(2,4)…are cordinates (x,y)

so put x=2 and y=4

now u have

4=2a+b—-> equation 1

if y=5x+1 and y=ax+b [or 4=2a+b] are perpendicular then

gradient of eq1 and gradient of y=5x+1 =-1

gradient of y=5x+1 is 5 bcz y=mx+c

n m=5 i.e. gradient =5

gradient of first eq. is a

ask me how!

since every line has eq. y=mx+c

and in this case line is y=ax+b

m is gradient so….m=a

so….we have

gradient 1 multiple gradient 2 =-1

5xa=-1

so a is -1/5

now lets find b

in eq 1. put value of a

4=2a+b

4=-2/5 +b

4+ 2/5=b

b=22/5

b=4 2/5

Since the slope of y = 5x + 1 is 5, the slope of y = ax + b will be -1/5.

Using point-slope form: y – y1 = m(x – x1), we get

y – 4 = -1/5(x – 2)

y = -1/5 * x + 4 + 2/5

y = -0.2x + 4.4

a = -0.2 = -1/5

b = 4.4 = 4 and 2/5 = 22/5

I hope this helps!

thus: y=1/5x+b

since the line passes through (2,4), plug those in to the equation for x and y,

4=1/5(2)+b

b=4-1/5(2)

b=3.6

a=0.2