Find the equation of the straight line which passes through the point of intersection of the straight lines x + y = 6 and x + 2y - 8 = 0 and is parallel to the straight line joining the points (3, 4) and (5, 6)?

Option 4 : None of these

As per the given data,

x + y = 6            ….. (1)

x + 2y - 8 = 0     ….. (2)

From equation (1) and equation (2), we get

⇒ x = 4 and y = 2

∴ Point intersection of the lines x + y = 6 and x + 2y - 8 = 0 is (4, 2)

We know that,

Two point form of a line through the points (x₁, y₁) and (x₂, y₂) is (y - y₁) = × (x - x₁)

Equation of the line with points (3, 4) and (5, 6)

⇒ (y - 4) = × (x - 3)

⇒ (y - 4) = (x - 3)

⇒ x - y + 1 = 0

We know that,

Slope of line ax + by + c = 0 is –b/a

⇒ Slope of the line x - y + 1 = 0 is - (-1)/1 = 1

∴ Required line has a slope of 1 and passes through the point (4, 2)

We know that,

Point - slope form of a line through the point (x₁, y₁) and slope ‘m’ is (y - y₁) = m(x - x₁)

Line passing through (4, 2) and with slope 1 is

⇒ (y - 2) = 1 × (x - 4)

⇒ (y - 2) = x - 4

⇒ x - y - 2 = 0