1. given eqn,
y= -8x+5
8x+y-5=0
slope of given eqn (m1) = 8
slope of line perpendicular to given eqn (m) = -1/8 (because perpendicular lines have opposite and reciprocal slopes)
Then,
Eqn of line having slope -1/8 and passing through points (3,5) is given by,
let, x1 = 3, y1 = 5
(note that this is the point slope form)
(y-y1) = m(x-x1)
or, y-5 = -1/8(x-3)
or, 8y-40 = -x+3
or, x+8y-3-40 = 0
Therefore, x+8y-43 = 0 is the required equation.
2. given eqn,
8x+y-5=0
Slope of given eqn, (m1) = 8
Slope of line parallel to given line (m) = 8 (because parallel lines have the same slope)
Then,
Eqn of line having slope 8 and passing through points (3,5) is given by,
let, x1 = 3, y1 = 5
(y-y1) = m(x-x1)
or, y-5 =8(x-3)
or, y-5 = 8x-24
Therefore, 8x-y-19=0 is the required equation.
