The point $\mathrm{P}$ lies on the line $\mathrm{A} \mathrm{B}$, where $\mathrm{A} \equiv(2,4,5)$ and…
The point $\mathrm{P}$ lies on the line $\mathrm{A} \mathrm{B}$, where $\mathrm{A} \equiv(2,4,5)$ and $\mathrm{B} \equiv(1,2,3)$. If $\mathrm{z}$ co-ordinate
of point $\mathrm{P}$ is 3, the its y co-ordinate is
2
$-2$
$-3$
3
Solution
Equation of line passing through $\mathrm{A}$ and $\mathrm{B}$ is
$\frac{x-2}{1-2}=\frac{y-4}{2-4}=\frac{z-5}{3-5}=k \quad \text {...(say) } \Rightarrow \frac{x-2}{-1}=\frac{y-4}{-2}=\frac{z-5}{-2}=k$
Hence coordinates of any point onthis line are
$(-k+2,-2 k+4,-2 k+5)$
As per condition given, we have
$-2 k+5=3 \Rightarrow k=1$
Hence $y$ coordinate $=-2+4=2$