Find the equation of line parallel to the line 3x - 2y = 4 and passing

1.	Find the equation of line parallel to the line 3x - 2y = 4 and passing through (1, 5)1. 3x - 2y + 7 = 02. 3x + 2y - 13 = 03. 2x + 3y - 17 = 04. 2x - 3y + 12 = 0
Answer» Correct Answer - Option 1 : 3x - 2y + 7 = 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 is 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 3x - 2y = 4 ⇒ y = \(3\over2\)x + 2 ⇒ Slope(m1) = \(3\over2\) and c1 = 2 Now for the slope of parallel line (m2) m1 = m2 = \(3\over2\) Parallel line has the slope \(3\over2\) and passes through (1, 5) ∴ Equation of the parallel line is(y - y1) = m (x - x1) ⇒ y - 5 = \(3\over2\) (x - 1) ⇒ 3x - 2y + 7 = 0

Find the equation of line parallel to the line 3x - 2y = 4 and passing through (1, 5)1. 3x - 2y + 7 = 02. 3x + 2y - 13 = 03. 2x + 3y - 17 = 04. 2x - 3y + 12 = 0

Answer» Correct Answer - Option 1 : 3x - 2y + 7 = 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 3x - 2y = 4

⇒ y = \(3\over2\)x + 2

⇒ Slope(m1) = \(3\over2\) and c1 = 2

Now for the slope of parallel line (m2)

m1 = m2 = \(3\over2\)

Parallel line has the slope \(3\over2\) and passes through (1, 5)

∴ Equation of the parallel line is(y - y1) = m (x - x1)

⇒ y - 5 = \(3\over2\) (x - 1)

⇒ 3x - 2y + 7 = 0