Write an equation of the line that is perpendicular to 3x + 9y = 7 and passes through the point (6, 4).

Question

asked Nov 27, 2019 136k views

2 Answers

Tinisha · Answer 1 · 2019-11-28T17:30:10+0000

Hi there!

The equation of any line perpendicular to

ax + by = c

is

Therefore, in this situation, the formula of the line would be

9x - 3y = c

Given that the line passes through the point (6, 4) we get

9 * 6 - 3 * 4 = c

Therefore, the equation of the line that is perpendicular to 3x + 9y = 7 and passes through (6, 4) is

9x - 3y = 44

Stdcall · Answer 2 · 2019-12-02T12:26:59+0000

Answer: The equation of the line is

Step-by-step explanation: We are given to find the equation of the line that is perpendicular to 3x + 9y = 7 and passes through the point (6, 4).

The slope-intercept form of the given equation is

$3x+9y=7\\\\\Rightarrow 9y=-3x+7\\\\\Rightarrow y=-(1)/(3)x+(7)/(9).$

So, slope will be given by

m=-(1)/(3).

If 'p' represents the slope of the perpendicular line, then we must have

$m* p=-1\\\\\Rightarrow -(1)/(3)* p=-1\\\\\Rightarrow p=3.$

Therefore, the equation of the line with slope p = 3 and passing thjrogh the point (6, 4) is given by

$y-4=p(x-6)\\\\\Rightarrow y-4=3(x-6)\\\\\Rightarrow y-4=3x-18\\\\\Rightarrow 3x-y=14.$

Thus, the required equation of the line is [tex]3x-y=14.[/tex]