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If two points (X₁, Y₁) and (X₂, Y₂) (i.) Distance √(x₂-x₂)² + (3₂-y)² (ii.) (ii.) (iv.) Нече So (ii.) (iii -) mid point Slope (m) = given distance equation of line & slope mid point = tano (x₁+x², 3, +1₂) у, туг) 2 2 = = y₂-y₁ P₁ = (√₂,5), √(√2+3)² + (2-5) ² -2-5 (√2-3, 5-2) -3-√2 7 3+√2 y = m(x-x₂) + y₂₁₂ P₂= (-3,-2) So (iv.) equation of line is = are given = -7 -(3+√2) then 2+9+ 6√2+49 y = 7 (√₂-3, ²-) AS As 7 0 = tan² (3+√₂) 3+√2 7 3+√2 = (x+3)-2 A 60 +6√2 te

If two points (X₁, Y₁) and (X₂, Y₂) (i.) Distance √(x₂-x₂)² + (3₂-y)² (ii.) (ii.) (iv.) Нече So (ii.) (iii -) mid point Slope (m) = given distance equation of line & slope mid point = tano (x₁+x², 3, +1₂) у, туг) 2 2 = = y₂-y₁ P₁ = (√₂,5), √(√2+3)² + (2-5) ² -2-5 (√2-3, 5-2) -3-√2 7 3+√2 y = m(x-x₂) + y₂₁₂ P₂= (-3,-2) So (iv.) equation of line is = are given = -7 -(3+√2) then 2+9+ 6√2+49 y = 7 (√₂-3, ²-) AS As 7 0 = tan² (3+√₂) 3+√2 7 3+√2 = (x+3)-2 A 60 +6√2 te

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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If two points (X₁, Y₁) and (x₂, Y₂₁) (i.) Distance = √√(x₂-x₁)² + (1₂-y₁) ² (५२-५,) 2 (15-) mid point (-) (iv.) Here So (ii) Slope (m) = given distance Н2-х, equation of line & (iii) slope = tano (x₁+x², y₁ + y₂) у, туг) 2 = = Уг-у, mid point = (√2-3, P₁ = (√₂,5), √(√2+3)² + (2-5)² -2-5 -3-√2 7 3+√2 y = = m(x-x4₂²) + y₂ So (iv.) equation of line is P₂= (-3,-2) are given = 5=2) (√2-3, 6-2)=(√2-3, 2) As 0= -7 -(3+√2) = then 2+9+ 6√2+49 y = 7 3+√2 = 7 3+√2 не 7 tan' (3+√₂) AS (x+3)-2 60+6√2
Transcribed Image Text:If two points (X₁, Y₁) and (x₂, Y₂₁) (i.) Distance = √√(x₂-x₁)² + (1₂-y₁) ² (५२-५,) 2 (15-) mid point (-) (iv.) Here So (ii) Slope (m) = given distance Н2-х, equation of line & (iii) slope = tano (x₁+x², y₁ + y₂) у, туг) 2 = = Уг-у, mid point = (√2-3, P₁ = (√₂,5), √(√2+3)² + (2-5)² -2-5 -3-√2 7 3+√2 y = = m(x-x4₂²) + y₂ So (iv.) equation of line is P₂= (-3,-2) are given = 5=2) (√2-3, 6-2)=(√2-3, 2) As 0= -7 -(3+√2) = then 2+9+ 6√2+49 y = 7 3+√2 = 7 3+√2 не 7 tan' (3+√₂) AS (x+3)-2 60+6√2
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