The diagram shows a quadrilateral ABCD. The point E lies on AD such that angle AEB = 90 degree. The Line EC is parallel to the x axis and the line CD is parallel to the y axis. The point A and E are (-1,6) and (3,4) respectively. Given that the gradient of AB is 1/3 .
Ok so i have found the coordinate of B i.e (5,8). In the next part they gave one more information, 'Arwa of triangle EBC is 24 units'. Find coordinate of C and of D? How do you find those two ?? PLS HELP
For other readers, entire exercise is as follows:
Consider a quadrilateral ABCS. The point E lies on AD such that the angle AEB is right. The line EC is parallel to the x-axis, and the line CD is parallel to the y-axis. The point A is at (-1, 6), and the point E is at (3, 4). The gradient (or slope) of AB is \(\displaystyle m=\frac{1}{3}\)
(i) Find the coordinate of the point B.
(ii) Given that the area of the triangle EBC is 24 square units, find the coordinates of points C and D.
You don't show your work. For the benefit of other students viewing this thread, I'll guess it was something along the lines of the following:
(i) The slope of the line AD, which contains E, is:
. . . . .\(\displaystyle m\, =\, \dfrac{4\, -\, 6}{3\, -\, (-1)}\, =\, \dfrac{-2}{4}\, =\, -\dfrac{1}{2}\)
Then the slope of the line EB, being perpendicular (due to the right angle) must be m = 2. Using this slope and the point E, we get the equation for the line through B and E as being:
. . . . .\(\displaystyle y\, -\, 4\, =\, 2(x\, -\, 3)\)
. . . . .\(\displaystyle y\, =\, 2x\, -\, 6\, +\, 4\)
. . . . .\(\displaystyle y_{EB}\, =\, 2x\, -\, 2\)
The line AB has slope m = 1/3, and we have the coordinates for point A, so the line AB is given by:
. . . . .\(\displaystyle y\, -\, 6\, =\, \dfrac{1}{3}\, (x\, -\, (-1))\)
. . . . .\(\displaystyle y\, -\, 6\, =\, \dfrac{x}{3}\, +\, \dfrac{1}{3}\)
. . . . .\(\displaystyle y_{AB}\, =\, \dfrac{x}{3}\, +\, \dfrac{19}{3}\)
The point B is at the intersection of these two points, so:
. . . . .\(\displaystyle 2x\, -\, 2\, =\, \dfrac{x}{3}\, +\, \dfrac{19}{3}\)
. . . . .\(\displaystyle 6x\, -\, 6\, =\, x\, +\, 19\)
. . . . .\(\displaystyle 6x\, -\, x\, =\, 19\, +\, 6\)
. . . . .\(\displaystyle 5x\, =\, 25\)
. . . . .\(\displaystyle x\, =\, 5\)
Then:
. . . . .\(\displaystyle y\, =\, 2(5)\, -\, 2\)
. . . . .\(\displaystyle y\, =\, 10\, -\, 2\)
. . . . .\(\displaystyle y\, =\, 8\)
So the coordinates for point B are (5, 8).
To proceed on part (ii), start with the fact that they've given you info on a triangle:
(ii) Given that the area of the triangle EBC is 24 square units, find the coordinates of points C and D.
Specifically, they've given you info on the point C. Where can this lead us?
You know that the area of a triangle is given by (1/2)bh, where b is the length of the base and h is the height. You know that one side of the triangle EBC is horizontal, becaue the line EC parallels the x-axis. Knowing the coordinates of E, what then is the equation for the line EC?
Knowing the coordinates for point B, how tall then must the triangle EBC be? Given this height, area, and area formula, what then must be the width of the base, b, along the line EC?
Once you've found the length of the base, and knowing that the other end of this base (being the point C) lies on the same line as does E, what then are the coordinates for point C?
You know that the point D is on the line CD and on the line AD. You know the equation for the line AD. The line CD is vertical, so you can easily find the equation of this line. What then are the coordinates of point D?
If you get stuck, please show your work and reasoning so far, as I have above. Thank you!
