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 = 0
2. 2x – y – 2 = 0
3. 2y – x – 5 = 0
4. 2y + x – 11 = 0
1. 2x + y – 10 = 0
2. 2x – y – 2 = 0
3. 2y – x – 5 = 0
4. 2y + x – 11 = 0
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