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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

\frac{-1}{4}

+ p2

100 = 16 – 16p + 4

p^2

+ 4 – 4p +

p^2

5p^2-20p-80=0

p^2-4p-16=0

, as required.

(edited 14 years ago)

Reply 1

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'.

(edited 14 years ago)

Reply 2

The Smeezington

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'.

Reply 3

ztibor

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 14 years ago)

Reply 4

emma363

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? :s-smilie: