# What is the equation of the line perpendicular to x + 5y + 3 = 0 and passing through (4, 8)?

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The equation x + 5y + 3 = 0 can be written as

=> 5y = -x – 3

=> y = (-1/5) x – 3/5

This is of the form y = mx + c, where m is the slope and c is the y-intercept.

The slope of x + 5y + 3 = 0 is -1/5. The slope of the line perpendicular to this is the inverse reciprocal or 5.

As the perpendicular passes through (4, 8), we have

y – 8 = 5 (x – 4)

=> y – 8 = 5x – 20

=> 5x – y – 12 = 0

**The required line is 5x – y – 12 = 0.**

To find the equation of the line perpendicular to x + 5y + 3 = 0 and passing through (4, 8)..

Any line perpendicular to ax+by+c = 0 and passing through a point (x1,y1) is given by:

**b(x-x1)-a(x-y1) = 0.**

So the equation of the line perpendicular to **x + 5y + 3 = 0** and passing through **(4,8)** is given by:

**5(x-4)-(y-8) = 0.**

5x-20-y+8 = 0

**Or 5x-y -12 = 0.**