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Examples on Distance between two Parallel Lines

Note:

1. Actually

d = + ax₁+by₁+c/√a₂+b₂

2. To find the distance of a point from the given line, in the left side of the equation (right side being zero) substitute the co-ordinates of the point, and divide the result by √(coefficient of x)² + (coefficient of y)²

3. When to use complete perpendicular Distance formula? The complete perpendicular Distance formula is used when the length of the perpendicular to the line is given.

Illustration:

Find the distance of the point P(–2, 3) from the line AB which is
x – y = 5.

Solution:

The equation of the line is

x – y – 5 = 0 [Making right side zero (note this step)]

∴ Perpendicular Distance of the point (2, 3)

= (–2)–3–5/√(1)²+(–1)² = –10/√2

= –5√2

∴ Changing the sign, perpendicular Distance in magnitude = 5√2.

Enquiry: We can now find the distance of a point form a line but how can we determine as to which side of the line does the point lie?

From the figure, we observe that

ax₀ + by₀ + c = 0 (? point (x₀, y₀) lies on the line) …… (1)

Consider, ax₁+ by₁ + c

= (ax₀ + c) + by₁ [x₀ = x₁ = x₂]

= b(y1 – y0)

= –ve

Consider, ax₂ + by₂ + c

= (ax₀ + c) + by₂

= b(y₂ – y₀)

= +ve

Thus we observe that the point is on one side of the line, if put in the expression of line is gives one sign, while the point is on the other side of the line, if put in the expression of line it gives opposite sign.

Illustration:

Final the condition so that the points (x₁, y₁) and (x₂, y₂) lie on the same side, of the line ax + by + c =

Solution:

Since, (ax₁ + by₁ + c) and (ax₂ + by₂ + c)

Should be of the same sign.

∴ Their product should be positive i.e.

(ax₁+by₁+c)(ax₂+by₂+c) > 0, which is the required condition.

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