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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.

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    $\begingroup$ Your question is about a right-triangle, not a rectangle. $\endgroup$ – uniquesolution Jul 10 at 21:40
  • $\begingroup$ Hint: Assuming that $B$ lies on the given line, what do you know about $\triangle{ABK}$? $\endgroup$ – amd Jul 10 at 22:41
  • $\begingroup$ $y=2x$. $y$ is the size of a cathetus. What is $x$? $\endgroup$ – the_candyman Jul 10 at 22:50
  • $\begingroup$ @the_candyman That’s the equation of the line on which the cathetus lies. $\endgroup$ – amd Jul 10 at 23:01
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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: $$ A = (0.4,0.8),\qquad B = (2.8,5.6) \qquad C = (5.2,-1.6) $$ $$ A = (2.8,5.6),\qquad B = (0.4,0.8) \qquad C = (7.6,3.2) $$

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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.

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