# Which has to be the value of the parameter a so the lines d1 and d2 to be parallels?d1: 2x-5y+3=0 d2: (a+1)x-3y-1=0

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### 3 Answers

For d1 to be parallel to d2, the slope of d1 has to be equal to the slope of d2:

m1 = m2

Let's find out the slope of d1:

2x-5y+3=0

We'll re-write the equation into the form: y=mx+n, where m is the slope of the line.

For this reason, we'll isolate -5y to the left side:

-5y = -2x - 3

We'll multiply by -1:

5y = 2x+3

We'll divide both sides by 5:

y = (2/5)*x + (3/5)

**So, the slope of d1 is m1 = 2/5**

Now, we'll find out the slope of d2:

(a+1)x-3y-1=0

We'll isolate -3y to the left side:

-3y = -(a+1)x + 1

We'll divide by -3, both sides:

y = (a+1)*x/3 - 1/3

**The slope of d2 is m2 = (a+1)/3**

But m1=m2, so we'll get:

2/5 = (a+1)/3

We'll cross multiply:

5(a+1) = 6

5a+5-6=0

5a-1=0

We'll add 1 both sides:

5a = 1

We'll divide by 5, both sides:

**a = 1/5**

Let:

Slope of line d1 = m1 and

Slope of line d2 = m2

To find the value of m1 and m2 we transform the equations of the lines in the form y = mx + c.

Thus for d1:

2x - 5y + 3 = 0

- 5y = -2x - 3

y = -2/-5x - 3/-5

= = (2/5)x + 3/5

Therefore: m1 = 2/5

Similarly for d2:

(a + 1)x - 3y - 1 = 0

- 3y = -(a + 1)x + 1

y = -[(a + 1)/-3]x + 1/-3

y = (a/3 + 1/3)x - 1/3

Therefore: m2 = a/3 + 1/3

For d1 and d2 to be parallel m1 = m2. Therefore:

2/5 = a/3 + 1/3

Multiplying both sides of above equation by 15 we get:

6 = 5a + 5

5a = 6 - 5 = 1

a = 1/5 = 0.2

Answer:

a = 0.2

d1:2x-5y+3=0

d2:(a+1)x-3y-1=0 To find the condtions for the lines to be parallel .

We know that the lines a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 are || iff a1/a2= b1/b2 Aplying this to the coefficients in the same manner, we get:

2/(a+1) = -5/(-3) = 5/3.

2 *(3) = (a+1)(5)

6=5a+5.

6-5 =5a. Or

1 =5a.

1/5 =a Or

a =1/5