Which is the value of the coefficient a if the lines are parallel? 3x+y-5=0 and ax-4y+7=0
3x + y -5 =0
The slope (m1) = -3
ax -4y +7 =0
The slope (m2)= a/4
Since both line are parallel, then :
m1 = m2
-3 = a/4
==> a= -12
Then the equation is:
-12x -4y + 7 =0
If the 2 given lines are parallel, then the values of their slopes are equal.
The given equations are 3x+y-5=0 and ax-4y+7=0, so we'll have to put them in the standard from, which is y = mx+n.
We'll start with 3x+y-5=0.
We'll isolate y to the left side:
y = -3x+5
So, the slope can be easily determined as m1 = -3.
That means that the second slope has the same value, m2 = -3.
We'll put the other equation into the standard form, isolating y to the left side:
-4y = -ax-7
We'll divide by -4:
y = (a/4)*x + 7/4
The slope is m2 = a/4
But m2 = -3, so a/4 = -3
a = -12
a1x+b1y+c1 and a2x+b2y+c2 are parallel , if a1/a2 = b1/b2.
Using the above principle for the given lines, 3x+y-5 = 0 and ax-4y+7 = 0,
3/a = 1/-4. Mutiply by 4a.
3*4 = -a. Or
a = -12