The lenght of a line segement is of 10units and the co-ordinate of one end point are (2,-3) if the abscissa of the other end is 10 find the ordinate of the other end??
Working plz plz
SORRY I DON'T KNOW
soory idk can u help me?
well, you know that
(10-2)^2 + (y+3)^2 = 10^2
So, now you can easily find y (the ordinate)
yyes ???????//
To find the ordinate of the other end of the line segment, we can use the distance formula. The formula for finding the distance between two points (x1, y1) and (x2, y2) is:
d = √((x2 - x1)^2 + (y2 - y1)^2)
In this case, we have one endpoint with coordinates (2, -3), and we know the length of the line segment is 10 units. We need to find the ordinate (y-coordinate) of the other endpoint when the abscissa (x-coordinate) is 10.
Let's assume the ordinate of the other endpoint is y. Using the distance formula, we have:
10 = √((10 - 2)^2 + (y - (-3))^2)
Simplifying the equation:
100 = (8^2 + (y + 3)^2)
100 = 64 + (y + 3)^2
(y + 3)^2 = 100 - 64
(y + 3)^2 = 36
Taking the square root of both sides:
y + 3 = ±√36
y + 3 = ±6
Now, we solve for y by subtracting 3 from both sides:
y = -3 ± 6
This gives us two possible values for y: -3 + 6 = 3 and -3 - 6 = -9. Therefore, the possible ordinates of the other endpoint are 3 and -9.