# Find t if d1 is parallel with d2. The equation of d1: 2x-ty+1=0 The equation of d2: -tx+8y-3=0

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d1= 2x-ty +1 =0

Rewrite in standard form:

==> y= (2x+1)/t = (2/t)x + 1/t

==> the slope (m1) = 2/t

d2= -tx+8y-3=0

Rewrite in standard form:

==> y= (tx+3)/8 = (t/8)x + 3/8)

==> the slope (m2)= t/8

Since d1 and d2 are parallel, then m1=m2

==> 2/t = t/8

Cross multiply:

==> t^2 = 16

==> t = +-4

Equation of di1 is:

2x - ty + 1 = 0

We change this equation in the form y = mx + c as follows:

-ty = -2x - 1

y = (-2/-t) x - 1/-t

y =(2/t) + 1/t

Similarly we change t the equation of d2 in the form y = mx + c as follows:

-tx + 8y -3 = 0

8y = tx + 3

y = (t/8)x + 3/8

In a equation of a straight line of the form y = mx + c, the slope of the line is given by m.

Therefore:

Slope of d1 = 2/t

Slope of d2 = t/8

As d1 and d2 are parallel their slopes are equal.

Therefore:

2/t = t/8

multiplying both side of above equation by 8t we get:

16 = t^2

Therefore:

t = 16^(1/2) = 4 and -4

Answer:

t = 4 and -4

If 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-ty+1=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 -ty to the left side:

-ty = -2x - 1

We'll multiply by -1:

ty = 2x+1

We'll divide both sides by t:

y = (2/t)*x + (1/t)

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

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

-tx+8y-3=0

We'll isolate 8y to the left side:

8y = tx + 3

We'll divide by 8, both sides:

y = tx/8 - 3/8

**The slope of d2 is m2 = t/8**

But m1=m2, so we'll get:

**2/t=t/8**

We'll cross multiply:

t^2=16

t1=sqrt16

**t1=4**

**t2=-4**

If a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 are the two lines and if they are || then

a1/a2 = b1/b2.

So here the lines are 2x+ty+1 = 0 and -tx+8y-3 = 0.

Therefore a1/a2 = b1/b2 gives: 2/-t = -t/8. Or

8*2 = (-t)(-t) =t^2.

t^2 = 16 . Or t = -4 Or t = 4 are the values of t for which d1 and d2 are ||/