Use the point-slope formula to write an equation of the line through the points (0,3)(11,4)

Points : (0,3)(11,4)

Slope formula : M= y2-y1/x2-x1

M= 4-3/11-0

M= 1/11

Equation of a line is : y= mx+b

Now get one pair of point to solve for b .

(11,4)

4= 1/11(11)+b

4= 1+b

4-1=b

3=b

Equation y=mx +b

Y= 1/11x+3

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'.