Line n contains points (2, 4) and (-4, -2). Find the distance between line n and point B (3, 1).
first find line m in the form y = m x + b
m = (-2-4)/(-4 -2) = 6/6 = 1
so
y = 1 x + b
4 = 2 + b
b = 2
so
y = x + 2
now we want a line perpendicular to that one through (3,1)
slope = -1/1 = -1
y = -x + b
1 = -3 + b
b = 4
y = -x + 4
now were does that perpendicular hit the first line?
y = -x + 4
y = x + 2
2 y = 6
y = 3
x = 1
so
we want distance from (1,3) to (3,1)
d = sqrt (-2^2 + 2^2) = sqt(4+4) = 2 sqrt 2
As shown above, the line is
y = x + 2
x-y+2 = 0
So, the distance from (3,1) to the line is
|3-1+2|/√(1+1) = 4/√2 = 2√2
Well, it seems like line n has already found a couple of points to hang out with. But now it's time to invite point B to the party! Let's find out how far it is from line n.
To calculate the distance between a line and a point, we can use a formula. Let's call the line n as AB, with A being (2, 4) and B being (-4, -2). And our point B is hanging out at (3, 1).
First, let's calculate the slope of line AB. We use the formula: m = (y2 - y1) / (x2 - x1) to calculate it. So, m = (-2 - 4) / (-4 - 2) = -6 / -6 = 1.
Now, we can use point-slope form to find the equation of the line:
y - y1 = m(x - x1), which becomes y - 4 = 1(x - 2), or simply y = x + 2.
Next, we need to find the distance between point B (3, 1) and the line y = x + 2. We'll use the formula: d = |Ax + By + C| / sqrt(A^2 + B^2), where A, B, and C are coefficients from the equation of the line.
Plugging in the values, we get d = |3 + 1 + 2| / sqrt(1^2 + 1^2), which simplifies to 6 / sqrt(2), or approximately 4.24 (rounded to two decimal places).
So, the distance between line n and point B is like 4.24 units. It's not too far, but not too close either. It's like a perfect distance to tell a good joke and break the ice at the party!
To find the distance between a line and a point, we can use the formula for the distance between a point and a line. The formula is:
distance = |Ax + By + C| / sqrt(A^2 + B^2)
where the line is represented by the equation Ax + By + C = 0.
First, let's find the equation of line n using the two given points (2, 4) and (-4, -2).
The slope (m) of the line can be found using the formula:
m = (y2 - y1) / (x2 - x1)
where (x1, y1) = (2, 4) and (x2, y2) = (-4, -2).
m = (-2 - 4) / (-4 - 2)
= -6 / -6
= 1
Using the point-slope form of a linear equation, we have:
y - y1 = m(x - x1)
Substituting (x1, y1) = (2, 4) and m = 1, we get:
y - 4 = 1(x - 2)
y - 4 = x - 2
y = x + 2
Therefore, the equation of line n is y = x + 2.
Next, we substitute the coordinates of point B (3, 1) into the equation of line n to find the perpendicular distance between line n and point B.
Substituting (x, y) = (3, 1) into y = x + 2, we have:
1 = 3 + 2
1 = 5
Since this equation is not true, point B does not lie on line n.
Therefore, the distance between line n and point B is the perpendicular distance from point B to line n.
To find the perpendicular distance, we can use the formula:
distance = |Ax + By + C| / sqrt(A^2 + B^2)
where (A, B) is the direction vector of line n, and (x, y) is the coordinates of point B.
The direction vector of line n is (1, 1). Therefore, A = 1 and B = 1.
Substituting A = 1, B = 1, x = 3, and y = 1 into the formula, we have:
distance = |1(3) + 1(1) + 2| / sqrt(1^2 + 1^2)
= |3 + 1 + 2| / sqrt(2)
= |6| / sqrt(2)
= 6 / sqrt(2)
= 6 / (√2)
Therefore, the distance between line n and point B is 6 / (√2), or approximately 4.24.
To find the distance between a line and a point, we can use the formula for the distance between a point and a line in a coordinate plane.
First, let's find the equation of line n using the given points (2, 4) and (-4, -2). We can use the formula for the equation of a line given two points:
Slope (m) = (y2 - y1) / (x2 - x1)
m = (-2 - 4) / (-4 - 2)
m = -6 / -6
m = 1
Using the point-slope form, we have:
y - y1 = m(x - x1)
y - 4 = 1(x - 2)
y - 4 = x - 2
y = x + 2
The equation of line n is y = x + 2.
Now, we can find the distance between the line and the point B (3, 1). To do this, we will use the formula for the distance between a point and a line:
Distance = |Ax + By + C| / sqrt(A^2 + B^2)
In this case, the equation of line n is y = x + 2, which can be rewritten as -x + y - 2 = 0.
So, A = -1, B = 1, and C = -2.
Plugging these values into the formula, we have:
Distance = |(-1)(3) + (1)(1) - 2| / sqrt((-1)^2 + (1)^2)
Distance = |-3 + 1 - 2| / sqrt(1 + 1)
Distance = |-4| / sqrt(2)
Distance = 4 / sqrt(2)
Distance = 4 / (√2)
Distance = 4√2 / 2
Distance = 2√2
Therefore, the distance between line n and point B (3, 1) is 2√2 units.