6.) The equation of base of an equilateral triangle is
x + y = 2 and vertex is (2,-1), then the length of
the side of the triangle is equal to
Answers
If the equation of the base of an equilateral triangle is x + y =2 and the vertex (2, −1), then the length of the side of the triangle is (in unit)?
Stabilize handhelp shots. Cute is tough to capture.
The altitude is perpendicular to the given side, through the given vertex. The perpendicular family to x+y=2 is gotten by swapping the coefficients on x and y , negating one of them, so is x−y=constant and the constant is given by the vertex,
x−y=2−−1=3
We intersect with
x+y=2
Adding,
2x=5
x=5/2
y=2−x=2−5/2=−1/2
So our altitude, between (52,−12) and (2,−1), has length
h:(5/2−2)2+(−1/2−−1)2−−−−−−−−−−−−−−−−−−−−−√=1/2–√
The sides of the 30/60/90 triangle are in ratio 1:3–√:2. That’s half an equilateral triangle i.e. 3–√ is the altitude when 2 is the side, so in our case
s=23–√12–√=26–√=6–√3
Answer: 6–√/3
Let’s try a different way. Slopes are tangents, so we can compute the slopes of the other two sides through the tangent sum angle formula. y=−x+2 has a slope of -1, tanA=−1. We add 60∘ to that, tanB=tan60∘=3–√
m=tan(A+B)=tanA+tanB1−tanAtanB=−1+3–√1+3–√=2−3–√
The other other slope is 2+3–√ but we won’t need it.
Now we have a point (p,q)=(2,−1) and a slope m=2−3–√ that intersects a line x+y=2 or generally ax+by=c. Let’s do the general case.
Find the squared distance d2 between (p,q) and (x,y), the intersection of the two lines ax+by=c and y−q=m(x−p).
y=m(x−p)+q
ax+b(mx+q−mp)=c
ax+bmx=c+bmp−bq
x=c+bmp−bqa+bm
We don’t need to solve for y=cm−amp+aqa+bm because
d2=(x−p)2+(y−q)2=(x−p)2+(m(x−p))2=(1+m2)(x−p)2
d2=(1+m2)(c+bmp−bq−ap−bmpa+bm)2
d2=(c−ap−bq)2(1+m2)(a+bm)2
Any chance this is right? Let’s try it. We have (p,q)=(2,−1) and m=2−3–√, and x+y=2 so a=1,b=1,c=2.
m2=(2−3–√)2=7−43–√
d2=(2−2−(−1))2(1+7−43–√)(3−3–√)2=8−43–√12−63–√=23
d=23−−√=6–√3✓
Can’t believe that checked!