Answer:
The distance from A(-1/4, 5) to the line -x + 2·y = 14 is approximately 1.7 units
Step-by-step explanation:
The shortest distance from a point to a line is the perpendicular from the line to the point
The equation of the given line is -x + 2·y = 14, which can be written in slope and intercept form as follows;
-x + 2·y = 14
2·y = 14 + x
y = 14/2 + x/2 = x/2 + 7
y = x/2 + 7
The slope, m₁ = 1/2 and the y-intercept = 7
Therefore, the slope, m₂, of the perpendicular line to the line -x + 2·y = 14 which was rewritten as y = x/2 + 7 is m₂ = -1/m₁ = -1/(1/2) = -2
Therefore, we have that the slope of the line from A(-1/4, 5) to the line -x + 2·y = 14, is -2
The equation of the line is therefore;
y - 5 = -2(x - (-1/4))
y - 5 = -2·x - 1/2
y = -2·x - 1/2 + 5 = -2·x + 9/2
y = -2·x + 9/2
The coordinates of the point on the line -x + 2·y = 14 (y = x/2 + 7) that coincides with the perpendicular from the point A(-1/4, 5) is therefore given as follows;
-2·x + 9/2 = x/2 + 7
9/2 - 7= x/2 + 2·x
-2.5 = 2.5·x
x = -2.5/2.5 = -1
x = -1
y = x/2 + 7 = -1/2 + 7 = 6.5
y = 6.5
The coordinates of the point on the line -x + 2·y = 14 (y = x/2 + 7) that coincides with the perpendicular from the point A(-1/4, 5) is (-1, 6.5)
The distance from A(-1/4, 5) to B(-1, 6.5), is given as follows;
[tex]l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}[/tex]
Where;
(x₁, y₁) = (-1/4, 5), and (x₂, y₂) = B(-1, 6.5)
Plugging in the values, we have;
[tex]l_{AB} = \sqrt{\left (6.5-5 \right )^{2}+\left ((-1)-(-1/4) \right )^{2}} = \sqrt{\left (-0.75 \right )^{2}+\left (1.5 \right )^{2}} = \sqrt{\dfrac{45}{16} } = \dfrac{3}{4} \cdot \sqrt{5}[/tex]
[tex]l_{AB} = \dfrac{3}{4} \cdot \sqrt{5} \approx 1.677[/tex]
Therefore, the distance from A(-1/4, 5) to the line -x + 2·y = 14 ≈ 1.7 units, rounding to the nearest tenth.
