1. ## MX1 Linear

Hey guys, really need some help with this question!

5F Q9 from Cambridge MX1

Ty!

cambridge.PNG

2. ## Re: MX1 Linear

Our specific example, given the lines 2x - 3y + 13 = 0 and x + y - 1 = 0, we know that the line that passes through their point of intersection must be given by:

a) (2x - 3y + 13) + k(x + y - 1) = 0

b) (2x - 3y + 13) + k(x + y - 1) = (2 + k)x - (3 - k)y + (13 - k) = 0 from which we can easily form an expression for the gradient as m = (2 + k)/(3 - k)

Since it must be parallel to: 4x+3y-1=0
then m = -4/3

So,
m = (2+k)/(3 - k)
-4/3 = (2 + k)/(3 - k)
k = 18

Sub k value into (2x - 3y + 13) + k(x + y - 1)=0

20x + 15y - 5 = 0

Simplify 20x + 15y - 5 = 0

20x + 15y - 5 = 0
4x + 3y - 1 = 0

Hence, l3

3. ## Re: MX1 Linear

Thank you so much!! That makes so much sense!

4. ## Re: MX1 Linear

Originally Posted by supR
Thank you so much!! That makes so much sense!
No problem. You're welcome!

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