Problem: line through intersection point of 4x+5y+7=0, 3x-2y-12=0 and (3,1) [ Calculator, Method and examples ]
Solution:
Your problem `->` line through intersection point of 4x+5y+7=0, 3x-2y-12=0 and (3,1)
The point of intersection of the lines can be obtainted by solving the given equations
`4x+5y+7=0`
`:.4x+5y=-7`
and `3x-2y-12=0`
`:.3x-2y=12`
`4x+5y=-7 ->(1)`
`3x-2y=12 ->(2)`
equation`(1) xx 2 =>8x+10y=-14`
equation`(2) xx 5 =>15x-10y=60`
Adding `=>23x=46`
`=>x=46/23`
`=>x=2/1`
`=>x=2`
Putting `x=2` in equation `(2)`, we have
`3(2)-2y=12`
`=>-2y=12-6`
`=>-2y=6`
`=>y=-3/1`
`=>y=-3`
`:.x=2" and "y=-3`
`:. (2,-3)` is the intersection point of the given two lines.
The given points are `A(3,1),(2,-3)`
`:. x_1=3,y_1=1,x_2=2,y_2=-3`
Using two-points formula, The equation of a line A is
`(y-y_1)/(y_2-y_1)=(x-x_1)/(x_2-x_1)`
`:. (y-1)/(-3-1)=(x-3)/(2-3)`
`:. (y-1)/(-4)=(x-3)/(-1)`
`:. -1(y-1)=-4(x-3)`
`:. -y +1=-4x +12`
`:. 4x-y-11=0`
Hence, The equation of line is `4x-y-11=0`
|
|
|
Solution provided by AtoZmath.com
|
