The equation of the line passing through the point $(3, 4)$ and having slope $5$ is

5 x - y - 11 = 0

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5 x + y - 11 = 0

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5 x + y + 11 = 0

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5 x - y + 11 = 0

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Solution

The correct option is A $5 x - y - 11 = 0$
Take the equation in slope-intercept form which is $y = m x + c$
If it passes through $P (p, q)$ and $m = a$ , then the value of $c$ can be obtained by putting
$x = p, y = q$ and $m = a$
Given slope of the equation $= 5$ and passes through $(3, 4)$ .
The general equation of line is : $y = m x + c$
Thus. line is $y = 5 x + c$
It passes through $(3, 4)$ .
$4 = 5 (3) + c$
$c = 4 - 15 = - 11$
Thus, the line is $y = 5 x - 11$ .

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n

y^{''} (0)

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