Show that the points A(0, 6), B(2, 1) and C(7, 3) are three corners of a square ABCD. Find (i) the slope of the diagonal BD and (ii) the coordinates of the fourth vertex D.
In a square, all sides are perpendicular to the adjacent side, so the product of slope of two adjacent sides is -1.
Let the position of point D(a,b).
Given points of the square are A(0, 6),B(2, 1),C(7, 3) and D(a,b).
The slope of line AB =
The slope of line BC =
The slope of line CD =
The slope of line DA =
The slope of diagonal AC =
The slope of diagonal BD = m5
(i) We know that in a square, two diagonals are perpendicular to each other, therefore
The slope of diagonal AC×slope of diagonal BD = -1
So the slope of diagonal BD is 7/3.
(ii) We know that midpoint of diagonal AC = midpoint of diagonal BD
and comparing x and y coordinates respectively.
So coordinate of the point D(5,8).