Write the equation of a line that is perpendicular to the line x +3y = 16 and passes through the point (-3, -4). please help :)
Answers
3x-y= -5
Step-by-step explanation:
Step 1: Finding the slope of the perpendicular line
Given, x+3y=16 => y = (16/3) - (1/3)*x
Since the required line will be perpendicular to this line, if we consider the slope of the perpendicular line to be m, then
-(1/3)*m = -1 => m = 3
Therefore, the equation of the perpendicular line should be of the form y = mx + c where m is the slope and c is a constant
i.e. y = 3x + c
Step 2: Using the given point's co-ordinates to determine the value of c and complete the equation
Since, (-3,-4) passes through that perpendicular line, it should satisfy y = 3x + c.
Therefore, -4 = 3*(-3) + c => c = 5
Therefore the equation of the perpendicular line is y = 3x + 5 => 3x-y= -5
Step 3: Answer
The required equation of the perpendicular line is 3x-y = -5
General formula for such type of maths:
Suppose an equation, ax+by= c was given. We needed to find out the equation of line which is perpendicular to that line. The perpendicular passes through point (p,q).
The equation of that perpendicular line will be,
(b/a)*x- y = (b/a)*p- q
For example in this math, a = 1, b= 3, c = 16, p = -3 and q = -4
Therefore, the equation of perpendicular line is
(3/1)x - y = (3/1)*(-3) - (-4)
=> 3x-y = -9+4
=> 3x-y = -5