Equation of a straight line passing through the point $(2, 3)$ and inclined at an angle of ${tan}^{- 1} \frac{1}{2}$ with the line $y + 2 x = 5$ , is:

y = 3

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x = 2

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3 x + 4 y - 18 = 0

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4 x + 3 y - 17 = 0

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Solution

The correct option is A $3 x + 4 y - 18 = 0$
when $θ$ is the angle between two line with slopes $m_{1}$ and $m_{2}$
$t a n θ = \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}$
Let the equation of the line be $y = m x + c$ , which is inclined at an angle
${tan}^{- 1} \frac{1}{2}$ with $y = 5 - 2 x$ .
Then,

$\frac{m - (- 2)}{1 + (- 2) m} = \frac{1}{2}$ [Where $(- 2)$ being the slope of the given line]
or, $2 m + 4 = 1 - 2 m$
or, $m = - \frac{3}{4}$ .
The required equation become $4 y + 3 x = c_{1}$ , which passes through

$(2, 3)$
So, $12 + 6 = c_{1}$ .
The required line is $3 x + 4 y = 18$ .

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Similar questions

(2, 3)

{tan}^{- 1} (\frac{1}{2})

y + 2 x = 5

Find the equation of the line passing through the points (2, 3) and the point of intersection of the lines $4 x - 3 y = 7$ and $3 x + 4 y + 1 = 0$ .

Find the equations to the straight lines passing through the point (2, 3) and inclined at an angle of $45^{\circ}$ to the line $3 x + y - 5 = 0$

x - y = 1

2 x - 3 y + 1 = 0

3 x + 4 y = 14

x + 2 y = 5

3 x + 7 y = 17

3 x + 4 y = 10

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