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Find the distance from point A to the given line.

1. A(−1, 7), y = 3x
and
2. A ( − 1 — 4 , 5 ), −x + 2y = 14

Sagot :

Answer:

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;

Where;

(x₁, y₁) = (-1/4, 5), and (x₂, y₂) = B(-1, 6.5)

Plugging in the values, we have;

Therefore, the distance from A(-1/4, 5) to the line -x + 2·y = 14 ≈ 1.7 units, rounding to the nearest tenth.