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
Jun 3, 2018

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

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 AB=AC. We have to consider both cases.

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:

(( x'),( y')) = ( (cos 45^circ, -sin 45^circ),(sin 45^circ, cos 45^circ) ) ((x),(y)) = 1/sqrt{2} ( (1, -1),(1, 1) ) ((x),(y))

Let's drop the square root of two and remember to scale the area appropriately.

x'=x-y

y'=x+y

x=(x'+y')/2 quad quad y=(y'-x')/2

x+y=5 becomes y'=5

7x-y=3 becomes 7 ((x'+y')/2) - (y'-x')/2 = 3

7x'+7y'-y'+x'=6

8x' + 6y' = 6

Substituting, they meet at A(-3,5) quad [this is in the transformed space; they meet at (1,4) in the original space.]

Let's translate the vertex we know to the center:

x''=x'+3

y''=y'-5

Let's drop the primes till the end for sanity. Our transformed sides are:

y=0 quad aka the x axis

4x+3y=0 quad a line through the origin#

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 P(p,0). We have one on the line 4x+3y=0 we can call Q(3q,-4q).

The area is |1/2(p(-4q)) |=|2pq|. We set it up so p and q have the same sign. We scaled by sqrt{2} so area scaled by 2, so 10 in the transformed space,

10 = 2 pq

pq = 5

We have three possibilities for isosceles triangles, OP=OQ, OP=PQ, OQ=PQ. Each of these has two solutions, second and fourth quadrant for Q. We'll work each out squared.

|OP|^2=|OQ|^2

p^2 = (3q)^2 + (4q)^2 = (5q)^2

We knew that Pythagorean Triple was coming.

p = pm 5q

We already determined p and q have the same sign.

p=5q

5=pq=5q^2

q=pm 1 quad p=5q=pm 5

That's about as a simple as we could hope for. This probably corresponds to the intended solution.

|OP|^2=|PQ|^2

p^2 = (p-3q)^2 + (4q)^2

(5/q)^2 = ((5/q)-3q)^2 + (4q)^2

25 = (5-3q^2)^2 + 16q^4

0 = -30q^2 + 25q^4

We can rule out q=0

q = pm sqrt(6/5), p = pm 5 \sqrt{5/6}

|OQ|^2=|PQ|^2

(5q)^2 = (p-3q)^2 + (4q)^2

(5q)^2 = (5/q-3q)^2 + (4q)^2

q = pm \sqrt{5/6}, p=pm 5 sqrt{6/5}

We can summarize our solutions as

q=pm 1, pm \sqrt{5/6}, pm \sqrt{6/5}

Let's figure out how to map each q back to an equation in the original space.

x''=x'+3 = x-y+3

y''=y'-5=x+y-5

x=(x'+y')/2 = (x''-3 + y''+5)/2 = (x''+y''+2)/2

y=(y'-x')/2 = (y''+5 - (x'' - 3))/2 = (y'' - x'' + 8)/2

We see A''(0,0) to A(1,4) so that worked

P''(p,0)=P''(5/q,0) to P( 5/{2q}+1, -5/{2q}+4 )

Q(3q,-4q) to Q(-q/2+1, -7/2 q + 4)

The line PQ is

(y -(-7/2 q + 4))(5/{2q} - (-q/2)) = (x - (-q/2+1))(-5/{2q} -(-7/2q))

(q^2 + 5) (2 y + 7 q - 8) = (7 q^2 - 5) (2 x + q - 2)

2 (7 q^2 - 5) x - 2 (q^2 + 5) y = (q^2 + 5)( 7 q - 8) -(7 q^2 - 5) (q-2)

Now we write our six equations for our six qs: pm 1, pm \sqrt{5/6}, pm \sqrt{6/5}

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

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.