Help with another stupid equation?

Question

Find the equation of the perpendicular to x+3y+1=0 and passes through P(1,2)

now what the hell has perpendicular got to do with algebra

Answer 1

a line that is perpindicular to another means that when they intersect, they will form two or four 90 degree angles (depending if its a cross or a T)

to find the perpindicular to a given equation, all you have to do is use the negative reciprocal of the slope:

x + 3y + 1 = 0
3y = -x - 1
y = -1/3x - 1/3
slope = -1/3
negative reciprocal of slope = 3/1

now just use that slope in the y intercept formula, plugging in the given coordinates for x and y to solve for b

y = mx + b
2 = 3(1) + b
-1 = b

now that you have the new b and the new slope, your equation should be:

y = 3x -1

Answer 2

first you need to solve the equation for y.
y=-(1/3)x - (1/3)

Perpendicular lines have slopes that are opposite reciprocals of each other. So the slope of the original equation is -(1/3), so the slope of the perpendicular line is 3. That means 3 is the coefficient of x in your new equation.

To make sure the line passes through (1,2) put those numbers into this equation for x and y, then solve for b
y=3x+b
2=3(1) +b
b=-1

So that means the new equation perpendicular to the original is y=3x-1

Answer 3

slope of the given line,let m1 = -(1/3)
{because slope of line= -(coefficient of x/coefficient of y)
now we know m1*m2= -1
{where m2 =slope of line perpendicular to it }
therefore m2=3/1
now equation of the line perpendicular to given line is
y-y1 =m2*(x-x1)
here y1=2 and x1=1
the point P(1,2) through which perpendicular line passes
equation of line is :
y-2=3*(x-1)
=>3x-y-1=0

Answer 4

There is nothing stupid about the equation. Are you referring to the person who has to solve it? Algebra can apply to any situation in which there are numerical relationships. Really confused now?