Question

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

Answer 1

Hi there!

The equation of any line perpendicular to
$ax + by = c$
is
$bx - ay = c$

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
$9 \times 6 - 3 \times 4 = c$
$54 - 12 = c$
$c = 44$

Therefore, the equation of the line that is perpendicular to 3x + 9y = 7 and passes through (6, 4) is
$9x - 3y = 44$

Answer 2

Answer: The equation of the line is $3x-y=14.$

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

$3x+9y=7\\\\\Rightarrow 9y=-3x+7\\\\\Rightarrow y=-\dfrac{1}{3}x+\dfrac{7}{9}.$

So, slope will be given by

$m=-\dfrac{1}{3}.$

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

$m\times p=-1\\\\\Rightarrow -\dfrac{1}{3}\times p=-1\\\\\Rightarrow p=3.$

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]