You can use the graph given to draw two similar triangles and then use them to find the slope.
The description is shown below.
The Pythagoras theorem:
If ABC is a triangle with AC as the hypotenuse and angle B with 90 degrees then we have:
[tex]|AC|^2 = |AB|^2 + |BC|^2[/tex]
How to use similar triangles in this case to show that slope of AB is same as slope of AC?
Refer to the diagram attached below.
For triangle ADB, we have:
[tex]|AD|[/tex] = 44 units,(on graph) (from y axis)
[tex]|BD|[/tex] = 1 units,
And from Pythagoras theorem, and angle [tex]ACF[/tex] is right angle(of 90 degrees) and thus [tex]ACF[/tex] being right angled triangle.
[tex]|AB|[/tex] = [tex]\sqrt{|AD|^2 + |BD|^2} = \sqrt{44^2 + 1^2 } \approx 44.01\: \rm units[/tex] (positive root since [tex]|AC|[/tex] is length, thus, non negative)
For triangle ACF:
[tex]|FA| = 44 + 44 = 88 = 2 \times |DA|[/tex]
[tex]|CF| = 2 \: \rm units= 2 \times |BD|[/tex]
And from Pythagoras theorem, and [tex]\angle ACF[/tex] is right angle(of 90 degrees) and thus [tex]ACF[/tex] being right angled triangle.
[tex]|AC| = \sqrt{|FA|^2 + |CF|^2} = \sqrt{88^2 + 2^2 } \approx 88.02\: \rm units[/tex] = twice of [tex]|AB|[/tex] (positive root since [tex]|AC|[/tex] is length, thus, non negative) (its exactly two times if you don't simplify roots)
Thus, all sides of ACF is twice of each corresponding sides of ADB. Thus, by SSS similarity(which needs all sides of each other triangle scaled (multiplied) by a constant number), we have ΔADB ~ ΔAFC
Slope of hypotenuse of a right angled triangle is the ratio of the base and its height.
As base of AFC is twice of base of ADB and height of AFC is twice of height og ADB, thus that 2 as a factor will be eliminated from the numerator and denominator while calculating slope and thus, both triangle's hypotenuse (AC and AB) will come to have same slope.
Learn more about SSS similarity here:
https://brainly.com/question/12836908
