Determining third vertex of the right angled isosceles triangle

Question

If A(9,-9), B(1,-3) are the vertices of a right angled isosceles triangle, then the third vertex is?? Here in this question i got stuck to the point that which side is taken as the given coordinates. Whether it will be equal sides or hypotenuse??? If we consider both then how many will be the solution???

Answer 1

There will be two cases.

1. AB as hypotenuse $\rightarrow$ two positions for the third vertex.
2. AB as an equal side $\rightarrow$ now there will be four positions for the third vertex. i.e. Taking either A or B as the $90^{\circ}$ angle.

Answer 2

The points $ \ A \ (9 \ , \ -9) \ $ and $ \ B \ (1 \ , \ -3) \ $ represent one side of the right isosceles triangle(s) we are concerned with. Owing to the symmetry of such triangles, we only need to consider the ways in which we can have either a right angle or a $ \ 45º \ $ angle at either of these points. The slope of the line $ \ \overline{AB} \ $ is $ \ \frac{(-3) \ - \ (-9)}{1 \ - \ 9} \ = \ -\frac34 \ \ , \ $ so one pair of perpendicular lines [in green in the graph above] on which vertices we seek may be found are

$$ y \ - \ (-3) \ \ = \ \ \frac43·(x \ - \ 1) \ \ \rightarrow \ \ y \ \ = \ \ \frac43·x \ - \ \frac{13}{3} \ \ \ \text{[line } \ L_1] \ \ , $$

$$ y \ - \ (-9) \ \ = \ \ \frac43·(x \ - \ 9) \ \ \rightarrow \ \ y \ \ = \ \ \frac43·x \ - \ 21 \ \ \ \text{[line } \ L_2] \ \ . $$

We also need to find the slopes of lines that make $ \ 45º \ $ angles to line $ \ \overline{AB} \ \ , \ $ which we can determine by applying the "angle-addition formula for tangent". The tangent of the angle that $ \ \overline{AB} \ $ makes to the $ \ x-$axis is the slope $ \ -\frac34 \ \ , \ $ so the slopes of these additional lines are found from $$ \tan \left(\theta \ \pm \ 45º \right) \ \ = \ \ \frac{-\frac34 \ \pm \ \tan 45º}{1 \ - \ \left(-\frac34 \right) · (\pm \tan 45º)} \ \ = \ \ \frac{-\frac34 \ \pm \ 1}{1 \ - \ \left(-\frac34 \right) · (\pm 1)} $$ $$ = \ \ \frac{\frac14 }{\frac74} \ \ \text{or} \ \ \frac17 \ \ \ , \ \ \ \frac{-\frac74 }{\frac14} \ \ \text{or} \ \ -7 $$

(these are perpendicular slopes, as would be expected).

The equations of these "45º-angle" lines are then

$$ y \ - \ (-3) \ \ = \ \ \frac17·(x \ - \ 1) \ \ \rightarrow \ \ y \ \ = \ \ \frac17·x \ - \ \frac{22}{7} \ \ \ \text{[line } \ L_3] \ \ , $$

$$ y \ - \ (-9) \ \ = \ \ \frac17·(x \ - \ 9) \ \ \rightarrow \ \ y \ \ = \ \ \frac17·x \ - \ \frac{72}{7} \ \ \ \text{[line } \ L_4] \ \ . $$

[lines marked in orange]

$$ y \ - \ (-3) \ \ = \ \ -7·(x \ - \ 1) \ \ \rightarrow \ \ y \ \ = \ \ -7·x \ + \ 4 \ \ \ \text{[line } \ L_5] \ \ , $$

$$ y \ - \ (-9) \ \ = \ \ -7·(x \ - \ 9) \ \ \rightarrow \ \ y \ \ = \ \ -7·x \ + \ 54 \ \ \ \text{[line } \ L_6] \ \ . $$

[lines marked in violet]

