Question 1025982
x+4y=12
4y=-x+12
y=(-x/4)+3
parallel to this and passing through (1,-2)
point slope formula with slope (-1/4)
y+2=(-1/4)(x-1)
y=(-x/4)-(7/4)
{{{graph(300,200,-10,10,-10,10,(-x/4)+3,(-x/4)-(7/4))}}} THIS IS PARALLEL
Perpendicular has slope 4, negative reciprocal.
passing through (0,1)  y-1=4x; y=4x+1.
{{{graph(300,200,-10,10,-10,10,(-x/4)+3,4x+1)}}}THIS IS PERPENDICULAR
X+4Y=12
2X-3Y=-9, rewriting
-2X-8Y=-24, multiplying top by -2.
-11Y=-33
Y=3
X=0 (0,3) IS THE INTERSECTION POINT.
From (-1,5) to L, the perpendicular distance must be along a line with slope 4
that line has an equation of y-5=4(x+1); y-5=4x+4 and y=4x+9
Find the intersection of that line and the original line
x+4y=12
-4x+y=9
4x+16y=48 multiplying the first by 4.
17y=57; y=57/17
x+4y=12
16x-4y=-36
17x=-24; x=-24/17
Do those points work in both lines?
-24/17 +228/17=204/17, which is 12.
y=4x+9 or 57/17=4(-24/17)+153/17.  This is (-96+153)/17=57/17.
Therefore, the intersection is (-24/17, 57/17) and we need to find the distance between that point at (-1.5)
That is the distance formula, with is the square root (sum of the square of the distances of x and y).
x distance is from -1 to -24/17, which is -7/17.  y distance is between 5 and 57/17, and that is (28/17).
Square those and add them, which is (49/289) and (784/289) or (833/289). The square root of that is sqrt(833)/17=28.86/17, or 1.698 or 1.7
{{{graph(300,200,-10,10,-10,10,(-x/4)+3, 4x+9)}}}