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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 restri

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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

#2
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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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#1
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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
+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

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