C1:straight lines
Maths and statistics discussion, revision, exam and homework help.
-
- Reputation:
- Thread Starter
- Adored and Respected Member
- Location: manchester
- Posts: 521
C1:straight linesHi everyone,
I know this is one of the basic chapters, but I've got a question here:
ABCD is a rectangle in which the coordinates of A and C are (0,4) and (11,1) respectively, and the gradient of the side AB is -5.
a)find the equations of side AB and BC. OK, so I've got AB, y=-5x+4. But for BC, am I supposed to assume it's perpendicular to AB? The coordinates of B are given in the next step of the question, so do I need the information for this or is it possible to work out BC without the coordinates of B?
b)Show the coordinates of B is (1,-1)
c)Calculate the area of the rectangle.
Find the coordinates of the point on the y-axis which is equidistant from points A and D. -
- Reputation:
- Overlord in Training
- Posts: 2,609
Re: C1:straight linesa)It's a rectangle ABCD hence AB and CD are parallel and BC and DA are parallel. AB and CD must also be perpendicular to BC and DA. You have the coordinates of C, you only need one coordinate of the line to find the equation.
b) many ways of doing this, my preference would be to note that B is where both line BC and AB intersect.
c) relatively straight forward here however if you have any problems with it just ask -
- Reputation:
- Thread Starter
- Adored and Respected Member
- Location: manchester
- Posts: 521
Re: C1:straight linesOK, so did b) fine, thanks
C) I just thought would be base x hight (AB = AC). So I used Pythagorus to do that. So forAB :
for AC:
Which totals
but the answer is 52! What did I do wrong?Last edited by gavinlee; 11-05-2012 at 11:06. -
- Reputation:
- Overlord in Training
- Posts: 2,609
Re: C1:straight linesAC isn't the height, it's the diagonal line throught the rectangle, use BC for the height.(Original post by gavinlee)
OK, so did b) fine, thanks
C) I just thought would be base x hight (AB = AC). So I used Pythagorus to do that. So forAB :
for AC:
Which totals
but the answer is 52! What did I do wrong? -
- Reputation:
- Overlord in Training
- Posts: 2,609
Re: C1:straight linesSorry was away. That won't be the answer since AD does not lie on the y axis.(Original post by gavinlee)
Thanks Roar, got that now.
For:
"Find the coordinates of the point on the y-axis which is equidistant from points A and D".
Is this just the y coordinate of the midpoint of line AD?
A point on the y axis has the equation (0,y)
distance from A to this point = ((0-0)2 + (4-y)2)1/2
You can do the same with B and use simultaneous equations to find y when distance from A to this point=distance from B to this point.Last edited by roar558; 11-05-2012 at 13:24. -
- Reputation:
- Overlord in Training
- Posts: 2,609
Re: C1:straight linessorry was a typo meant do the same with D it's the same principles though A and D aren't both on the y axis.(Original post by gavinlee)
Hi Roar - thanks for your reply. But wouldn't what you suggested give us the point equidistant from A and from B, when the question asks for equidistance from A and D? -
- Reputation:
- Thread Starter
- Adored and Respected Member
- Location: manchester
- Posts: 521
Re: C1:straight linesOK thanks very much for your help today(Original post by roar558)
sorry was a typo meant do the same with D it's the same principles though A and D aren't both on the y axis.
