Let vec{u} = 2 hat{i} + hat{j} and vec{v} = 3 | vedclass

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(A) The position vectors of points $u$ and $v$ are given as $(2, 1)$ and $(3, -5)$ respectively. The equation of the line passing through points $(x_1

Vedclass · Accepted Answer

Only $P$ and $Q$

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(A) The position vectors of points $u$ and $v$ are given as $(2, 1)$ and $(3, -5)$ respectively. The equation of the line passing through points $(x_1, y_1) = (2, 1)$ and $(x_2, y_2) = (3, -5)$ is given by $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$. Substituting the values: $y - 1 = \frac{-5 - 1}{3 - 2}(x - 2) \Rightarrow y - 1 = -6(x - 2) \Rightarrow y - 1 = -6x + 12 \Rightarrow 6x + y = 13$. Now,we check which points satisfy the equation $6x + y = 13$: For $P\left(\frac{5}{2}, -2ight)$: $6\left(\frac{5}{2}ight) + (-2) = 15 - 2 = 13$. (Satisfies) For $Q\left(\frac{7}{3}, -1ight)$: $6\left(\frac{7}{3}ight) + (-1) = 14 - 1 = 13$. (Satisfies) For $R\left(\frac{9}{4}, 0ight)$: $6\left(\frac{9}{4}ight) + 0 = \frac{27}{2} = 13.5 
eq 13$. (Does not satisfy) Thus,only points $P$ and $Q$ lie on the line.

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