Question

Write the equation of a line that is perpendicular to the line y=13x+4 and passes through the point (2,−5).

The equation of the perpendicular line in slope-intercept form is:

Answer 1

Answer:

$y = ( - 1 \div 13)x + ( - 63 \div 13)$

Step-by-step explanation:

Given line

$y = 13x + 4 \: \: \: \\ so \: \: \: \: slope \: of \: given \: line \: = 13$

Slope of perpendicular line is (-1÷13)

So equation of perpendicular line is

$y = ( - 1 \div 13)x + c$

Since line passes through point (2,-5). So

$- 5 = ( - 1 \div 13) \times 2 + c$

So,

$c = 63 \div 13$

So equation of line in slope intercept form is

$y = ( - 1 \div 13)x + ( - 63 \div 13)$