Find the distance from P and L
These have zero slope and unidentified slope and im not sure what to do after that..
Line L contains points (-2,1) and (4,1). Point P has coordinates (5,7)
Line L contains points (4,-1) and (4,9). Point P has coordinates (1,6)
These are very simple questions.
Good that you notice that in the first question the line is horizontal, namely it runs along at 1 unit above the x-axis. So isn't (5,7) just 6 above that line?
In the second, you have a vertical line. Tell me how far (1,6) is from that vertical line.
Making a diagram will make your answer very obvious.
So your x would be 1 and your y 6? And then i can do the distance formula uses those point and the points of p (5,7) to find the distance..?
Did you not follow my explanation for the first question?
Ok , for the 2nd , plot the two points (4,-1) and (4,9) and join them to form a line. Is that line not vertical?
now plot the (1,6) and draw a horizontal line from that point to your vertical line. I bet it hits it ast (4,6)
now count how many spaces from (1,6) to (4,6)
notice this will only work with horizontal and vertical lines.
You will probably learn shortly how to find the distance from a point (a,b) to a line with equation Ax + By + C = 0
To find the distance between Point P and Line L, you can use the formula for the distance between a point and a line.
First, we need to determine the equation of Line L.
For the first set of points on Line L: (-2,1) and (4,1), we can see that the y-coordinates are the same, indicating that Line L is a horizontal line. Since the y-coordinate is constant at 1, the equation of Line L becomes y = 1.
For the second set of points on Line L: (4,-1) and (4,9), we can also see that the x-coordinate is constant at 4, indicating that Line L is a vertical line. Since the x-coordinate is constant at 4, the equation of Line L becomes x = 4.
Now, we can find the distance between Point P and Line L using the formula:
distance = |Ax + By + C| / sqrt(A^2 + B^2),
where the equation of the line is Ax + By + C = 0.
For Line L:y = 1, the equation in the form of Ax + By + C = 0 is y - 1 = 0. Therefore, A = 0, B = 1, and C = -1. Plugging these values into the formula, we have:
distance = |0(5) + 1(7) - 1| / sqrt(0^2 + 1^2)
= |0 + 7 - 1| / sqrt(1)
= |6| / 1
= 6.
So the distance from Point P(5,7) to Line L y = 1 is 6 units.
For Line L:x = 4, the equation in the form of Ax + By + C = 0 is x - 4 = 0. Therefore, A = 1, B = 0, and C = -4. Plugging these values into the formula, we have:
distance = |1(1) + 0(6) - 4| / sqrt(1^2 + 0^2)
= |1 - 4| / sqrt(1)
= |-3| / 1
= 3.
So the distance from Point P(1,6) to Line L x = 4 is 3 units.