Answer to Question #139986 in Analytic Geometry for kys

Question

Determine the equation of the line perpendicular to 3y + 4x = 1 and passing through the point A (-1, 4)

Answer 1

$\bold {Answer}$

$4y -3x = 19$

$\bold {Solution}$

The the products of the gradients of perpendicular lines is $-1$ , that is, $mm' = -1$ where $m$ is the gradient of one line and $m'$ is the gradient of a line perpendicular to the first line.

The general form of a straight line equation is $y = mx + c$ . Reducing $3y +4x = 1$ to standard form gives:

$3y + 4x = 1$

$=> 3y = -4x + 1$

$=> \dfrac {3y}{3} = \dfrac {-4x}{3} + \dfrac {1}{3}$

$=> y = -\dfrac {4}{3}x + \dfrac {1}{3}$

Now, the gradient is $-\dfrac {4}{3}$

Since $mm' = -1$

$=> -\dfrac {4}{3}m' = -1$

$=> -4m' = -1 × 3$

$=> m' = \dfrac {-3}{-4}$

$= \dfrac {3}{4}$

Thus, the perpendicular line has gradient $\dfrac {3}{4}$

The equation of the perpendicular line through $\bold A$ is given by:

$y - y_{1} = m'(x-x_{1})$

$=> y - 4 = \dfrac {3}{4} (x - (-1))$

$=> y = \dfrac {3}{4}(x+1)+4$

$=>4y = 3(x+1) + 16$

$=> 4y = 3x + 3+16$

$=> 4y -3x = 19$