Question #ce20f

1 Answer
Dec 16, 2017

Draw a triangle diagram to be played around with, or in this case, simply average each coordiante to get (9.5, 16).

Explanation:

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):

enter image source here

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.

enter image source here

From there, we can see that point C has coordinates (2 + 15, 22 - 12) = (17, 10):

enter image source here

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.

enter image source here

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:

enter image source here

Similarly, a horizontal line from point D to the right, landing on point F on line BC:

enter image source here

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.

enter image source here

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.