Notes On Distance Between Parallel Lines - CBSE Class 11 Maths
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.

#### 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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