SOLUTION: Find the equation in standard form, of the line perpendicular to 3x

SOLUTION: Find the equation in standard form, of the line perpendicular to 3x - 6y = 9 and passing through (-2, -1)

Question 30569: Find the equation in standard form, of the line perpendicular to 3x - 6y = 9 and passing through (-2, -1)
Answer by sdmmadam@yahoo.com(530) (Show Source): You can put this solution on YOUR website!
Find the equation in standard form, of the line perpendicular
to 3x - 6y = 9 and passing through (-2, -1)
The given line is 3x - 6y = 9
(can divide through out by 3 and proceed)
That is 3x-6y-9 =0 ----(1) (putting the line in the general form)
Any line perpendicular to (1) is given by
6x +3y +k =0 ----(2)(please see the foot note for this consideration of (2) )
Given that P(-2, -1)is a point on (2)
Therefore putting x= -2 and y = -1 in (2)
6X(-2)+3X(-1)+k = 0
-12-3+k =0
-15+k = 0
Therefore k = 15 ----(*)
Putting k = 15 in (2)we get our line
6x+3y+15 =0
Dividing by 3
2x+y+5 = 0
Answer: The required line in the general form is 2x+y+5 = 0
Verification: We check if P(-2, -1)is a point on the answer line.
LHS = 2x+y+5 = 2X(-2)+(-1) + 5 = -4-1+5 =0 =RHS
Therefore our line is the required line
Note:(how do we consider (2) perpendicular to (1), looking at (1)?
Interchange the coefficients of x and y in (1) and (2) numerically and
change the sign of the coefficient of y in (2)(that is if it is negative in (1)
then positive in (2) and if positive in (1) then negative in (2))
Why should you do this ? How does it establish perpendicularity?
The principle is: when two lines are perpendicular,the product of their slopes is equal to (-1) That is if one line has slope = m, then a line perpendicular to it has slope =(-1/m)
In (1) we observe that the slope is (6/3) = 2
And therefore we should get slope of perpendicular line as (-1/2)
and hence the interchange of the coefficients(numerically)
and changing the sign of coefficient of y
How do you remember it in the form of a formula?
If (A)x+(B)y+C = 0 ----(1) is the given line
slope = (-A/B)----(*)
any line perpendicular to (1) is taken as
(Interchange the coefficients of x and y in (1) and (2) numerically and
change the sign of the coefficient of y in (2))
(B)x -(A)y +k = 0 ----(2)
slope = (B/A)-----(**)
From (*) and (**) the product of the slopes = (-A/B)X(B/A) = -1
Recall: that when two lines are parallel slopes are equal
and hence any line parallel to (1) should be of the form
(A)x+(B)y+k'= 0 ---(3)
slope= (-A/B)
The x and y-terms the same in both (1) and (3)

What more do you observe?
A very important observation is that (2) is not a single line but a family of lines with every member of the family perpendicular to (1). Giving different values to k we get different lines each perpendicular to (1)
Then how is it that we are asked to give the equation to a particular line perpendicular to (1)?
Every value of k determines a line perpendicular to (1)
Our required line is identified by the fact that P(-2, -1)is a point on (2) which determines that unique k responsible for giving our unique line.
Similarly different values to k' in (3)
would fetch different lines parallel to (1)

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