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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

Best Answer 

 #2
avatar+12 
+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
avatar+19597 
+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
avatar+12 
+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

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