Anonymous

using vector concepts : 4 + 5( a - 2 ) = 0 ===> a = 6 / 5

Anonymous

Line through A(3, 5) and B(7, 10): Slope: m₁ = (By - Ay)/(Bx - Ax) = (10 - 5)/(7 - 3) = 5/4. Line through C(0, 2) and D(1, a): Slope: m₂ = (Dy - Cy)/(Dx - Cx) = (a - 2)/(1 - 0) = a - 2. Two lines are perpendicular if the product of their slopes is equal to - 1. m₁ * m₂ = - 1. Hence: (5/4) * (a - 2) = -1 <=== Solve the equation for a. Multiply both sides by 4. 5 * (a - 2) = -4. 5a - 10 = -4 <=== Add 10 to both sides. 5a = 6 <=== Divide both sides by 5. ===> a = 6/5 (ANSWER).

Anonymous

The typical equation of a line is: y = mx + b → where m: slope and where b: y-intercept How to get the slope of the line that passes through A (3 ; 5) B (7 ; 10) ? m₁ = (yB - yA) / (xB - xA) = (10 - 5) / (7 - 3) = 5/4 ← this is the slope of the line (AB) How to get the slope of the line that passes through C (0 ; 2) D (1; a) ? m₂ = (yD - yC) / (xD - xC) = (a - 2) / (1 - 0) = a - 2 ← this is the slope of the line (CD) Two lines are perpendicular if the product of their slope is - 1. m₁ * m₂ = - 1 m₂ = - 1/m₁ a - 2 = - 1/(5/4) a - 2 = - 4/5 a = - (4/5) + 2 a = - (4/5) + (10/5) → a = 6/5

Anonymous

Slope of AB = (10-5)/(7-3) = 5/4 Slope of CD = (a-2)/(1-0) = a-2 AB and CD are perpendicular, so a-2 = -(5/4)⁻¹ = -⅘ a = 2-⅘ = 6/5

Anonymous

The equation of the line through (x₀,y₀), perpendicular to the line through (x₁,y₁) and (x₂,y₂) is (x₂-x₁)x+(y₂-y₁)y=(x₂-x₁)x₀+(y₂-y₁)y₀ So (7-3)×1 + (10-5)×a = (7-3)×0 + (10-5)×2, from which 4 + 5a = 10 and finally a = 6/5. To check, the line through AB is (10-5)x - (7-3)y = 10*3 - 7*5 which is 5x - 4y = -5, which is perpendicular to 4x + 5y = 4*0 + 5*2 (which passes through C) which simplifies to 4x + 5y = 10, and so 4*1 + 5*a = 10, from which 4 + 5a = 10 and finally a = 6/5.