Closest Point Ebe264

1. **State the problem:** We need to find which point on the line segment $JW$ is closest to the point $G(5,3)$. The points given are $J(0,0)$, $W(-5,-3)$, and $G(5,3)$. The options are points $(5,3)$, $(0,0)$, $(10,6)$, and $(-5,-3)$. 2. **Understand the line $JW$:** The line passes through $J(0,0)$ and $W(-5,-3)$. The vector from $J$ to $W$ is $$\vec{JW} = (-5 - 0, -3 - 0) = (-5, -3).$$ 3. **Find the projection of $G$ onto line $JW$:** The vector from $J$ to $G$ is $$\vec{JG} = (5 - 0, 3 - 0) = (5, 3).$$ 4. The projection scalar $t$ is given by $$t = \frac{\vec{JG} \cdot \vec{JW}}{\|\vec{JW}\|^2} = \frac{(5)(-5) + (3)(-3)}{(-5)^2 + (-3)^2} = \frac{-25 - 9}{25 + 9} = \frac{-34}{34} = -1.$$ 5. Since $t = -1$, the projection point lies on the line extended beyond $J$ opposite to $W$. The closest point on the segment $JW$ to $G$ is the endpoint $J(0,0)$ because $t < 0$ means the projection is outside the segment. 6. **Check distances to given points:** - Distance $G$ to $J(0,0)$: $$\sqrt{(5-0)^2 + (3-0)^2} = \sqrt{25 + 9} = \sqrt{34}.$$ - Distance $G$ to $W(-5,-3)$: $$\sqrt{(5+5)^2 + (3+3)^2} = \sqrt{10^2 + 6^2} = \sqrt{100 + 36} = \sqrt{136}.$$ - Distance $G$ to $(10,6)$: $$\sqrt{(5-10)^2 + (3-6)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}.$$ - Distance $G$ to $(5,3)$: $$\sqrt{(5-5)^2 + (3-3)^2} = 0.$$ 7. However, $(5,3)$ is not on the line $JW$ because it does not satisfy the parametric form of $JW$. The line $JW$ is parameterized as $$\vec{r}(t) = (0,0) + t(-5,-3) = (-5t, -3t).$$ For $(5,3)$, we would need $-5t = 5$ and $-3t = 3$, which gives $t = -1$ and $t = -1$ respectively, but $t = -1$ corresponds to $(-5(-1), -3(-1)) = (5,3)$ which is the point $(5,3)$, so it is on the line extended beyond $J$ opposite to $W$. 8. Since the segment $JW$ is between $t=0$ and $t=1$, the closest point on the segment is $J(0,0)$. **Final answer:** The point on $JW$ closest to $G$ is $(0,0)$.