Question

Pls answer this .. i will mark u brainliest pls pls pls

Answer 1

SOLUTION

Let the equation of a straight line lerpendicularto y -axis be 3x+4y=C.....(1)

Therefore,

Intercepted form of equation (1),

$\frac{x}{ \frac{c}{3} } + \frac{y}{ \frac{c}{4} } = 1$

Therefore,

$\frac{c}{3} + \frac{c}{4} = - 4 \\ = > \frac{4c + 3c}{12} = - 4 \\ = > 7c = - 48 \\ = > c = - \frac{ 48}{7}$

By substituting value of c in 1,

$3x + 4y = - \frac{48}{7} \\ = > 3x + 4y + \frac{48}{7} = 0 \\ = > 21x + 28y + 48 = 0$

Therefore,

Required equation is 3x+4y+48/7=0 or 21x+28y+48=0