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MX1 Linear (1 Viewer)
- Thread starter supR
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OkDen
Active Member
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
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
Last edited:
OkDen
Active Member
No problem. You're welcome!