Question

Find the equation of a line drawn perpendicular to the line $\frac{x}{4} +\frac{ y}{6} = 1$ through the point where it meets the y-axis

Answer 1

The equation of the given line is $\frac{ x}{4} +\frac{ y}{6} = 1$
This equation can also be written as $3x + 2y - 12 = 0$

$y =\frac{ -3}{2} x + 6$, which is of the form $y = mx + c$

∴ Slope of the given line $=-\frac{3}{2}$

∴ Slope of line perpendicular to the given line $=\frac{-1}{(\frac{-3}{2})} = \frac{2}{3}$
Let the given line intersect the y-axis at (0, y).
On substituting x with 0 in the equation of the given line, we obtain
$\frac{y}{6} = 1$
$⇒y = 6$

∴ The given line intersects the y-axis at (0, 6).
The equation of the line that has a slope of $\frac{2}{3}$ and passes through point (0, 6) is
$(y – 6) = \frac{2}{3} (x – 0)$

$3y – 18 = 2x$
$2x – 3y + 18 = 0$

Thus, the required equation of the line is $2x – 3y + 18 = 0$