Let u = hat{i} - 2hat{j} and v = -3hat{i} + 5 | vedclass

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(A) The line passing through points $u(1, -2)$ and $v(-3, 5)$ has the equation: $\frac{x - 1}{-3 - 1} = \frac{y - (-2)}{5 - (-2)} \Rightarrow \frac{x

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Only $P$ and $Q$

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(A) The line passing through points $u(1, -2)$ and $v(-3, 5)$ has the equation: $\frac{x - 1}{-3 - 1} = \frac{y - (-2)}{5 - (-2)} \Rightarrow \frac{x - 1}{-4} = \frac{y + 2}{7} \Rightarrow 7x - 7 = -4y - 8 \Rightarrow 7x + 4y + 1 = 0$.For point $P(-\frac{1}{7}, 0)$: $7(-\frac{1}{7}) + 4(0) + 1 = -1 + 0 + 1 = 0$. Thus,$P$ lies on the line.For point $Q(0, -\frac{1}{4})$: $7(0) + 4(-\frac{1}{4}) + 1 = 0 - 1 + 1 = 0$. Thus,$Q$ lies on the line.For point $R(-2, 3)$: $7(-2) + 4(3) + 1 = -14 + 12 + 1 = -1 
eq 0$. Thus,$R$ does not lie on the line.Therefore,only points $P$ and $Q$ lie on the line segment.

Let $u = \hat{i} - 2\hat{j}$ and $v = -3\hat{i} + 5\hat{j}$. Consider three points $P, Q,$ and $R$ having position vectors $-\frac{1}{7}\hat{i}$,$-\frac{1}{4}\hat{j}$,and $-2\hat{i} + 3\hat{j}$ respectively. Among these,the points on the line segment passing through $u$ and $v$ are

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