Find the equation of the line that passes through (5,-3) and perpendicular to 3x+2y=3
The equation for the line is:
y-y1= m(x-x1) where (x1,y1) is any point passes through the line and m is the slope>
==> y-(-3) = m(x-5)
==> y= m(x-5) -3
Since the line is perpendicular to 3x+2y=3, then the product of the slopes equals (-1)
The slope for 3x+2y =3 is (-3/2)
Then the slope for the line is (2/3)
Then the equation is:
y= (2/3)(x-5) -3
y= (2/3)x -10/3 -3
y= (2/3)x -19/3
Let's recall the fact that 2 line are perpendicular if the product of their slopes is -1.
For this reason, we'll put the given equation of the line into the standard form: y = mx + n.
We'll isolate the term in y to the left side:
2y = 3 - 3x
We'll divide by 2 both sides:
y = (-3/2)x + 3/2
The slope of this line is m1 = -3/2.
That means that the slope of the perpendicular line will be found from the relation:
m1*m2 = -1
m2 = -1/m1
m2 = -1/(-3/2)
m2 = 2/3
The equation of the line that passes through the point (5,-3) and is pepedicular to the line 3x+2y=3 is:
y - (-3) = m2*(x - 5)
y + 3 = (2/3)(x - 5)
y + 3 = (2/3)x - 10/3
We'll add -3 both sides:
y = (2/3)x - 10/3 - 3
y = (2/3)x - 19/3
We find the slope of the given line in the form y = mx + c to find its slope:
3x + 2y = 3
2y = - 3x + 3
y = (-3/2)x = 3
Therefore the slope of the given line is -3/2.
Slope of the a line perpendicular to another is inverse of the slope of first line, multiplied by -1
Therefore slope of perpendicular to given line = -1/(-3/2) = 2/3
Therefore the equation of the perpendicular may be given as:
y = (2/3)x + c
This perpendicular line passes through the point (5, -3). Therefore to get the value of c in its equation we substitute the values of the coordinates of the point in the equation.
- 3 = (2/3)5 + c
- 3 = 10/3 + c
c = - 3 - 10/3 = - 19/3
Substituting this value in equation of perpendicular:
y = (2/3)x - 19/3
Multiplying this equation by -3, and taking all the terms on left hand side we get:
2x - 3y - 19 = 0
An equation of a line perpendicular to a line ax+by+c= 0 is bx-ay+k = 0 where k is a constant to be determined by the condtion that it passes through aparticular point.
So 3x+2y = 3 or 3x+2y-3 = 0 has the perpendicular 2x-3y+k = 0.This passes through the point(5,-3).So (5,-3) should satisfy 32x-3y + k =0:
2(5)-3(-2)+k = 0.
10+6+k = 0
k = -16.
Therefore , the line 2x-3y = -16 or 2x-3y+16 is the line passing through the point(5 ,-2) and perpendicular to 3x+2y=3.