Question

Find the value of $\frac{Q}{P}$ if (a, a), (-a, -a) and (Pa, Qa) are the vertices of an equilateral triangle

• A. 0
• B. -1
• C. 1
• D. 2

Answer 1

The correct option is B -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
Q=-P
Or, $\frac{Q}{P}=-1$