# A few Maths questions

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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.

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.

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#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?

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(Original post by

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?

**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?

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#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?

So the distance of the line joining the two points must be ((y2 - y1)^2 + (x2 - x1))^0.5 no?

2

(Original post by

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?

**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?

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#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.

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(Original post by

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.

**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.

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#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?

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(Original post by

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?

**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?

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#10

(Original post by

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.

**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.

0

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