# Find the slope of the line 3x-(t+1)y-1=0 if is parallel with the line 5x-2y+3=0?

krishna-agrawala | Student

Please note that slope of any line parallel to the line:

5x - 2y + 3 = 0   ...   (1)

will be same as the slope of this line. If we know the slope of the original line, we do not need to know anything about the parallel line to know its slope.

To find the slope of original line represented by equation (1) we transform it in the form y = mx + c as follows.

5x - 2y + 3 = 0

-2y = -5x - 3

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

y = 2.5x + 1.5

The slope of line in the form y = mx + c is equal to m.

Therefore:

Slope of the original line = 2.5

Therefore slope of all lines parallel to it including the line represented by the equation 3x - (t +1)y - 1 = 0

Slope = 2.5

Note:

Though the question does not ask for value of t, we can find it by finding the value of its slope independently in terms of t, and then equating that value of slope to 2.5. Then solving that equation for t will give us its value.

neela | Student

To find the slope of the line 3x-(t+1)y-1 = 0 || with the line 5x-2y+3= 0.

Solution:

If the two lines a1x+b1y+c= 0 and a2x+b2y+c2 = 0 are ||, then   the slopes of the first line -a1/b1 and the slope of the latter line -a2/b2 are equal.

Since 3x-(t+1)y-1 = 0 and 5x-2y+3 are the given  ||  lines, their slopes are equal. So,

-3/(-(t+1)) = -5/(-2). So the slope of the line 3x-(t+1)-1 is 5/2 = 2.5.

If you further wish to find the value of t:

3/(t+1) = 5/2.

3*2= 5(t+1) = 5t+5. Ot

6-5 = 5t Or

t = 1/5.

giorgiana1976 | Student

For lines to be parallel, the slope of d1:3x-(t+1)y-1=0 has to be equal to the slope of d2:5x-2y+3=0.

m1 = m2

Let's find out the slope of d2:

5x-2y+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 -2y to the left side:

-2y = -5x - 3

We'll multiply by -1:

2y = 5x + 3

We'll divide both sides by 2:

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

So, the slope of d2 is m2 = 5/2

Now, we'll find out the slope of d1:

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

We'll isolate -(t+1)y to the left side:

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

We'll divide by -(t+1), both sides:

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

The slope of d1 is m1 = 3/(t+1)

But m1=m2, so we'll get:

5/2 =  3/(t+1)

We'll cross multiply:

5(t+1) = 6

5t+5-6=0

5t-1=0

5t = 1

We'll divide by 5, both sides:

t = 1/5

So, the slope of 3x-(t+1)y-1=0 is m1 = 3/(t+1)

m1 = 3/(1/5 + 1) = 3*5/6 = 5/2

m1 = 5/2