We now construct the third vertex for all of the permitted right isosceles triangles. Two of these [in orange] have $ \ \overline{AB} \ $ as the hypotenuse of length $ \ \sqrt{ \ ( \ 9 - 1 \ )^2 \ + \ ( \ [-9] - [-3] \ )^2 } \ = \ \sqrt{ \ 8^2 \ + \ 6^2 } \ = \ 10 \ \ , \ $ so the lengths of their legs are $ \ 5 \sqrt2 \ \ . \ $ These vertices are the intersection points of lines $ \ L_3 \ $ and $ \ L_6 \ \ , \ $ $$ \frac17·x \ - \ \frac{22}{7} \ \ = \ \ -7·x \ + \ 54 \ \ \Rightarrow \ \ \frac{50}{7}·x \ \ = \ \ \frac{7·54 \ + \ 22}{7} \ \ = \ \ \frac{400}{7} $$ $$ \Rightarrow \ \ x \ \ = \ \ 8 \ \ , \ \ y \ \ = \ \ -2 \ \ , $$

and of lines $ \ L_4 \ $ and $ \ L_5 \ \ , \ $ $$ \frac17·x \ - \ \frac{72}{7} \ \ = \ \ -7·x \ + \ 4 \ \ \Rightarrow \ \ \frac{50}{7}·x \ \ = \ \ \frac{7·4 \ + \ 72}{7} \ \ = \ \ \frac{100}{7} $$ $$ \Rightarrow \ \ x \ \ = \ \ 2 \ \ , \ \ y \ \ = \ \ -10 \ \ . $$

(We can quickly verify that the distance-squared of these points from either $ \ A \ $ or $ \ B \ $ is $ \ 1^2 + 7^2 \ = \ [5 \sqrt2]^2 \ \ . \ ) $

There will be two other "third vertices" each on lines $ \ L_1 \ $ and $ \ L_2 \ \ , \ $ for the triangles which have $ \ \overline{AB} \ $ as one of their legs; the hypotenuses of these triangles thus have length $ \ 10 \sqrt2 \ \ . \ $ Lines $ \ L_4 \ $ and $ \ L_6 \ $ meet $ \ L_1 \ $ at

$$ \frac43·x \ - \ \frac{13}{3} \ \ = \ \ \frac17·x \ - \ \frac{72}{7} \ \ \Rightarrow \ \ \frac{7·4 \ - \ 3·1}{21}·x \ \ = \ \ \frac{7·13 \ - \ 3·72}{21} $$ $$ \Rightarrow \ \ 25·x \ \ = \ \ - 125 \ \ \Rightarrow \ \ x \ \ = \ \ -5 \ \ , \ \ y \ \ = \ \ -11 \ \ , $$

$$ \frac43·x \ - \ \frac{13}{3} \ \ = \ \ -7·x \ + \ 54 \ \ \Rightarrow \ \ \frac{25}{3}·x \ \ = \ \ \frac{3·54 \ + \ 13}{3} \ \ = \ \ \frac{175}{3} $$ $$ \Rightarrow \ \ x \ \ = \ \ 7 \ \ , \ \ y \ \ = \ \ 5 $$

[marked in green].

We could carry out similar computations to find the intersections of $ \ L_3 \ $ and $ \ L_5 \ $ with $ \ L_2 \ \ , \ $ but we may simply note that $ \ L_2 \ $ is parallel to $ \ L_1 \ $ and that these remaining vertices are just translations of the two we found on $ \ L_1 \ $ by the vector $ \ \overrightarrow{BA} \ = \ \langle \ 9 - 1 \ , \ (-9) - (-3) \ \rangle \ = \ \langle \ 8 \ , \ -6 \ \rangle \ \ . \ $ Hence, the vertices on $ \ L_2 \ $ [in violet] are $$ (-5 \ , \ -11) \ \rightarrow \ (3 \ , \ -17) \ \ \ , \ \ \ (7 \ , \ 5) \ \rightarrow \ (15 \ , \ -1) \ \ \ . $$

(As before, we can check that the distance-squared of these four vertices from $ \ A \ $ for the first pair and from $ \ B \ $ for the second is $ \ 2^2 + 14^2 \ = \ [10 \sqrt2]^2 \ \ . \ ) $

From the line slopes we found, we would have anticipated that all of these vertices would have rational coordinates; it is a pleasant surprise to find that the coordinates are in fact all integers.

Answer 3

To make a right isosceles triangle, $\space DB\space $ must be half of $\space AB\space $ and $\space BC=\sqrt{DB^2+DC^2} =\sqrt{50}\approx 7.071067812. \quad$

In the diagram below, this seems to place the vertex at $\space C(8,-2)\space$ as does the $DC$ delta of $(3,4)$.