A line and three points are shown on the grid. The line on the grid contains the points (–1, 2) and (5, –2). The point is located at (–4, –4) on the grid. Find the coordinates of points three points that lay on a line, which passes through point P and is perpendicular to line c.
slope of line c = (-2-2)/(5+1) = -4/8 = -2/3
slope of perpendicular is then 3/2
so
y = 3/2 x + b
point (-4, -4)
-4= (3/2)(-4) + b
-4 = -6 + b
b = 2
so y = (3/2) x + 2
go ahead, find 3 points
try using desmos. Its an online graphing calculator that works wonders
Why did the line go to the therapist?
Because it had too many points of intersection and couldn't handle the pressure!
Now, back to your question. To find the coordinates of the points on a line that passes through point P (-4, -4) and is perpendicular to line c, we need to determine the slope of line c and then find the negative reciprocal of that slope to get the slope of the perpendicular line.
Using the formula for slope, m = (y2 - y1)/(x2 - x1), for points (x1, y1) = (-1, 2) and (x2, y2) = (5, -2):
m = (-2-2) / (5-(-1)) = -4 / 6
m = -2/3
Now, since the perpendicular line has a negative reciprocal slope to line c, the perpendicular line will have a slope of 3/2.
Using the point-slope form of a linear equation, we can write the equation of the perpendicular line. Let (x, y) be the coordinates of a point on the perpendicular line:
y - (-4) = (3/2)(x - (-4))
y + 4 = (3/2)(x + 4)
2y + 8 = 3x + 12
3x - 2y = -4
Now, we can choose any three values for x and solve for the corresponding y to get three points on the line:
Let's choose x = 0:
3(0) - 2y = -4
-2y = -4
y = 2
Point 1: (0, 2)
Let's choose x = 2:
3(2) - 2y = -4
6 - 2y = -4
-2y = -10
y = 5
Point 2: (2, 5)
Let's choose x = 4:
3(4) - 2y = -4
12 - 2y = -4
-2y = -16
y = 8
Point 3: (4, 8)
Therefore, the coordinates of three points that lie on a line passing through point P (-4, -4) and perpendicular to line c are (0, 2), (2, 5), and (4, 8).
To find the coordinates of the points that lie on a line passing through point P and are perpendicular to line c, we need to determine the equation of line c and then find the equation of the perpendicular line passing through point P.
First, let's find the equation of line c using the given points (–1, 2) and (5, –2). We can use the slope-intercept form of a linear equation, which is y = mx + b, where m is the slope of the line and b is the y-intercept.
To find the slope (m) of line c, we can use the slope formula:
m = (y2 - y1) / (x2 - x1)
In this case, (x1, y1) = (–1, 2) and (x2, y2) = (5, –2).
m = (-2 - 2) / (5 - (-1))
= (-4) / (5 + 1)
= -4 / 6
= -2/3
So, the slope of line c is -2/3.
Now, we can substitute the slope (-2/3) and one of the given points (e.g., (-1, 2)) into the slope-intercept form of the equation (y = mx + b) to find the value of b:
2 = (-2/3)(-1) + b
2 = 2/3 + b
2 - 2/3 = b
6/3 - 2/3 = b
4/3 = b
So, the value of b is 4/3.
The equation of line c in slope-intercept form is y = (-2/3)x + 4/3.
Now that we have the equation of line c, we can find the equation of the perpendicular line passing through point P (–4, –4). Since the two lines are perpendicular, the slopes of the lines are negative reciprocals of each other. The negative reciprocal of -2/3 is 3/2.
Using the point-slope form of a linear equation, which is y - y1 = m (x - x1), we can substitute the slope (3/2) and the coordinates of point P (–4, –4) into the equation to find the equation of the perpendicular line:
y - (-4) = (3/2)(x - (-4))
y + 4 = (3/2)(x + 4)
y + 4 = (3/2)x + 6
y = (3/2)x + 6 - 4
y = (3/2)x + 2
So, the equation of the perpendicular line passing through point P is y = (3/2)x + 2.
Now, to find the coordinates of the points lying on this perpendicular line, we can choose any x-coordinate and substitute it into the equation to find the corresponding y-coordinate.
For example, if we let x = 0, then we can find the y-coordinate:
y = (3/2)(0) + 2
y = 2
So, one point on the perpendicular line when x = 0 is (0, 2).
Similarly, we can find other points by substituting different x-coordinates into the equation.
For instance, if x = 2:
y = (3/2)(2) + 2
y = 3 + 2
y = 5
So, another point on the perpendicular line when x = 2 is (2, 5).
In this manner, you can select different x-values and find their corresponding y-values to determine the coordinates of other points lying on the perpendicular line passing through point P.