7= -x+3. consider the line y = find the equation of the line that is parallel to this line and passes through the point (-5, 6). find the equation of the line that is perpendicular to this line and passes through the point (-5, 6)

Answer with explanation:

The equation of line is, y= -x +3

→x+y-3=0(1)

⇒Equation of line Parallel to Line , ax +by +c=0 is given by, ax + by +K=0.

Equation of Line Parallel to Line 1 is

  x+y+k=0

The Line passes through , (-5,6).

→ -5+6+k=0

→ k+1=0

→k= -1

So, equation of Line Parallel to line 1 is

x+y-1=0

⇒Equation of line Perpendicular  to Line , ax +by +c=0 is given by, bx - a y +K=0.

Equation of Line Perpendicular to Line 1 is

  x-y+k=0

The Line passes through , (-5,6).

→ -5-6+k=0

→ k-11=0

→k= 11

So, equation of Line Parallel to line 1 is

x-y+11=0

first 15 divide by 2 = 7.5

100 divided by 7.5 is 13.

15 / 2 = 7.5

100 / 7.5 = 13.