The least distance from point P to line l is approximately equal to 3.536 units.
There are several ways to calculate the least distance, in this case we decided to use the approach based on vectors and projections. First, we generate two auxiliary vectors:
Vector parallel to the line l
[tex]\overrightarrow {AB} = (5, 1) - (2, 4)[/tex]
[tex]\overrightarrow{AB} = (3, - 3)[/tex]
A vector from one point on the line l to the point P
[tex]\overrightarrow{AP} = (1, 1) - (2, 4)[/tex]
[tex]\overrightarrow {AP} = (- 1, - 3)[/tex]
Now we proceed to find the vector distance by using the following vectorial formula:
[tex]\vec d = \frac{\overrightarrow{AP} \,\bullet\, \overrightarrow {AB}}{\overrightarrow {AB} \,\bullet \,\overrightarrow {AB}} \cdot \overrightarrow{AB} - \overrightarrow {AP}[/tex]
[tex]\vec d = \frac{(-1)\cdot (3) + (- 3) \cdot (- 3)}{3^{2}+ (-3)^{2}} \cdot (3, - 3) - (-1, - 3)[/tex]
[tex]\vec d = - \frac{1}{6}\cdot (3, - 3) - (- 1, - 3)[/tex]
[tex]\vec d = \left(-\frac{1}{2}, \frac{1}{2} \right) + (1, 3)[/tex]
[tex]\vec d = \left(\frac{1}{2}, \frac{7}{2} \right)[/tex]
The magnitude of the vector distance is determined by Pythagorean theorem:
[tex]d = \sqrt{\left(\frac{1}{2} \right)^{2}+\left(\frac{7}{2} \right)^{2}}[/tex]
d ≈ 3.536
The least distance from point P to line l is approximately equal to 3.536 units.
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