1. Find the equation of a line passing through the point of intersection of lines `x+y+1=0` and `3x+y-5=0` and point `A(1,-3)`

Solution:
The point of intersection of the lines can be obtainted by solving the given equations

`x+y+1=0`

`:.x+y=-1`

and `3x+y-5=0`

`:.3x+y=5`

`x+y=-1 ->(1)`

`3x+y=5 ->(2)`

Substracting `=>-2x=-6`

`=>2x=6`

`=>x=6/2`

`=>x=3`

Putting `x=3` in equation `(1)`, we have

`3+y=-1`

`=>y=-1-3`

`=>y=-4`

`:.x=3" and "y=-4`

`:. (3,-4)` is the intersection point of the given two lines.


The given points are `A(1,-3),(3,-4)`

`:. x_1=1,y_1=-3,x_2=3,y_2=-4`

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+3)/(-4+3)=(x-1)/(3-1)`

`:. (y+3)/(-1)=(x-1)/(2)`

`:. 2(y+3)=-1(x-1)`

`:. 2y +6=-x +1`

`:. x+2y+5=0`

Hence, The equation of line is `x+2y+5=0`

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2. Find the equation of a line passing through the point of intersection of lines `4x+5y+7=0` and `3x-2y-12=0` and point `A(3,1)`

Solution:
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`

Putting `x=2` in equation `(2)`, we have

`3(2)-2y=12`

`=>-2y=12-6`

`=>-2y=6`

`=>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`