Answer:
The equation in point slope form for the line perpendicular to y=3x+5 that passes through (4,-1) is y = -⅓x + ⅓
Step-by-step explanation:
Given
Let P represent the lind
Equation of P; y = 3x + 5
Let Q represent the other point.
Coordinates of Q; Q(4,-1)
It is said that the line P is perpendicular to point Q.
So, the first thing to do is to calculate the slope of Q;
Since both lines are perpendicular, then we make use of formula for calculating the condition of perpendicularity
This is given as m1m2 = -1
Where m1 = slope of P
m2 = slope of Q
To get m1;
m1 is the coefficient of x in equation of line P.
So, m1 = 3.
Now we can solve for m2
m1.m2 = -1. ---- make m2 the subject of formula
m2 = -1/m1
Substitute 3 for m1
m2 = -1/3
m2 = -⅓
Recall that the coordinates of Q is 4 and -1.
To calculate the equation of Q; we make use of the following
m = (y - y1)/(x - x1)
Where m = m2 = -⅓
x1 = 4 and y1 = -1
By substituton, we have
-⅓ = (y - (-1))/(x - 4)
-⅓ = (y + 1)/(x - 4) --- multiply both sides by 3(x - 4)
-⅓ * 3(x - 4) = 3(x - 4) * (y + 1)/(x - 4)
-(x - 4) = 3(y + 1)
-x + 4 = 3y + 3 --- make y the subject of formula
3y = -x + 4 - 3
3y = -x + 1 --- divide through by 3
3y/3 = (-x + 1)/3
y = -x/3 + 1/3
y = -⅓x + ⅓
Hence, the equation in point slope form for the line perpendicular to y=3x+5 that passes through (4,-1) is y = -⅓x + ⅓
