Closest Point Ded971

1. **State the problem:** We need to find which point on the line segment $KL$ is closest to the point $M$. 2. **Identify points:** Given points are $K(-10,-1)$, $L(6,-5)$, and $M(2,-4)$. 3. **Formula for distance from a point to a line:** The shortest distance from a point $M(x_0,y_0)$ to a line passing through points $K(x_1,y_1)$ and $L(x_2,y_2)$ is given by: $$d = \frac{|(y_2 - y_1)x_0 - (x_2 - x_1)y_0 + x_2 y_1 - y_2 x_1|}{\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}}$$ 4. **Calculate components:** - $x_1 = -10$, $y_1 = -1$ - $x_2 = 6$, $y_2 = -5$ - $x_0 = 2$, $y_0 = -4$ Calculate numerator: $|(y_2 - y_1)x_0 - (x_2 - x_1)y_0 + x_2 y_1 - y_2 x_1|$ $= |(-5 + 1) \times 2 - (6 + 10) \times (-4) + 6 \times (-1) - (-5) \times (-10)|$ $= |(-4) \times 2 - 16 \times (-4) - 6 - 50|$ $= |-8 + 64 - 6 - 50|$ $= |0| = 0$ Calculate denominator: $$\sqrt{(-5 + 1)^2 + (6 + 10)^2} = \sqrt{(-4)^2 + 16^2} = \sqrt{16 + 256} = \sqrt{272}$$ 5. **Distance:** $$d = \frac{0}{\sqrt{272}} = 0$$ 6. **Interpretation:** Distance zero means point $M$ lies exactly on the line $KL$. 7. **Check which given point on $KL$ is closest to $M$:** Since $M$ lies on the line, the closest point on $KL$ to $M$ is the projection of $M$ onto $KL$. 8. **Check if $M$ lies between $K$ and $L$:** - Vector $KL = (6 - (-10), -5 - (-1)) = (16, -4)$ - Vector $KM = (2 - (-10), -4 - (-1)) = (12, -3)$ Calculate dot product: $$KL \cdot KM = 16 \times 12 + (-4) \times (-3) = 192 + 12 = 204$$ Calculate $|KL|^2$: $$16^2 + (-4)^2 = 256 + 16 = 272$$ Calculate projection scalar: $$t = \frac{KL \cdot KM}{|KL|^2} = \frac{204}{272} \approx 0.75$$ Since $0 < t < 1$, $M$ lies between $K$ and $L$ on the segment. 9. **Conclusion:** The shortest distance from $M$ to $KL$ is zero, and $M$ lies on the segment $KL$. Therefore, the closest point on $KL$ to $M$ is $M$ itself. 10. **Check given options:** The point $(2,-4)$ matches $M$. **Final answer:** The point on $KL$ closest to $M$ is $(2,-4)$.