# Equations of rectangle's lines

In right-triangle $$ABC$$ is known that it is isosceles, $$\hat A = 90^o$$, equation of one cathetus is $$y=2x$$, and the middle of hypotenuse is $$K(4,2)$$. The problem asks for other two remaining equations of lines of the rectangle.

I canot find some solving method, I tried with heights, and to use the isosceles property, but no result.

• Your question is about a right-triangle, not a rectangle. – uniquesolution Jul 10 at 21:40
• Hint: Assuming that $B$ lies on the given line, what do you know about $\triangle{ABK}$? – amd Jul 10 at 22:41
• $y=2x$. $y$ is the size of a cathetus. What is $x$? – the_candyman Jul 10 at 22:50
• @the_candyman That’s the equation of the line on which the cathetus lies. – amd Jul 10 at 23:01

Let's first ignore the isoceles condition and find all right triangles that satisfy the remaining conditions.

We have $$(4,2) = K$$. Let us denote line $$y=2x$$ as $$l$$

1. Find any two points $$A, B \in l$$ such that $$|AK|=|BK|$$. You can find them by drawing a line orthogonal to $$l$$ passing through $$K$$, and then choosing two points of $$l$$ that have equal distance from this line.
2. Draw a line going through $$B$$ and $$K$$. Draw a line orthogonal to $$l$$ passing through point $$A$$. Define point $$C$$ as the intersection of these two lines.

You can check that points $$A$$, $$B$$, $$C$$ defined in this way create a right triangle, with point $$K$$ being the middle of hypotenuse.

The formuale for $$A$$,$$B$$,$$C$$ are $$A = (1.6-t,3.2-2t)$$ $$B = (1.6+t,3.2+2t)$$ $$C = (6.4-t,0.8-2t)$$ for some $$t\in\mathbb R$$.

Since we want the triangle be an isosceles, we need $$|AB|=|AC|$$. Solving this equation gives us $$t=\pm1.2$$ and we have two solutions: $$A = (0.4,0.8),\qquad B = (2.8,5.6) \qquad C = (5.2,-1.6)$$ $$A = (2.8,5.6),\qquad B = (0.4,0.8) \qquad C = (7.6,3.2)$$

The second cathetus is perpendicular to the first (right triangle) and equidistant to $$K$$ (isosceles). The first constraint tells us that its equation is of the form $$x+2y=c$$, while using the point-line distance formula with the second generates the quadratic equation $$(c+8)^2 = 6^2,$$ so there are two solutions. The intersections of these two lines with $$y=2x$$ are two of the vertices of the triangle. The two possible hypotenuses are then the lines through these vertices and $$K$$. I’ll leave it to you to work out their equations. If you’re familiar with homogeneous coordinates, you can compute them directly with a few cross products.