Math Q's Coordinate Geometry Help! (1 Viewer)

[x.Exhaust.x]

Hey guys . These questions are on the topic of co-ordinate geometry. I tried posting this thread up before which took a lot of time, but then it said: 'Invalid Link: Please notify the administrator' .

Ok back on topic. Here are the q's (Or if you happen to have Year 11 Prelim. Cambridge, refer to Exercise 5E pages 178,179 , q6, q7b), q7d), q11b), q13)a), q13)b). Please show all working out. Thank you.

1. The line y-x+h=0 is more than 1/root 2 units from the point (2,7). What range of values may h take?

2. Use the perpendicular distance formula to determine how many times each line intersects the given circle:

a) 7x+y-10=0, x^2+y^2 = 2

b) x+2y+3=0, (x+2)^2 + (y-1)^2 = 6

3. Please refer to Question 11b) on the diagram of Cambridge. If you don't have cambridge, then I'll explain what I did:

Well I found the perpendicular distance, radius and centre of the circle which was the first part of the question. Then it tells me to determine the length of the chord cut off from the line by the circle. Is the perpendicular distance of a circle equal to the chord of the circle?

4. The point P(x,y) is equidistant from the lines 2x+y-3=0 and x-2y+1=0 which intersect at A.

a) use the distance formula to show that |2x+y-3| = |x-2y+1|
b) hence find the equations of the lines that bisect the angles at A

 

[Aerath]

1. The line y-x+h=0 is more than 1/root 2 units from the point (2,7). What range of values may h take?

Rewrite eqn as x-y+h = 0

Using perpendicular distance formula which is: d = |Ax+By+C|/(root[A^2 + B^2])
Sub in all appropriate numbers:
d = |x-y+h|/(root[1^2 + 1^2]) > 1/root2
d = |2-7+h|/(root2) > 1/root2
|2-7+h| > 1
|h-5| > 1

If h-5 is >0
h > 6

If h-5 <0
h-5 < -1
h < 4

Therefore h<4, h>6 are solutions, I think.

2. Use the perpendicular distance formula to determine how many times each line intersects the given circle:

a) 7x+y-10=0, x^2+y^2 = 2

b) x+2y+3=0, (x+2)^2 + (y-1)^2 = 6

For this question, if the radius = distance, then it intersects at one point, if radius > distance, intersects at zero points, radius < distance, then intersects at two points.

So, for a):
Circle: Centre = (0,0) r = root2

Perpendicular distance from (0,0) to 7x+y-10 = 0
d = |Ax+By+C|/(root[A^2 + B^2])
= |0+0-10|/(root[49+1]
= 10/2root5
= 5root5.

Since 5root5 > root2, therefore , intersects at no points.

Sorry - going to get back to homework now. Sorry if it's a little rushed. =\

 

[x.Exhaust.x]

Thanks Aerath. Could anyone do the rest of the q's?

P.S For the first question, isn't it meant to be x-y-h=0? Still unsure . When I looked at the answers, your answers are wrong .

 

[Aerath]

Yeah, I screwed it up. But I think you get the general gist of it now. I'll finish the rest of the questions when I get home.

 

[lyounamu]

3.

C = chord length

From Q11(a), we already found that r = 2 & d = r/(square root of 5)

Using the Pythagoras' theorem we can find the half of the chord length

(1/2 C)^2 + (4/(square root of 5))^2 = 2^2
(1/2 C)^2 = 4 - 16/5
1/2 C = square root of (4/5)
C = 2 . 2/(square root of 5)
= 4/(square root of 5)

4(a).
(x,y) a=2, b=1 and c=-3
d1 = absolute value of (ax1 + by1 + c)/(square root of (a^2 + b^2))
= absolute value of (2x + y - 3)/(square root of (2^2) + 1^2))
= absolute value of (2x + y - 3)/(square root of 5)
(x,y) a=1, b=-2 and c = 1
d2 = absolute value of (ax1+ by1 + c)/(square root of (a^2 + b^2))
= absolute value of (x - 2y + 1)/(square root of (1^2 + (-2)^2))
= absolute value of (x- 2y + 1)/(square root of 5)

It was given that d1 = d2
Therefore, (absolute value of (2x+y-3))/square root of 5) = (absolute value of (x-2y+1))/(square root of 5)
Therefore, absolute value of (2x+y-3) = absolute value of (x-2y+1) (square root of 5 cancelled at both sides).

b) 2x+y-3 = 0 ...(1)
x-2y+1 = 0 ...(2)
2x + y - 3 = 0 ...(1)
2x - 4y + 2 = 0 ...(3) from (2)
5y -5 = 0
5y = 5
y=1
Substitute that to find x.
x = 1

Now, find the angle that bisects the angle. (Angle is 90 degrees because both lines are negative reciprocal of each other = perpendicular).
Therefore, the angle is 45 degrees since 90/2 = 45 degrees.

Using the formula tan @ = (m1 - m2))/(1+m1m2) (I will chosoe m = -2)
1 = m1 +2 / 1-2m
1- 2m1 = m1 +2
3m1 = -1
m1 = -1/3
When I choose m=1/2, 1 = (m1 - 1/2)/(1+1/2m1)
1+1/2m1 = m1 - 1/2
1/2m1 = 3/2
m1 = 3
Now, you find two gradients which are m=3 & m=-1/3

Now find the two equations using two gradients for each equation (and hence two equations).
Equation 1 (with m =3):y - 1 = 3(x-1)
y = 3x - 2
3x -y - 2 = 0

Equation 2 (with m=-1/3): y-1 = -1/3(x-1)
-3y +3 = x -1
3y + x -4 =0

 

[lolokay]

