Calculate the coordinates of T

Question

A (-2;-5), B,C and D are the vertices of quadrilateral ABCD such that diagonals AC is perpendicular to diagonal BD at T. The equation of BTD is given by 2y+x=18 and AB =15units.

Answer 1

The slope of BTD is -1/2 so the slope of the perpendicular ATC is 2. 

The equation of ATC, assuming T is the intersection of the diagonals AC and BD, is y+5=2(x+2), so y=2x-1.

The diagonals intersect when 2x-1=9-x/2. 5x/2=10 so x=4, making y=2*4-1=9-4/2=7.

Therefore T is the point (4,7). We don’t need the length of AB. [In any case, the location of B can be one of two points: (-2,10) or (10,4).]

The locations of vertices C and D can be anywhere along the diagonals.

Answer 2

I think the only condition where AC and BD are perpendicular is: when the quadrilateral is a square. One need to draw the shape of the quadrilateral with having the left bottom corner of it  at the origin of X and Y coordinate system.

After you draw the shape with (AB=15 unites), you will figure out that the other 3 sides are also 15 units because it’s a square. From the figure you can easily say that the coordinates for T is (7.5, 7.5) , but since the question asked you to calculate, then use the midpoint equation for the line BDT to calculate the coordinate of point T as: X = (X1+X2)/2 and Y = (Y1+Y2)/2

You can get the X1, Y1, X2, and Y2 from the two corners of equilateral.

I guess this will be the answer.