find the equation of the line containing the points (2, -3) and has a slope perpendicular to the line 3x-5y+6=0
If two lines are perpendicular, then their slopes are the negative reciprocals of each other.
If the equation of one line is Ax + By + C = 0, then a perpendicular line would look like Bx - Ay + K = 0
So your new line would be 5x + 3y + K = 0
sub in the given point (2,-3), solve for K and you are done.
This is the easiest way to solve this type of question.
whats 5+5
To find the equation of a line containing a given point and having a slope perpendicular to another line, you need to follow a few steps:
1. Determine the slope of the given line.
2. Find the negative reciprocal of the given slope to get the perpendicular slope.
3. Use the perpendicular slope and the given point to form the equation of the line using the point-slope form.
Let's go through each step in detail:
1. Determine the slope of the given line:
The given line is in the form Ax + By + C = 0. We can rewrite it in the slope-intercept form (y = mx + b) by solving for y:
3x - 5y + 6 = 0
-5y = -3x - 6
y = (3/5)x + 6/5
The slope of this line is 3/5.
2. Find the negative reciprocal of the given slope:
The negative reciprocal of 3/5 is -5/3. Flipping the fraction and changing the sign gives us the perpendicular slope.
3. Use the perpendicular slope and the given point to form the equation of the line using the point-slope form:
We have the point (2, -3) and the slope -5/3. The point-slope form of a line is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
Plugging in the values:
y - (-3) = (-5/3)(x - 2)
y + 3 = (-5/3)(x - 2)
Expanding and simplifying:
y + 3 = (-5/3)x + (10/3)
3y + 9 = -5x + 10
3y = -5x + 1
This is the equation of the line passing through the point (2, -3) with a slope perpendicular to the line 3x - 5y + 6 = 0.