Find the equation of a line having a slope of -2 and passes through th

1.	Find the equation of a line having a slope of -2 and passes through the intersection if 2x - y = 1 and x + 2y = 3 1. y - 2x + 1 = 0 2. y + 2x - 3 = 03. 2y - x - 3 = 04. 2y + x + 1 = 0
Answer» Correct Answer - Option 2 : y + 2x - 3 = 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 the 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 lines are: 2x - y = 1 ...(i) x + 2y = 3 ...(ii) Adding 2 × (i) to (ii) ⇒ 5x = 5 ⇒ x = 1 Putting value of x in (i) 2(1) - y = 1 ⇒ y = 1 The intersection is (1, 1) So line has the slope -2 and passes through (1, 1) ∴ Equation of the perpendicular line is (y - y1) = m (x - x1) ⇒ y - 1 = -2 (x - 1) ⇒ y + 2x - 3 = 0

Find the equation of a line having a slope of -2 and passes through the intersection if 2x - y = 1 and x + 2y = 3 1. y - 2x + 1 = 0 2. y + 2x - 3 = 03. 2y - x - 3 = 04. 2y + x + 1 = 0

Answer» Correct Answer - Option 2 : y + 2x - 3 = 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 lines are:

2x - y = 1 ...(i)

x + 2y = 3 ...(ii)

Adding 2 × (i) to (ii)

⇒ 5x = 5

⇒ x = 1

Putting value of x in (i)

2(1) - y = 1

⇒ y = 1

The intersection is (1, 1)

So line has the slope -2 and passes through (1, 1)

∴ Equation of the perpendicular line is

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

⇒ y - 1 = -2 (x - 1)

⇒ y + 2x - 3 = 0