Determine a if the lines (a-3)*x + 2y-7=0 and x+(a+1)*y+2=0 are parallel.

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

Two parallel lines have the same slope.

So we have to find the slope of (a-3)*x + 2y-7=0 and x+(a+1)*y+2=0

(a-3)*x + 2y-7=0

=> 2y = -(a - 3)x + 7

=> y = -(a - 3)/2* x + 7/2

The slope is (3 - a)/2

x+(a+1)*y+2=0

=> (a + 1)*y = -x - 2

=> y = -x / (a +1) - 2 / (a+1)

The slope is -1/(a+1)

As the line are parallel: (3 - a)/2 = -1/(a+1)

=> (3 - a)(a +1) = -2

=> 3a - a^2 - a + 3 = -2

=> a^2 - 2a - 5 = 0

a1 = [2 + sqrt (4 + 20)]/2

= 1 + sqrt 6

a2 = 1 - sqrt 6

Therefore a can be (1 + sqrt 6) and (1- sqrt 6)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Two lines are parallel if and only if the ratios of the corresponding coefficients are the same.

(a-3)/1 = 2/(a+1)

We'll cross multiply:

(a-3)(a+1) = 2

We'll remove the brackets:

a^2 + a - 3a - 3 = 2

We'll move all terms to the left side:

a^2 - 2a - 3 - 2 = 0

We'll combine like terms:

a^2 - 2a - 5 = 0

We'll apply quadratic formula:

a1 = [2 + sqrt(4 + 20)]/2

a1 = (2 + 2sqrt6)/2

a1 = 1 + sqrt6

a2 = 1 - sqrt6

The values of a are: {1 - sqrt6; 1 + sqrt6}.

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