Question

find the equation of straight lines which are perpendicular to the line 3x+4y-6=0 and are at a distance of 4 units from(2,1)​

Answer 1

Answer:

Step-by-step explanation:

The equation of line perpendicular to 3x +4y -6 = 0 will be in format of 4x -3y +k = 0

Also,

The perpendicular line is at distance of 4 units from(2,1)​

$d = \frac{ \pm ax + by + c}{\sqrt{{a}^{2} + {b}^{2}}} \\ \\ 4 = \frac{ \pm 4 \times 2 - 3 \times 1 + k }{\sqrt{16 + 9}} \\ \\ 4 = \frac{ \pm 8 - 3 + k }{\sqrt{25}} \\ \\ 4 = \frac{ \pm 5 + k }{5} \\ \\ 20 = \pm (5+k) \\ \\ Case 1 : \\ 20 = 5 + k \\ \\ k = 15 \\ \\ Case 2 : \\ 20 = -5 - k \\ \\ k = -25 \\ \\ So, \: equation \: of \: perpendicular \: line \: can \: be \\ \\ \huge{\boxed{4x - 3y +15 = 0 \: or 4x - 3y -25 = 0}}$