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Question
A line is drawn perpendicular to 5x = y + 7. Find the equation of the line if the area of the triangle formed by this line with co-ordinate axes is 10 sq.units
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Solution
Let the given line be PQ whose equation is
5x = y + 7
5x – y – 7 = 0 ......(1)
Let AB be the line perpendicular to the line PQ such that the area of the triangle OAB is 10 units.
The equation of the line AB is
– x – 5y + k = 0
x + 5y – k = 0
x + 5y = k ......(2)
`x/"l" + (5y)/"k"` = 1
`x/"k" + y/("k"/5)` = 1
∴ A is (k, 0) and B is `(0, "k"/5)`
OA = k and OB = `"k"/5`
Area of ∆OAB = `1/2 xx "OA" xx "OB"`
= `1/2 xx "k" xx "k"/5`
Given area of ∆OAB = 10
∴ `("k"^2)/10` = 10
k2 = 100
⇒ k = ±10
∴ The required equation of the straight line is
x + 5y = ±10
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