Need help on part d. I got a-c. but how do you get d and why.

10. The line L1 passes through the point A(2, 5) and has gradient $\frac{-1}{2}$

a). Find an equation of L1, giving your answer in the form y = mx + c. (3)

The point B has coordinates (–2, 7).

b). Show that B lies on L1. (1)

c). Find the length of AB, giving your answer in the form k$\sqrt 5$, where k is an integer. (3)

The point C lies on L1 and has x-coordinate equal to p.

The length of AC is 5 units.

d). Show that p satisfies

$p^2-4p-16=0$

MARK SCHEME METHOD for d (that i don't understand)

x = p, so y = $\frac{-1}{2}$ p + 6.

AC2 = (2 – p)2 + ($\frac{-1}{2}$ + 6 – 5)2 = 52

25 = 4 – 4p + $p^2$ + (1 $\frac{-1}{2}$)2

25 = 4 – 4p + $p^2$ +1 $\frac{-1}{4}$ + p2

100 = 16 – 16p + 4$p^2$ + 4 – 4p + $p^2$

$5p^2-20p-80=0$

or $p^2-4p-16=0$ , as required.

10. The line L1 passes through the point A(2, 5) and has gradient $\frac{-1}{2}$

a). Find an equation of L1, giving your answer in the form y = mx + c. (3)

The point B has coordinates (–2, 7).

b). Show that B lies on L1. (1)

c). Find the length of AB, giving your answer in the form k$\sqrt 5$, where k is an integer. (3)

The point C lies on L1 and has x-coordinate equal to p.

The length of AC is 5 units.

d). Show that p satisfies

$p^2-4p-16=0$

MARK SCHEME METHOD for d (that i don't understand)

x = p, so y = $\frac{-1}{2}$ p + 6.

AC2 = (2 – p)2 + ($\frac{-1}{2}$ + 6 – 5)2 = 52

25 = 4 – 4p + $p^2$ + (1 $\frac{-1}{2}$)2

25 = 4 – 4p + $p^2$ +1 $\frac{-1}{4}$ + p2

100 = 16 – 16p + 4$p^2$ + 4 – 4p + $p^2$

$5p^2-20p-80=0$

or $p^2-4p-16=0$ , as required.

(edited 13 years ago)

Plug x = p into the equation of L1 to find the corresponding y value in terms of p.

Length of a line = $\sqrt{x^2+y^2}$.

Squaring both sides gives the somewhat easier equation: $x^2 + y^2 = l^2$, where l is the length.

Note that the equation is pythagoras, you're essentially finding the hypotenuse of the triangle with horizontal side 'x' and vertical side 'y'.

Length of a line = $\sqrt{x^2+y^2}$.

Squaring both sides gives the somewhat easier equation: $x^2 + y^2 = l^2$, where l is the length.

Note that the equation is pythagoras, you're essentially finding the hypotenuse of the triangle with horizontal side 'x' and vertical side 'y'.

(edited 13 years ago)

would it be possible to explain it using a diagram. this question is currently ruining my life

Original post by Spungo

Plug x = p into the equation of L1 to find the corresponding y value in terms of p.

Length of a line = $\sqrt{x^2+y^2}$.

Squaring both sides gives the somewhat easier equation: $x^2 + y^2 = l^2$, where l is the length.

Note that the equation is pythagoras, you're essentially finding the hypotenuse of the triangle with horizontal side 'x' and vertical side 'y'.

Length of a line = $\sqrt{x^2+y^2}$.

Squaring both sides gives the somewhat easier equation: $x^2 + y^2 = l^2$, where l is the length.

Note that the equation is pythagoras, you're essentially finding the hypotenuse of the triangle with horizontal side 'x' and vertical side 'y'.

Original post by The Smeezington

Need help on part d. I got a-c. but how do you get d and why.

10. The line L1 passes through the point A(2, 5) and has gradient – .

a). Find an equation of L1, giving your answer in the form y = mx + c. (3)

The point B has coordinates (–2, 7).

b). Show that B lies on L1. (1)

c). Find the length of AB, giving your answer in the form k?5, where k is an integer. (3)

The point C lies on L1 and has x-coordinate equal to p.

The length of AC is 5 units.

d). Show that p satisfies

p2 – 4p – 16 = 0.

