Find the gradient of line AC = ((3-2))/((5-2))
= 1/3
The diagonal BD is perpendicular to line AC (property of diagonal of a square)
The product of the gradient of the two lines is -1
Hence gradient of the other line is -3
The two lines meet the midpoint of the line
That of AC =[ (2 +5)/2, (2+3)/2]
= (3.5, 2.5)
The equation of the line BD
(y- 2.5) = -3(x-3.5)
Y =-3x -0.5
joseph rimiru answered the question on October 1, 2017 at 10:17
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