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Question 566004: Find the equation of the line through (2,1) and the point of intersection of 3x-5y-10=0 and x+y+1=0.
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! the point of intersection of 3x-5y-10=0 and x+y+1=0.
the point of intersection can be found by solving the linear system
3 x -5 y = 10 .............1
1 x + 1 y = -1 .............2
Eliminate y
multiply (1)by 1
Multiply (2) by 5
3 x -5 y = 10
5 x 5 y = -5
Add the two equations
8 x = 5
/ 8
x = 5/8
plug value of x in (1)
3 x -5 y = 10
15/8 -5 y = 10
-5 y = 65/8 -5 y = 65/8
y = 65/40=13/8
The coordinates of intersection are (5/8 , 13/8)
The other point is (2,1)
x1 y1 x2 y2
5/ 8 13/8 2 1
slope m = (y2-y1)/(x2-x1)
( 1 - 13/8 )/( 2 - 5/8
( -5/8 / 11/8 )
m= - 5/11
Plug value of the slope and point ( 5/8 , 13/8 ) in
Y = m x + b
13/8 = - 2/7 + b
b= 13/8 + 2/7
b= 21/11
So the equation will be
Y = - 4/9 x + 17/9
m.ananth@hotmail.ca
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