In mathematics given that x=L2-L1/L1(O2-O1). 1.make L1 the subject formula
2.make O2the subject formula​

1. L1 =

2. O2 =

Step-by-step explanation:

Given: x =

1. To make L1 the subject of formula;

x =

cross multiply to have;

L2 - L1 = xL1(O2 - O1)

collect like terms,

L2 = xL1(O2 - O1) + L1

factorize the right hand side;

L2 = L1[x(O2 - O1)]

L1 =

2. To make O2 the subject of formula;

x =

cross multiply to have;

L2 - L1 = xL1(O2 - O1)

open the bracket to have;

L2 - L1 = xL1O2 - xL1O1

⇒ xL1O2 = L2 - L1 + xL1O1

O2 =

     =

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.