Find the equation of the line which passes through the point of intersection of line x - y - 5 = 0 and x - 3y - 1 = 0 and is perpendicular to the line x - 3y + 21 = 0.?

Question

Find the equation of the line which passes through the point of intersection of line x - y - 5 = 0 and x - 3y - 1 = 0 and is perpendicular to the line x - 3y + 21 = 0.?

x + y + 8 = 0
3x + y - 23 = 0
3x + y = 0
None of these

LohaHridoy

5 views

2025-01-04 04:42

1 Answers

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AhmedNazir · Accepted Answer · 2022-09-04T04:29:57+06:00

Option 2 : 3x + y - 23 = 0

As per the given data, we get

x - y - 5 = 0 ….. (1)

x - 3y - 1 = 0 ….. (2)

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

⇒ x = 7 and y = 2

∴ Point of intersection of the lines x - y - 5 = 0 and x - 3y - 1 = 0 is (7, 2)

We know that,

Slope of line ax + by + c = 0 is –a/b

⇒ Slope of the line x - 3y + 21 = 0 is - 1/(-3) = 1/3

We know that,

Product of slopes of two perpendicular lines = -1

∴ Required slope of the line = -3

Hence, the required line passes through the point (7, 2) and has a slope of -3.

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 (7, 2) and with slope (-3) is

⇒ (y - y₁) = m (x - x₁)

⇒ (y - 2) = -3(x - 7)

⇒ y - 2 = -3x + 21

⇒ 3x + y - 23 = 0

∴ Required line is 3x + y - 23 = 0