Co Ordinate Geometry Qs

Take Your Pants Off

Really struggling with this whole qs.
Any tips or advice on how to approach

Moody_mona

The first question is asking you to find the perpendicular distsnce from a point to a line, which you have a formula for. . .
To use that you need to have the point and the line. You have a point so you need to find the line using a combination of the slope and the line formula.

The second part asks you to find the area. You can use the formula 1/2 (base)(height) since you already know the height from the previous part and now you just need to find the length of the base.

RoRo979

first part spot on, second part can be done that way or move one point to (0,0) and i believe the formular is 1/2|x1y2 + x2y1|. It is something like that

MathsManiac

RoRo979 wrote: »

first part spot on, second part can be done that way or move one point to (0,0) and i believe the formular is 1/2|x1y2 + x2y1|. It is something like that

But the question says "hence", so you cannot use a method that does not rely on the result from part (i). If this was a Leaving Cert question, you would be penalised if the question said "hence" and you didn't do it by a "hence" method. On the other hand, if it says "Hence or otherwise...", then using the result of the first part is optional.

evolving_doors

A longer work around might be to find the co-ordinate of N first
(If you've forgotten the formula for perpendicular dist from pt. to a line!)

Could you get the equation of the line AN? (Considering you know that AN's slope relates to CB in some way).

Then find the equation of CB and see where the 'equations simultaneously' intersect. This will give the co-ordinate of N

It's interesting that O is set up as the midpoint of AC too! I'd have to think about it for a while but is there another way using the right angled triangle inscribed in a circle (or OA being equal to AN).

Also similar right angled triangles using the distances!!

evolving_doors

Armelodie wrote: »

A longer work around might be to find the co-ordinate of N first
(If you've forgotten the formula for perpendicular dist from pt. to a line!)

Could you get the equation of the line AN? (Considering you know that AN's slope relates to CB in some way).

Then find the equation of CB and see where the 'equations simultaneously' intersect. This will give the co-ordinate of N

It's interesting that O is set up as the midpoint of AC too! I'd have to think about it for a while but is there another way using the right angled triangle inscribed in a circle (or OA being equal to AN).

Also similar right angled triangles using the distances!!

Anyone do it using the right angled triangle inscribed in a circle with centre o and radius OA (or ON !). I just thought it was interesting that it cuts through the centre (0,0)

Kremin

i) Find equation of CB..
1-(-2)/3-(-1)
= 3/4.. this is the slope..
y-1=3(x-3)
y-1=3x-9

3x-y-8=0
Now use the formula for perpendicular distance from a point (x1,y1) to a line ax+by+c=0
a=3 b=-1 c=-8
x1:1 y1:2

Which is, d= |ax1+by1+c|/sqrt(a^2+b^2)

So |3(1)+(-1)(2)-8|/sqrt(9+1)
|3-2-8|/root10

7/root10 or 7root10/10

that's the perpendicular height, for part ii, I'd use the distance formula to get the distance between C and B and then use 1/2 base by height. This way you're using the result from part i.

Kremin

Gebgbegb wrote: »

Anyone do it using the right angled triangle inscribed in a circle with centre o and radius OA (or ON !). I just thought it was interesting that it cuts through the centre (0,0)

Could work out the slope of CB, then flip the fraction and reverse the sign.. then you have the slop of the perpendicular, put it in the y-y1=mx-mx1 formula and you could solve and find the intersection, then find the distance between N and A, seems like a fair bit more work than just using the perpendicular distance formula though .