10. The line L1 passes through the point A(2, 5) and has gradient – .

a). Find an equation of L1, giving your answer in the form y = mx + c. (3)

The point B has coordinates (–2, 7).

b). Show that B lies on L1. (1)

c). Find the length of AB, giving your answer in the form k?5, where k is an integer. (3)

The point C lies on L1 and has x-coordinate equal to p.

The length of AC is 5 units.

d). Show that p satisfies

p2 – 4p – 16 = 0.

Calculating from 'bottom up' the given gradient of the line - '-.' - should have been -0.5 =-1/2

L1: y=-1/2x+6

MARK SCHEME METHOD for d (that i don't understand)

x = p, so y = – p + 6.

AC2 = (2 – p)2 + (– p + 6 – 5)2 = 52

25 = 4 – 4p + p2 + (1 – p)2

25 = 4 – 4p + p2 +1 – p + p2

100 = 16 – 16p + 4p2 + 4 – 4p + p2

5p2 – 20p – 80 = 0

or p2 – 4p – 16 = 0, as required.

AT the first line: y=-1/2p+6

at the 2nd and 3rd line: there should to be (1-p/2)^2 term

from this you will get a (p^2)/4 term in the equation

then multiply by 4.

(edited 13 years ago)

Original post by ztibor

Original post by ztibor

Calculating from 'bottom up' the given gradient of the line - '-.' - should have been -0.5 =-1/2

L1: y=-1/2x+6

AT the first line: y=-1/2p+6

at the 2nd and 3rd line: there should to be (1-p/2)^2 term

from this you will get a (p^2)/4 term in the equation

then multiply by 4.

L1: y=-1/2x+6

AT the first line: y=-1/2p+6

at the 2nd and 3rd line: there should to be (1-p/2)^2 term

from this you will get a (p^2)/4 term in the equation

then multiply by 4.

not my thead but can I borrow you a second? because this questions lost me aswell!

I got to 4 + p^2 - 4p + 1 + (p^2)/4 = 25

and this lead me to 5p^2 - 16p - 80

where does the other -4p come from?

Original post by The Smeezington

would it be possible to explain it using a diagram. this question is currently ruining my life

Brace for bad paint skills.

Spoiler

a = x2 - x1

b = y2 - y1

Original post by emma363

not my thead but can I borrow you a second? because this questions lost me aswell!

I got to 4 + p^2 - 4p + 1 + (p^2)/4 = 25

and this lead me to 5p^2 - 16p - 80

where does the other -4p come from?

I got to 4 + p^2 - 4p + 1 + (p^2)/4 = 25

and this lead me to 5p^2 - 16p - 80

where does the other -4p come from?

You lost one term, because (1-p/2)^2=1+(p^2)/4-p

then multiply by 4

Where does the 52 come from though?

Original post by lunaticjordanmg

Where does the 52 come from though?

I WISH SOMEONE ANSWERED YOU :'))) 9 years ago damnn

Original post by lunaticjordanmg

Where does the 52 come from though?

IT MUST BE A TYPO OF 25 ACTUALLY (I know this is probably of no help 9 years later

- C1 January 2002?
- AQA GCSE Media Studies Paper 2 (8572/2) - 20th May 2024 [Exam Chat]
- Isaac Senior Physics Challenge 2024
- Maths for economists - constrained maximisation questoin
- Learning a 2nd language
- Does my crush like me?
- best way to revise maths
- Energy dissipated
- circle question help
- Isaac physics SPC 2022
- IB or A levels
- Edexcel old spec papers (maths)
- Let's Play Daily Squirdle Together!
- OCR GCSE English Literature Paper 1 (J352/01) - 17th May 2023 [Exam Chat]
- Learning a 2nd language
- WJEC Unit 3 Biology (5/6/2024)
- Senior Physics Challenge 2022
- Senior Physics Challenge Isaac Physics
- Can't enter UK as my country in enrolment?
- IGCSE Physics Focal Length

Last reply 4 days ago

STEP 2 in 2024: Sharing Your Story! [PLUS WITH SOME SOLUTIONS AND PREDICTION]Maths

80

Last reply 6 days ago

A level maths paper 2 (pure and statistics) and paper 3 (pure and mechanics) ocrLast reply 4 days ago

STEP 2 in 2024: Sharing Your Story! [PLUS WITH SOME SOLUTIONS AND PREDICTION]Maths

80

Last reply 6 days ago

A level maths paper 2 (pure and statistics) and paper 3 (pure and mechanics) ocr