# C1 WTF moment type of question

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.
(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'.
(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'.
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.

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.

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

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?

You lost one term, because (1-p/2)^2=1+(p^2)/4-p
then multiply by 4
i happened to do the same question for my trials exam and i am wondering where this question was from?
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