The equations of two sides AB and AC of an isosceles triangle ABC are x + y = 5 and 7x - y = 3 respectively. Find the equations of the side BC is the area of the triangle is 5 sq. units?
1 Answer
Explanation:
This one looks fun; wonder why no one did it. It's probably easier with trigonometry but I'll avoid that.
The vague wording is annoying. We're told isosceles but not told definitively it's
This would be easier to visualize (and hopefully to solve) if one of the lines is parallel to an axis. Let's try it in a transformed space that's a 45 degree rotation:
Let's drop the square root of two and remember to scale the area appropriately.
Substituting, they meet at
Let's translate the vertex we know to the center:
Let's drop the primes till the end for sanity. Our transformed sides are:
That's easier to think about. We have one vertex at the origin O. We have one on the x axis, let's call it
The area is
We have three possibilities for isosceles triangles,
We knew that Pythagorean Triple was coming.
We already determined
That's about as a simple as we could hope for. This probably corresponds to the intended solution.
We can rule out
We can summarize our solutions as
Let's figure out how to map each
We see
The line PQ is
Now we write our six equations for our six
Check:
graph{0=(x+y -5)(7x-y -3)(x-3y - -1)(x-3y - -21)(x - 7y -( 4 sqrt(30) - 27))( x -7y -( - 4 sqrt(30) - 27))( 17 x -31 y - ( - 20 sqrt(30) - 107))( 17 x - 31y - ( 20 sqrt(30) - 107 )) [-8.045, 11.955, -1.12, 8.88]}
x+y=5, 7x-y=3, x-3y=-1, x-3y = -21, x - 7y = 4 sqrt(30) - 27, x -7y = - 4 sqrt(30) - 27, 17 x -31 y = - 20 sqrt(30) - 107, 17 x - 31y = 20 sqrt(30) - 107
Interesting intersections. Looks right.