What is the solution to the system of equations shown on the graph?
1 answer:
You might be interested in
9514 1404 393
Answer:
3x + 2y = 0 or y = -3/2x
Step-by-step explanation:
The relationship between a line and its perpendicular is that their slopes are opposite reciprocals of each other. So, the thing you need to know about the line through the two given points is its slope. That is found using the slope formula:
m = (y2 -y1)/(x2 -x1) . . . . . . where the two give points are (x1, y1) and (x2, y2)
Putting your point coordinates in this formula gives ...
m = (5 -3)/(5 -2) = 2/3
Then the perpendicular line will have a slope that is the opposite reciprocal:
slope of perpendicular = -1/m = -1/(2/3) = -3/2
You have not said whether the perpendicular line must meet any other requirement, so the simplest one we can write is ...
y = mx + b . . . . . m is the slope and b is the y-intercept
y = -3/2x . . . . . . has a y-intercept of 0 (goes through the origin)
Multiplying by 2 and adding 3x gives ...
3x + 2y = 0 . . . . . standard form equation
_____
<em>Comment on the attachment</em>
The geometry program I used to create the attachment wrote the equation of the line through the given points as ...
2x -3y = -5 . . . . . standard form equation
The equation of a perpendicular line can be created from this form by swapping the x- and y-coefficients and negating one of them. Simply swapping the coefficients would give ...
-3x +2y = <something>
In standard form, we want the leading coefficient to be positive, so we can choose to negate the x-coefficient. The <something> can be zero to make the line go through the origin.
3x +2y = 0 . . . . . a perpendicular in standard form
__
If you need to have the line go through a particular point, you can determine the constant value (on the right of the = sign) by putting the point coordinates in for x and y.
_____
Method summary:
- find the slope of the given line
- determine its opposite reciprocal to find the slope of the perpendicular line
- use that new slope in an appropriate form to have the line go where you want it to go. (Point-Slope Form is often useful)
