The answer in my textbook is totally different.

Question

Thank you for answering my previous question and this one. I totally understand how you got the answers above, however in my book the answers are as follows: (i) The y coordinate of D is 6 because 1/2(-2+14). (ii) Gradient of AD in terms of h is 8/h and gradient of CD in terms of h is 8/12-h. The x coordinates of D and B have been given as 16 and -4 respectively. I do not understand how they got these answers but I thought maybe if you showed me the steps of the calculation I would understand. But I'm not sure why the answers are different. Sorry for bothering you but I am studying the AS level Maths syllabus myself so don't really have a teacher/tutor that I can refer to. Thank you so much.

Comments

I have revised my solution assuming A(0,-2) rather than A(0,2). See answer below. I have used a different method from the one implied in your textbook. There are other geometrical solutions, such as finding the midpoints of CD and BA using the isosceles triangles CMD and BMA, and dropping a perpendicular from the vertices onto the sides of the rectangle. From the midpoints you can calculate the coords of B and D. If you are having problems with your mathematical problems I may be able to help, if you wish. You can send me a private message.

Answer 1

Above is the revised picture.

It’s different because you gave A(0,2) as the point, when it probably should have been A(0,-2). If you make the necessary changes in the solution you should get the answer in your textbook.

Feel free to send me a private message if you still need help. In the meantime I can revise my solution on the assumption that A is the point (0,-2).

REVISED SOLUTION

The diagonals of a rectangle bisect one another, so first we find the midpoint M of diagonal AC, which is the average of the endpoints: ((0+12)/2,(-2+14)/2)=M(6,6). Since diagonal BD passes through M and is parallel to the x-axis, it must have a y coord of 6, since the horizontal line has the equation y=6.

We have the centre of a circle M and radius MC so we can draw a circle passing through A and C, with AC (or BD) as diameter. Any point P on the circumference forms a right angle APC, because the angle subtended by a diameter is a right angle, so the positions of B and D are determined by the intersection of y=6 with the circle. 

Using Pythagoras’ Theorem we can work out the radius of the circle=√((6-(-2))²+(6-0)²)=√(8²+6²)=√100=10. The equation of the circle is (x-6)²+(y-6)²=100. When y=6, x-6=±10, which gives us the x coords of B and D.

B(6-10,8) and D(6+10,8)=B(-4,8) and D(16,8). So h=16.

The gradients of AD (or BC) and CD (or BA) are (6-(-2))/(16-0)=½ and (6-(-2))/(-4-0)=-2. So we have the gradients as 8/h and -h/8. The product of the gradients of two lines perpendicular to one another is -1, and ½×(-2)=-1, and AD and CD are perpendicular to each other.

Note that 8/(12-h)=8/(-4)=-2=-h/8 when h=16.