4. The point P(x,y) is equidistant from the lines 2x+y-3=0 and x-2y+1=0 which intersect at A.

a) use the distance formula to show that |2x+y-3| = |x-2y+1|
b) hence find the equations of the lines that bisect the angles at A

a) the x's and y's refer to P(x,y) I assume. As P from the lines are equal,
|2x+y-3|/sqrt.5 = |x-2y+1|/sqrt.5


so those equations are equal at point P.

b) solve simultaneously to find A,
1)2x+y-3=0
2)x-2y+1=0
3)2x-4y+2=0
1-3)5y-5=0
y=1
x-2+1=0
x=1

as P is equidistant to the lines at those points, it is the centre of them, and the line passing through P and A(1,1) will therefore bisect the angle.

the gradients of the two line are -2 and 1/2, so they are perpendicular, so the angle bisecting them has an angle of 45*
Pick either gradient, to find gradient of a line at 45* to it, eg for the m1 = -2
|(m+2)/(1-2m)| = tan45 = 1, (formula for angle between 2 lines if you've done it)
m = 3, -1/3

y - 1 = 3(x - 1)
y-1 = 3x-3
3x - y - 2 = 0

and, y-1 = -1/3x + 1/3
x + 3y - 4 = 0

are hopefully the two bisecting lines. Not sure if I used the 'hence' though so there's probably a simpler way ?

 

[lyounamu]

1)
a = -1, b=1 and c=h with (2, 7)
((absolute value of (ax + by + c))/(square root of (a^2 + b^2))) > 1/square root of 2

absolute value of (-2 + 7 + h) / (square root of ((-1)^2 + 1^2)) > 1/square root of 2

absolute value of (5 +h) / square root of 2 > 1/square root of 2
absolute value of (5 + h) > 1
Therefore, 5+h > 1 or 5+h < -1
Therefore, h > -4 or h < -6

I don't feel like doing 2. It's just pointless & I am sick of writing asldjasldas^asldkjasldjas (you know what I mean). I hope that you can solve Q2 based on Aerath's working. Good luck.

 

[Aerath]

If you need help, just say it. I so wish I had a scanner. =\

 

[lyounamu]

Rewrite eqn as x-y+h = 0

Using perpendicular distance formula which is: d = |Ax+By+C|/(root[A^2 + B^2])
Sub in all appropriate numbers:
d = |x-y+h|/(root[1^2 + 1^2]) > 1/root2
d = |2-7+h|/(root2) > 1/root2
|2-7+h| > 1
|h-5| > 1

If h-5 is >0
h > 6

If h-5 <0
h-5 < -1
h < 4

Therefore h<4, h>6 are solutions, I think.


For this question, if the radius = distance, then it intersects at one point, if radius > distance, intersects at zero points, radius < distance, then intersects at two points.

So, for a):
Circle: Centre = (0,0) r = root2

Perpendicular distance from (0,0) to 7x+y-10 = 0
d = |Ax+By+C|/(root[A^2 + B^2])
= |0+0-10|/(root[49+1]
= 10/2root5
= 5root5.

Since 5root5 > root2, therefore , intersects at no points.

Sorry - going to get back to homework now. Sorry if it's a little rushed. =\

a= -1 and b = 1

You mixed up your a & b values.

 

[lyounamu]

Rewrite eqn as x-y+h = 0

Using perpendicular distance formula which is: d = |Ax+By+C|/(root[A^2 + B^2])
Sub in all appropriate numbers:
d = |x-y+h|/(root[1^2 + 1^2]) > 1/root2
d = |2-7+h|/(root2) > 1/root2
|2-7+h| > 1
|h-5| > 1

If h-5 is >0
h > 6

If h-5 <0
h-5 < -1
h < 4

Therefore h<4, h>6 are solutions, I think.


For this question, if the radius = distance, then it intersects at one point, if radius > distance, intersects at zero points, radius < distance, then intersects at two points.

So, for a):
Circle: Centre = (0,0) r = root2

Perpendicular distance from (0,0) to 7x+y-10 = 0
d = |Ax+By+C|/(root[A^2 + B^2])
= |0+0-10|/(root[49+1]
= 10/2root5
= 5root5.

Since 5root5 > root2, therefore , intersects at no points.

Sorry - going to get back to homework now. Sorry if it's a little rushed. =\

Shouldn't it be 2/root2 = root 2? Therefore it intersects

 

[Aerath]

Yeah - sorry - horribly bad mistake.

 

[lolokay]

Shouldn't it be 2root2 > 2? It doesn't really matter as you already proved that they don't intersect.

I thought it would be just sqrt.2, so it has 1 intersection

 

[lyounamu]

I thought it would be just sqrt.2, so it has 1 intersection

I was referring to a) where the d > r. (and hence no intersection).
For b) d = r, (hence one intersection).

 

[lolokay]

I was referring to a) where the d > r. (and hence no intersection).
For b) d = r, (hence one intersection).

I was referring to [2]a) too. Isn't b) 2 intersections?


+ in a), the working goes 10/(5sqrt2) not 10/5 sqrt2, right?

 

[lyounamu]

I was referring to [2]a) too. Isn't b) 2 intersections?


+ in a), the working goes 10/(5sqrt2) not 10/5 sqrt2, right?

I thought the questions were a) & b) on the Cambridge. I just realised that I was looking at totally different questions. I only looked at the Aerath working out anyway.From that I deduced that d (2root2)> r (root2) (and hence no intersection, which wasn't true). Then I realised that the question that Razizi was referring to was c) & d) on the Cambridge.

On Cambridge question, b) only has one intersection. But the b) that Razizi is referring to is obviously not the b) in Cambridge.

EDIT: I did c) and d), you are right. One intersection for c) and two intersections for d).

 

[x.Exhaust.x]

Sorry for the confusion. But thanks for your answers Aerath, Lyounamu and lolokay .