The distance from (x1, y1) to (x2, y2) is found by applying the Pythagorean Theorem to the fundamental triangle.

• The distance between the points is the hypotenuse of the triangle.
• The legs of the triangle are x2 - x1 and y2 - y1.

The midpoint between (3, 7) and (17, 13) will have x coordinate halfway between 3 and 17 and y coordinate halfway between 7 and 13.

• The x coordinate will therefore be (3+17) / 2 and the y coordinate will be (7 + 13) / 2, as indicated below.

The midpoint can also be found using similar triangles as in the figure below.

• The x coordinate of the midpoint will be x1 + 1/2 (x2 - x1) = (x1 + x2) / 2, and the y coordinate will be y1 + 1/2 (y2 - y1) = (y1 + y2) / 2.

We now consider the problem of finding an equation for the set of points (x,y) that are equidistant from the points (3 , 10) and (8, 2).

• The distances are as indicated in red in the figure below.
• We set these distances equal and square both sides of the equation.

In the third line below we expand the squares.

• In the fourth line we see that the squared terms x^2 and y^2 disappear when subtracted from both sides, leaving us a linear equation.
• We solve the linear equation for y to obtain the slope-intercept form y = 5/8 x + 41/16.

We note that the slope between the two points is -8/5, while the slope of the straight line we just obtained was 5/8.

• These slopes are negative reciprocals of one another, showing us that the line we found is perpendicular to the segment between the points.
• Since the set of points equidistant from the two points is the perpendicular bisector of the segment between the points, this confirms at least the slope of our resulting equation.

We now find a formula for the condition that the sum of the distances from (-3, 0) and from (3,0) to point (x,y) is 7.

• The two distances d1 and d2 are as indicated in the figure below.

Setting the sum of these distances equal to 7, we obtain the equation in the second line below.

• We wish to express this equation without the square roots. We will therefore square both sides of the equation.
• Using the fact that (a + b) ^ 2 = a^2 + 2 a b + b^2, we square both sides of the equation.
• Somewhat inconveniently, we see that the 2 a b term still leaves us with square roots. So it might seem like we haven't made any progress toward getting rid of the square roots.
• However, if as in the second-to-last line below we isolate the square root factors on the left side, and then simplify the right side, we obtain an equation we can again square; this time the square roots will disappear, though we will get some fourth-power terms (which we hope will cancel out).

Taking care to square the trinomial on the right-hand side carefully, we obtain the equation in the second line below.

• We expand the expression on the left-hand side and note that all fourth-order terms are identical on both sides and hence can be subtracted out of the equation.
• We are left with the equation 196 y^2 + 52 x^2 = 637, which for reasons that will become apparent later we express in the form y^2 / b^2 + x^2 / a^2 = 1.