Question

What is perpendicular to y= 3x -9 and passes through the point (3,1)

Answer 1

To find the equation of a line that is perpendicular to y = 3x - 9 and goes through the point (3,1), we use the negative reciprocal of the original slope (3) which is -1/3 and apply the point-slope equation, resulting in y = (-1/3)x + 2 as the perpendicular line's equation.

Step-by-step explanation:

The question asks for the equation of a line that is perpendicular to the given line y = 3x - 9 and passes through the point (3,1). First, we identify the slope of the given line, which is 3. Since perpendicular lines have slopes that are negative reciprocals of each other, the slope of the line we're seeking is -1/3. Using the point-slope form of a line, which is y - y1 = m(x - x1), we plug in the slope and the point (3,1), resulting in y - 1 = (-1/3)(x - 3). Simplifying this, we get the equation of the line that is perpendicular, which is y = (-1/3)x + 2.

Answer 2

`(3,1) , \ \ y= 3x -9 \\ \\ The \ slope \ is :m _(1) =3 \\ \\ If \ m_(1) \ and \ m _(2) \ are \ the \ gradients \ of \ two \ perpendicular \\ \\ lines \ we \ have \ m _(1) \cdot m _(2) = -1 \\ \\3\cdot m_(2)=-1 \ \ /:3\\ \\ m_(2)=-(1)/(3)`


`\Now \ your \ equation \ of \ line \ passing \ through \ (3,1) would \ be: \\ \\ y=m_(2)x+b \\ \\1=-(1)/( 3) \cdot3 + b`


`1=-1+b\\ \\ b=1+1\\ \\b=2 \\ \\ y =- (1)/(3)x+2`