Question 255730
point (-1,2) to the line x-4y+1=0
step 1 - write equation as y = mx + b as
(i) {{{Y = (1/4)x  + 1/4}}}
step 2 - identify the slope. 
The slope,m, is 1/4
step 3 - find the perpendicular slope. This means the negative flip of the original slope. So,
1/4 - - > -4/1
step 4 - using our new slope find equation of line passing through (-1,2) using y = mx + b. We get
(ii) {{{y = mx + b}}}
(iii) {{{2 = (-4)*(-1) + b}}}
(iv) {{{-2 = b}}}
new equation is
(v) {{{y = -4x - 2}}}
step 5 - set equation (i) = (v) and solve for x to get
{{{(1/4)x + 1/4 = -4x - 2}}}
multiply by 4 to get
{{{x + 1 = -16x - 8}}}
{{{17x = -9}}}
{{{x = -9/17}}}
step 6 find y. We get Y = 2/17
which is the crossing point at (-9/17, 2/17)
step 7 find distance between (-1,2) and (-9/17, 2/17)
{{{d = sqrt((y2-y1)^2 + (x2-x1)^2)}}}
{{{d = sqrt((2/17-2)^2 + (-9/17-1)^2)}}}
{{{d = sqrt(3.543 + 2.3391)}}}
{{{d = 2.425}}}