Question

For what value of $\frac{P}{Q}$ does A (a,a), B (-a,-a) and C (Pa,Qa) form the vertices of an equilateral triangle?

Answer 1

Let A = (-a,-a), B = (a,a) and C = (Pa,Qa)
Since ABC is an equilateral triangle
AB = BC
${AB}^2={BC}^2$
${(a-(-a\left)\right)}^2+{(a-(-a\left)\right)}^2={(Pa-a)}^2+{(Qa-a)}^2$
${\left(2a\right)}^2+{\left(2a\right)}^2=a^2(P^2-2P+1+Q^2-2Q+1)$
$8a^2=a^2(P^2+Q^2-2P-2Q+2)$
$8=(P^2+Q^2-2P-2Q+2)$
$P^2+Q^2-2(P+Q)=6$ ---------- (1)
AB = AC
${AB}^2={AC}^2$
${(a-(-a\left)\right)}^2+{(a-(-a\left)\right)}^2={(Pa-(-a\left)\right)}^2+{(Qa-(-a\left)\right)}^2$
${\left(2a\right)}^2+{\left(2a\right)}^2=a^2(P^2+2P+1+Q^2+2Q+1)$
$8a^2=a^2(P^2+Q^2+2P+2Q+2)$
$8=(P^2+Q^2+2P+2Q+2)$
$P^2+Q^2+2(P+Q)=6$ ---------- (2)
Subtracting (2) from (1), we get
4(P+Q)=0
P+Q=0
P=-Q
Or, $\frac{P}{Q}=-1$