Find the equation of a line perpendicular to the line 4x + 2y - 11 = 0

1.	Find the equation of a line perpendicular to the line 4x + 2y - 11 = 0 and passes through a point (3, 4)1. 2x + y - 10 = 02. 2x - y - 2 = 03. 2y - x - 5 = 04. 2y + x - 11 = 0
Answer» Correct Answer - Option 3 : 2y - x - 5 = 0 Concept: The general equation of a line is y = mx + c where m is the slope and c is any constant Slope of parallel lines are equal. Slope of perpendicular line have their product = -1 Equation of a line with slope m and passing through (x₁, y₁) (y - y₁) = m (x - x₁) Equation of a line passing through (x1, y1) and (x2, y2) is: \(\rm {y-y_1\over x-x_1}={y_2-y_1\over x_2-x_1}\) Calculation: Given line 4x + 2y - 11 = 0 ⇒ y = -2x + 5.5 ⇒ Slope(m₁) = -2 and c₁ = 5.5 Now for the slope of perpendicular line (m₂) m₁ × m₂ = -1 ⇒ -2 × m₂ = -1 ⇒ m₂ = \(1\over2\) Perpendicular line has the slope \(1\over2\) and passes through (3, 4) ∴ Equation of the perpendicular line is (y - y1) = m (x - x1) ⇒ y - 4 = \(1\over2\) (x - 3) ⇒ 2y - x - 5 = 0

Find the equation of a line perpendicular to the line 4x + 2y - 11 = 0 and passes through a point (3, 4)1. 2x + y - 10 = 02. 2x - y - 2 = 03. 2y - x - 5 = 04. 2y + x - 11 = 0

Answer» Correct Answer - Option 3 : 2y - x - 5 = 0

Concept:

The general equation of a line is y = mx + c

where m is the slope and c is any constant

Equation of a line with slope m and passing through (x₁, y₁)

(y - y₁) = m (x - x₁)

Equation of a line passing through (x1, y1) and (x2, y2) is:

\(\rm {y-y_1\over x-x_1}={y_2-y_1\over x_2-x_1}\)

Calculation:

Given line 4x + 2y - 11 = 0

⇒ y = -2x + 5.5

⇒ Slope(m₁) = -2 and c₁ = 5.5

Now for the slope of perpendicular line (m₂)

m₁ × m₂ = -1

⇒ -2 × m₂ = -1

⇒ m₂ = \(1\over2\)

Perpendicular line has the slope \(1\over2\) and passes through (3, 4)

∴ Equation of the perpendicular line is

(y - y1) = m (x - x1)

⇒ y - 4 = \(1\over2\) (x - 3)

⇒ 2y - x - 5 = 0