Question

which line is perpendicular to the line passing through the points (-4,2) and (4,-4)
a. 3x+4y=16
b. 3x-4y=4
c. 4x+3y=15
d. 4x-3y=9
e. 3x+4y=-4

Answer 1

To find c, put the values of x, y and m as 3, 4, -1.33. Thus 4 = -1.33*3 + c, or 4 = -4+c, or c = 8. Hence the equation of the line is 4x+3y=8 and that will be parallel to 4x+3y=12.

Answer 2

Given,

Two points $\[\left( { - 4,2} \right),\left( {4, - 4} \right)\]$

Solution,

Consider two points are $\[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)\]$.

The slope of line passing through these points will be,

$\[{m_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]$

The slope of the perpendicular line will be,

$\[ \Rightarrow {m_2} = \frac{{ - 1}}{{{m_1}}}\]$

The equation of perpendicular line will be,

$\[y - {y_1} = {m_2}\left( {x - {x_1}} \right)\]$

Calculate the slope of the line,

$\[\begin{array}{l}{m_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\\ \Rightarrow {m_1} = \frac{{ - 4 - 2}}{{4 - \left( { - 4} \right)}}\\ \Rightarrow {m_1} = \frac{{ - 6}}{8}\end{array}\]$

Slope of the perpendicular line,

$\[\begin{array}{l}{m_2} = \frac{{ - 1}}{{{m_1}}}\\ \Rightarrow {m_2} = \frac{{ - 1}}{{\left( {\frac{{ - 6}}{8}} \right)}}\\ \Rightarrow {m_2} = \frac{8}{6}\end{array}\]$

Equation of the perpendicular line,

$\[\begin{array}{l}y - {y_1} = {m_2}\left( {x - {x_1}} \right)\\ \Rightarrow y - 2 = \left( {\frac{8}{6}} \right)\left( {x + 4} \right)\\ \Rightarrow 6y - 12 = \left( {8x + 32} \right)\\ \Rightarrow 8x - 6y = - 44\\ \Rightarrow 4x - 3y = - 22\end{array}\]$

Hence the perpendicular line will be $\[4x - 3y = - 22\]$