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).


1