What is the orthocentre of a triangle given the points A(-1,5) B(7,2) C(-1,-4)

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The orthocentre of a triangle is the point where the altitude of the three sides intersect. For the triangle with vertices at A(-1,5), B(7,2) C(-1,-4) and first determine the altitude of any two sides and then the point at which they intersect.

The slope of the line AB is `(5 - 2)/(-1-7) = -3/8` . The slope of the altitude is `8/3` . As it passes through C(-1, -4) the equation is `(y + 4)/(x + 1) = 8/3`

=> 3(y + 4) = 8(x + 1)

=> 8x - 3y - 4 = 0

The slope of the line AC is `(-4 - 5)/(-1 + 1) = oo` . The altitude is a vertical line. As it passes through B(7, 2) the equation is y = 2.

The point of intersection of 8x - 3y - 4 = 0 and y = 2 is (1.25, 2)

The required orthocentre is (1.25, 2)

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