4), C equiv (4, | vedclass

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(C) The vertices of $\Delta ABC$ are $A(4,4), B(8,4), C(4,8)$.Since $AB$ is horizontal and $AC$ is vertical,$\Delta ABC$ is a right-angled triangle at

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$10$

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(C) The vertices of $\Delta ABC$ are $A(4,4), B(8,4), C(4,8)$.Since $AB$ is horizontal and $AC$ is vertical,$\Delta ABC$ is a right-angled triangle at $A(4,4)$.The midpoints are $P = \frac{A+B}{2} = (6,4)$,$Q = \frac{B+C}{2} = (6,6)$,and $R = \frac{C+A}{2} = (4,6)$.$\Delta PQR$ is also a right-angled triangle with the right angle at $P(6,4)$ because $PQ$ is vertical and $PR$ is horizontal.The orthocentre of a right-angled triangle is the vertex where the right angle is located.Thus,the orthocentre of $\Delta PQR$ is $P(6,4)$.Therefore,$\alpha = 6$ and $\beta = 4$.The value of $\alpha + \beta = 6 + 4 = 10$.

Let $A \equiv (4,4), B \equiv (8,4), C \equiv (4,8)$. If $P, Q, R$ are the midpoints of sides $AB, BC, CA$ respectively and $(\alpha, \beta)$ are the coordinates of the orthocentre of $\Delta PQR$,then the value of $\alpha + \beta$ is

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