Question

If the equation $a(b-c)x^2+b(c-a)xy+c(a-b)y^2=0$ represents exactly one real line in $x-y$ plane, then the value of $\frac{ln(c+a)+ln(a-2b+c)}{ln(a-c)}$ (assuming all the logarithms are defined) is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

$a(b-c)x^2+b(c-a)xy+c(a-b)y^2=0$

It represents only one line on X-Y plane.

Therefore, $b^2({c-a)}^2-4ac(b-c\left)\right(a-b)=0\Rightarrow b^2({c+a)}^2-{4b}^{2}ac-4ac(ab-b^2+cb-ac)=0\Rightarrow b^2({c+a)}^2-4(ac\left)b\right(a+c)+{4a}^2{c}^2=0\Rightarrow {\left[b\right(c+a)-2ac]}^2=0\Rightarrow b(c+a)-2ac=0\Rightarrow 2b(c+a)-4ac=0\Rightarrow 2b(c+a)+{(c-a)}^2-(c+a{)}^2=0\Rightarrow {(c+a)}^2-2b(c+a)={(c-a)}^2\Rightarrow (c+a)(c+a-2b)=({a-c)}^2$

Apply $ln$ on both sides

$ln(c+a)+ln(c+a-2b)=2ln(a-c)\frac{ln(c+a)+ln(a-2b+c)}{ln(a-c)}=2$