# A few Maths questions

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#1
Hey everyone,
Please could someone help me with the following maths questions:

(1) The curve C has equation y=x^2(x-6) + 4/x
The points P and Q lie lie on C and have x-coordinates 1 and 2 respectively.

(a) Show that the length of PQ is √170.

These are the types of question that I always struggle on and I have no idea how to do it.

(b) Show that the tangents to C at P and Q are parallel.

I have a vague idea how to do this I think. Would you differentiate the equation and then find the gradients of the two points?

(2) The line l1 has equation y = 3x + 2 and the line l2 has equation 3x + 2y − 8 = 0 .

(a) Find the gradient of the line l2.
I know how to do this one, the answer is -1.5 (I have included this as I may need it to answer the question which I don't understand).

(b) The point of intersection of l1 and l2 is P.

Find the coordinate of P.

I also know the answer to this (the coordinates are (4/9,10/3) I think).

Now this is the question I don't get:

(c) The lines l1 and l2 cross the line y = 1 at the points A and B respectively.

Find the area of triangle ABP.

Again, I have no idea how to do this question.

Thanks.
0
10 years ago
#2
Right, one at a time here, 1a, if you have two points how would you go about determining the length of the lines that joins them? Both points have a x value and a y value remember, if one was (0,0) and another (3,4) would you know what to do?
0
#3
(Original post by Jodin)
Right, one at a time here, 1a, if you have two points how would you go about determining the length of the lines that joins them? Both points have a x value and a y value remember, if one was (0,0) and another (3,4) would you know what to do?
errr....Would you use Pythagoras's theorem or something? So would it be the square root of 3^2+4^2 = the square root of 25 = 5?
0
10 years ago
#4
Yes, that's correct, now consider if we had a point (1,0) and (4,4), if you plot these two points it becomes clear that in order to use pythagoras we need to consider how the distance between the x values is in fact only 3. In general if we have two points (x1,y1) and (x2,y2) then the difference in y values is y2 - y1 and the difference in x is x2 - x1 no?

So the distance of the line joining the two points must be ((y2 - y1)^2 + (x2 - x1))^0.5 no?
2
#5
(Original post by Jodin)
Yes, that's correct, now consider if we had a point (1,0) and (4,4), if you plot these two points it becomes clear that in order to use pythagoras we need to consider how the distance between the x values is in fact only 3. In general if we have two points (x1,y1) and (x2,y2) then the difference in y values is y2 - y1 and the difference in x is x2 - x1 no?

So the distance of the line joining the two points must be ((y2 - y1)^2 + (x2 - x1))^0.5 no?
Ah yes thanks a lot I've finally got it now + rep
0
10 years ago
#6
Right oh, part b then, you have exactly the right idea, follow it through and you should notice the derivatives are equal. Be back in just a bit to tackle the last problem, try drawing it all out, you will at least then be able to see the triangle you're trying to work out the area of.
0
#7
(Original post by Jodin)
Right oh, part b then, you have exactly the right idea, follow it through and you should notice the derivatives are equal. Be back in just a bit to tackle the last problem, try drawing it all out, you will at least then be able to see the triangle you're trying to work out the area of.
Thanks a lot again 0
10 years ago
#8
Excellent, now for the last one hopefully you've drawn the triangle out. If I were to suggest to you to use the fact that the area of a triangle is 0.5 * length of base * perpendicular height would you know what to do?
0
#9
(Original post by Jodin)
Excellent, now for the last one hopefully you've drawn the triangle out. If I were to suggest to you to use the fact that the area of a triangle is 0.5 * length of base * perpendicular height would you know what to do?
Thanks! I've got it now! I think it's 49/18. Thanks a lot for your help! I would rep again but I'm not allowed to unfortunately.
0
5 years ago
#10
(Original post by usycool1)
Hey everyone,
Please could someone help me with the following maths questions:

(1) The curve C has equation y=x^2(x-6) + 4/x
The points P and Q lie lie on C and have x-coordinates 1 and 2 respectively.

(a) Show that the length of PQ is √170.

These are the types of question that I always struggle on and I have no idea how to do it.

(b) Show that the tangents to C at P and Q are parallel.

I have a vague idea how to do this I think. Would you differentiate the equation and then find the gradients of the two points?

(2) The line l1 has equation y = 3x + 2 and the line l2 has equation 3x + 2y − 8 = 0 .

(a) Find the gradient of the line l2.
I know how to do this one, the answer is -1.5 (I have included this as I may need it to answer the question which I don't understand).

(b) The point of intersection of l1 and l2 is P.

Find the coordinate of P.

I also know the answer to this (the coordinates are (4/9,10/3) I think).

Now this is the question I don't get:

(c) The lines l1 and l2 cross the line y = 1 at the points A and B respectively.

Find the area of triangle ABP.

Again, I have no idea how to do this question.

Thanks.
We both struggled on this question! Luckily I managed to get there in the end!
0
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