Equation that passes through the point (5,1) and is perpendicular to the line x-3y+16=0

Answer 1

Our line is

x - 3y + 16 = 0.

We solve for y and find:

y = 1/3x + 16/3.

The slope of the line is 1/3.

The slopes of perpendicular lines are negative reciprocals. Let m is the unknown slope.

m (1/3) = -1.

m = -3.

We can use the point slope formula to find the equation:

y - y_o = m(x - x₀).

In our case, x₀=5 and y_o=1, and the slope is m=-3.

The equation:

y - 1 = -3(x - 5).

Answer 2

Here, y = x/3 + 16/3

slope of the line, m = 1/3

slope of the perpendicular line is the negative inverse of m.

-m^-1 = -3

New line

y = -3x + b

Plugging in the points (5, 1)

1 = -3 * 5 +b

b = 16

Therefore, new line y = = -3x + 16 (y = mx +b)