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Distance Between Parallel Lines

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Distance Between Parallel Lines - Lesson Summary

Distance of a Point from a Line

Consider line L and point P in a coordinate plane.

The distance from point P to line L is equal to the length of perpendicular PM drawn from point P to line L. Let this distance be D.

Let line L be represented by the general equation of a line AX plus BY plus C is equal to zero.

Let the given line intersect the X- and the Y-axis at points Q and R, respectively.

The coordinates of point Q are obtained by solving the equation of the line and the equation of the X-axis, i.e. y=0.

On solving, the coordinates of point Q are (-C/A, 0).

The coordinates of point R are obtained by solving the equation of the line and the equation of the Y-axis, i.e. x=0.

On solving, the coordinates of point R are (0, -C/B).

Join PQ and PR to form triangle PQR.

Area of ∆ = ½ Base x Height

Area of ∆PQR = ½ QR x PM …… (1)

⇒ PM = (2 x Area of ΔPQR)/QR ….. (2)

If A (x1, y1), B (x2, y2) and C (x3, y3) form ∆ABC:

Area of ∆ABC = ½ ∣x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)∣

Given, P (x1, y1), Q (-C/A, 0) and R(0, -C/B)

Area of ∆PQR = ½ ∣x1{0 - (-C/B)} + (-C/A){(-C/B) - y1} + 0{y1 - 0}∣ 

⇒ Area of ∆PQR = ½ ∣x1(C/B) + (-C/A) (-C/B) + y1C/A + 0∣ 

= ½ ∣x1C/B + C2/AB + y1C/A∣

= ½ ∣C/AB∣ x ∣Ax1 + By1 + C∣ ……(3)

∴ 2 x Area of ∆PQR = ∣C/AB∣ x ∣Ax1 + By1 + C∣ ……(4)

Given, points A (x1, y1) and B (x2, y2)

Distance AB = √(x2 - x1)2 + (y2 - y1)2

Given, Q (-C/A, 0) and R (0, -C/B)

QR = √{ (-C/A-0)}2 + {0-(-C/B) }2

= √C2/A2 + C2/B2

= √{(C2/A2 B2 )(B2 + A2)

⇒ QR = ∣C/AB∣ √A2 + B2 …… (5)

⇒ PM = [∣C/AB∣ x ∣Ax1 + By1 + C∣]/[∣C/AB∣ √[A2 + B2]

⇒ PM = ∣Ax1 + By1 + C∣/√[A2 + B2]

⇒ Distance (d) of a point P (x1, y1) from a line Ax + By + C = 0 is:

d = ∣Ax1 + By1 + C∣/√[A2 + B2]

Distance between Two Parallel Lines

The distance between two parallel lines L1 and L2 in a coordinate plane:

If l1 ⃦ l2: ⇒ Slope of l1 = Slope of l2 = m.

Represent line L1 as y = mx + c1, and line L2 as y = mx + c2 in the slope-intercept form, respectively.

Let line L1 intersect the X-axis at point P.

Since point P lies on the X-axis, its y coordinate is equal to zero.

Substituting this value of y in the equation of line L1, the x-coordinate of point P is -c1/m.

Therefore, the coordinates of P are (-c1/m, 0).

The distance between lines L1 and L2 is equal to the length of the perpendicular drawn from point P to line L2. Let this distance be 'd'.

Distance (d) of a point P (x1, y1) from a line Ax + By + C = 0 is

d = ∣Ax1 + By1 + C∣/√A2 + B2]

y = mx + c2 ⇒ -mx + y - c2 = 0

Comparing -mx + y - c2 = 0 and Ax + By + C = 0

A = -m, B = 1 and C = -c2

Distance d between point P (-c1/m, 0) and line L2 is

d = |(-m)(-c1/m)+ 1 x 0 + (-c2)|/√((-m)2 + 12)

d = ∣c1 - c2∣/√(1 + m2) ……(1)

Distance between parallel lines y = mx + c1 and y = mx + c2 is

d = ∣c1 - c2∣/√(1 + m2)

Suppose the equation of the parallel lines L1 and L2 are given in general form.

Ax + By + C1 = 0 ⇒ y = (-A/B)x +(- C1/B) ...(2)

Ax + By + C2 = 0

y = (-A/B)x +(- C2/B) …..(3)

Comparing equations 2 and 3 with y = mx + c, we get

⇒ Slope (m) of lines L1 and L2 = -A/B

Also, c1 = -C1/B and c2 = -C2/B

Distance between parallel lines y = mx + c1 and y = mx + c2 is

d = ∣c1 - c2∣/√(1 + m2)

d = ∣(-C1/B ) - (-C2/B )∣/√(1 + (-A/B)2)

d = ∣(C2-C1)/B)∣/√(1 + A2/B2)

= ∣(C2-C1)/B ∣/√(B2 + A2)/B2)

= ∣(C2-C1)/B ∣/{√(B2 + A2)}B

d = ∣(C1-C2)∣/√ A2 + B2 ……(4)

Distance between parallel lines Ax + By + C1 and Ax + By + C2 is

d = ∣(C1-C2)∣/√ A2 + B2.

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