Given d1|| d2, calculate t? d1: x+y-1=0 d2:3x-ty-2=0
d1: x+y -1= 0
==> y= -x +1 ....(1)
d2: 3x -ty -2 =0
==> y= (3/t)x -2/t
Since d1 and d2 are parallel, then the slop of d1 (m1) = the slope of d2 (m2)
m2 = 3/t
But m1= m2
==> -1 = 3/t
==> t= -3
d1: x + y -1 =0
d2: 3x +3y -2 =0
The a1x+b1y+c1 =0 and a2x+b2y+c2 = 0 are || if,
a1/a2 = b1/b2
Here the || lines are x+y-1=0 and 3x-ty-2 = 0.
a1 = 1, b1 =1 . a2 = 3 and b2 = -t. Therefore,
1/3 = 1/(-t) . Multiply by 3t.
t = -3
If the 2 given lines are parallel, then the values of their slopes are equal.
The given equations are x+y-1=0 and 3x-ty-2=0, so we'll have to put them in the standard from, which is y = mx+n.
We'll start with x+y-1=0.
We'll isolate y to the left side:
y = -x+1
So, the slope can be easily determined as m1 = -1.
That means that the second slope has the same value, m2 = -1.
We'll put the other equation into the standard form, isolating y to the left side:
-ty = -3x + 2
We'll divide by -t:
y = 3x/t - 2/t
The slope m2 = 3/t
But m2 = -1, so 3/t = -1
t = -3
The line d2 is: 3x+3y-2=0