Question

$A=(2,4,5)$ and $B=(3,5,-4)$ are two points. If the $xy$-plane, $yz$-plane divide $AB$ in the ratios $a:b,p:q$ respectively then $\frac{a}{b}+\frac{p}{q}$=

• A. $\frac{7}{15}$
• B. $\frac{-7}{12}$
• C. $\frac{7}{12}$
• D. $\frac{22}{25}$

Answer 1

The correct option is C $\frac{7}{12}$

$A(2,4,5),B(3,5,-4)$

xy plane $\Rightarrow z=0$

yz plane $\Rightarrow x=0$

line through $A$ & $B$ $\Rightarrow \frac{x-2}{1}=\frac{y-4}{1}=\frac{z-5}{-9}=t$

$\therefore$ let $P(t+2,t+4,-9t+5)$ be point of intersection of line with xy plane.

$\therefore P=(\frac{23}{9},\frac{41}{9},0)$

Similarly,for $Q(t+2,t+4,9t+5)$ be point of intersection of line with yz plane $\Rightarrow t=-2$.

$Q(0,2,23)$

(i)let $P$ divide $AB$ in $1:\lambda (a:b)$

$\therefore (\frac{23}{9},\frac{41}{9},0)=(\frac{2\lambda +3}{\lambda +1},\frac{4\lambda +5}{\lambda +1},\frac{5\lambda -4}{\lambda +1})\Rightarrow \lambda =\frac{4}{5}\therefore \frac{a}{b}=\frac{5}{4}$

(ii)let $Q$ divide $AB$ in $\lambda :1(p:q)$

$\therefore (0,2,23)=(\frac{2+3\lambda }{\lambda +1},\frac{4+5\lambda }{\lambda +1},\frac{5-4\lambda }{\lambda +1})\Rightarrow \lambda =\frac{-2}{3}\therefore \frac{p}{q}=\frac{-2}{3}\therefore \frac{a}{b}+\frac{p}{q}=\frac{5}{4}-\frac{2}{3}=\frac{7}{12}$