Assume that C(doesn't equal)D and a and b are not both zero. show that Ax + By = C and Ax + By = D are parallel lines explain why the restrictions on a,b,c and d are necessary.

Guest Sep 9, 2014

#2**+5 **

Parallel lines have the same slope.

So that,

for Ax + By = C, we have y = (-A/B)x + C/B, where the slope is -A/B

for Ax + By = D, we have y = (-A/B)x + D/B, where the slope is -A/B

Hence, the two lines are parallel.

Note that:

if A = B = 0, y = 0/0 + C/0 (or D/0). The lines are undefined.

if C = D, it is a single line crossing the ordinate axis at the same point.

Demogorgon Sep 9, 2014

#1**+5 **

**show that Ax + By = C and Ax + By = D are parallel lines explain why the restrictions on a,b,c and d are necessary.**

(1) Ax+By=C

(2) Ax+By=D

y-intercept(x=0): $$(1)

\begin{array}{rcl}

\quad A*0 + By &=& C \\

By&=&C\\

y_{(x=0)}&=&\frac{C}{B}

\end{array}

\qquad

(2)

\begin{array}{rcl}

\quad A*0 + By &=& D \\

By&=&D\\

y_{(x=0)}&=&\frac{D}{B}

\end{array}$$

x-intercept (y=0): $$(1)

\begin{array}{rcl}

\quad Ax + B*0 &=& C \\

Ax&=&C\\

x_{(y=0)}&=&\frac{C}{A}

\end{array}

\qquad

(2)

\begin{array}{rcl}

\quad Ax + B*0 &=& D \\

Ax&=&D\\

x_{(y=0)}&=&\frac{D}{A}

\end{array}$$

$$slope = -\frac{y_{(x=0)}}{x_{(y=0)}}$$

$$\text{(1) slope} = -\dfrac{

\frac{C}{B}

}{\frac{C}{A}} = -\frac{C}{C}*\frac{A}{B} = -\frac{A}{B}$$

$$\text{(2) slope} = -\dfrac{

\frac{D}{B}

}{\frac{D}{A}} = -\frac{D}{D}*\frac{A}{B} = -\frac{A}{B}$$

slope (1) = slope (2) = $$-\frac{A}{B}$$ , line 1 and line 2 are parallel

heureka Sep 9, 2014

#2**+5 **

Best Answer

Parallel lines have the same slope.

So that,

for Ax + By = C, we have y = (-A/B)x + C/B, where the slope is -A/B

for Ax + By = D, we have y = (-A/B)x + D/B, where the slope is -A/B

Hence, the two lines are parallel.

Note that:

if A = B = 0, y = 0/0 + C/0 (or D/0). The lines are undefined.

if C = D, it is a single line crossing the ordinate axis at the same point.

Demogorgon Sep 9, 2014