Q88 / 180 · COMEDK 2023 (Evening Shift) · Mathematics
$\mathrm{P} \text { is a point on the line segment joining the points }(3,2,-1) \text { and }(6,2,-2) \text {. If the } x \text { co ordinate of } \mathrm{P} \text { is } 5 \text {, then its } \mathrm{y} \text { coordinate is }$
- A$-1$
- Correct answerB2
- C$-2$
- D1
Solution
Let the points be $A(3, 2, -1)$ and $B(6, 2, -2)$. Let $P(x, y, z)$ be a point on the line segment $AB$ such that $P$ divides $AB$ in the ratio $k:1$.
Using the section formula, the coordinates of $P$ are given by:
$x = \dfrac{k(6) + 1(3)}{k+1}$, $y = \dfrac{k(2) + 1(2)}{k+1}$, $z = \dfrac{k(-2) + 1(-1)}{k+1}$
Given that the $x$-coordinate of $P$ is $5$, we have:
$5 = \dfrac{6k + 3}{k+1}$
$5(k+1) = 6k + 3$
$5k + 5 = 6k + 3$
$k = 2$
Now, substitute $k = 2$ into the expression for the $y$-coordinate:
$y = \dfrac{2(2) + 1(2)}{2+1} = \dfrac{4 + 2}{3} = \dfrac{6}{3} = 2$
Answer: 2
Using the section formula, the coordinates of $P$ are given by:
$x = \dfrac{k(6) + 1(3)}{k+1}$, $y = \dfrac{k(2) + 1(2)}{k+1}$, $z = \dfrac{k(-2) + 1(-1)}{k+1}$
Given that the $x$-coordinate of $P$ is $5$, we have:
$5 = \dfrac{6k + 3}{k+1}$
$5(k+1) = 6k + 3$
$5k + 5 = 6k + 3$
$k = 2$
Now, substitute $k = 2$ into the expression for the $y$-coordinate:
$y = \dfrac{2(2) + 1(2)}{2+1} = \dfrac{4 + 2}{3} = \dfrac{6}{3} = 2$
Answer: 2
Hence, the correct answer is option: 'B'.