Question

If a ¹ p, b ¹ q, c ¹ r and $\left| \begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix} \right|$= 0, then $\frac{p}{p - a}$+ $\frac{q}{q - b}$+ $\frac{r}{r - c}$is equal to-

• A. 0
• B. 1
• C. –1
• D. 2

Answer 1

Answer: (D)

2

Answer 2

Apply R₁ – R₂, R₂ – R₃

D = $\left| \begin{matrix} p - a & b - q & 0 \\ 0 & q - b & c - r \\ a & b & r \end{matrix} \right|$= 0

= (p – a) [(q – b) r – b (c – r)] + a (b – q) (c – r) = 0

Dividing by (p – a) (q – b) (c – r), we get

$\frac{r}{c - r}$– $\frac{b}{q - b}$–$\frac{a}{p - a}$= 0 or $\frac{a}{p - a}$+$\frac{b}{q - b}$+$\frac{r}{r - c}$= 0

or $\frac{a}{p - a}$+ 1 +$\frac{b}{q - b}$+ 1 + $\frac{r}{r - c}$= 1 + 1

or $\frac{p}{p - a}$+ $\frac{q}{q - b}$+$\frac{r}{r - c}$= 2.