We'd like to know the steepness of the line so that we can visualize this geometrically without actually plotting points. Calculate the slope:
#y = (Deltay)/(Deltax) = (y_2 - y_1)/(x_2 - x_1) = (22 - 10)/(17 - 2) = 12/15 = 4/5#
Ah, so it's a positive but low steepness. Here's a sketch of what a line between those two points should look like (as a right triangle, with the right angle on a point #C#):
Ultimately, line #BC# is of length #y_2 - y_1 = 12# units, and line #AC# is of length #x_2 - x_1 = 15# units, both of which are from the calculation of the slope above.
From there, we can see that point #C# has coordinates #(2 + 15, 22 - 12) = (17, 10)#:
Yes, the picture does not exactly match the slope, but that's not the point. Here, what we're looking for is a point #D# on the hypotenuse (longest side) such that the distance (we'll call lowercase #d#) between point #A# and point #D# is equal to the distance between point #B# and point #D#.
So, what are the coordinates of point #D#? Let's draw a vertical line from point #D# downwards, landing on point #E# on line #AC#:
Similarly, a horizontal line from point #D# to the right, landing on point #F# on line #BC#:
This may seem useless, but this will help us find the coordinates of point #D# as you'll see. One thing that you could notice, is that these two smaller triangles formed, are congruent. Why? Have a look at their angles.
Since line #AE# is parallel to line #DF#, and they both touch line #AB#, then angle #EAB# is congruent to angle #FDB#. Similarly, line #ED# is parallel to line #FB#, and they both touch line #AB#, so angle #ADE# is congruent to angle #DBF#.
Finally, because the length of line #AD# is the same as that of line #DB#, which is #d#... by the ASA postulate (angle-side-angle), these two triangles are congruent!
And since these two triangles are congruent, all of their sides and angles are equal!
So #AC = AE + EC# where #AE = EC# so #AC = 2(AE) = 2(EC)#... and similarly #BC = BF + FC# where #BF = FC# so #BC = 2(BF) = 2(FC)#.
Here, to solve for the coordinates of point #D#, we have:
#D_x = A_x + AE#
#D_y = A_y + FC#
We already know the coordinates of point #A#, #(2, 10)#:
#D_x = 2 + AE#
#D_y = 10 + FC#
Then to solve for #AE#, we could use #AC = 2(AE)# and divide by #2# to get #AE = (AC)/2 = 15/2 = 7.5# units,
#D_x = 2 + 7.5 = 9.5#
#D_y = 10 + FC#
And then for #FC#, we have that #BC = 2(FC)# so #FC = (BC)/2 = 12/2 = 6# units.
#D_x = 2 + 7.5 = 9.5#
#D_y = 10 + 6 = 16#
So the coordinates of point #D# is #(9.5, 16)#, and that's our answer!
... Wait, isn't that the same as averaging the coordinates of the two points (point #A# at #(2, 10)# and point #B# at #(17, 22)#)?
#D_x = (A_x + B_x)/2 = (2 + 17)/2 = 19/2 = 9.5#
#D_y = (A_y + B_y)/2 = (10 + 22)/2 = 32/2 = 16#
OK, sure it is, but that rule only applies if the point happens to be exactly in the middle. What I have demonstrated coould be used and slightly tweaked (using AAA instead of ASA, showing that the smaller triangles are similar, and using a proportionality constant before getting to the answer) in cases where it isn't.
