Let u and v be two parallel lines passing through the points A=(5,0) and

B=(-5,0) respectively.Let the line 4x+3y=25 meet u at P and v at Q.

If the length of PQ is 5 units, show that there are two possibilities for the pair of parallel lines u and v.

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Since u and v are parallel lines there gradient will be same.

So we can say the two lines are;

`u rarr y = mx+c`

`v rarr y = mx+d`

It is given that u and v passing through the points A=(5,0) and B=(-5,0) respectively

`u rarr 0 = 5m+c`

`c = -5m`

`v rarr 0 = -5m+d`

`d = 5m`

`urarr y = m(x-5)`

`v rarr y = m(x+5)`

These two lines u and v meet 4x+3y=25 at P and Q.

For the point P;

`4x+3y=25`

`y = m(x-5)`

Once you solve this you will get;

`x = (5(5+3m))/(4+3m)`

`y = (5m)/(4+3m)`

`P = ((5(5+3m))/(4+3m),(5m)/(4+3m))`

For the point Q;

`4x+3y=25`

`y = m(x+5)`

When you solve the above you will get;

`x = (5(5-3m))/(4+3m)`

`y = (45m)/(4+3m)`

`Q = ( (5(5-3m))/(4+3m),(45m)/(4+3m))`

It is given that the length of PQ is 5 units.

`5^2 = [(5(5+3m))/(4+3m)-(5(5-3m))/(4+3m)]^2+[(5m)/(4+3m)-(45m)/(4+3m)]^2`

`25 = (900m^2+1600m^2)/(4+3m)^2`

`5^2 = (2500m^2)/(4+3m)^2`

`5^2 = ((50m)/(4+3m))^2`

`+-5 = (50m)/(4+3m)`

`+5 = (50m)/(4+3m)`

`m = 5/7`

`-5 = (50m)/(4+3m)`

`m = -4/13`

We have two answers for the gradient m. This means we have two lines of u and two lines of v that satisfies the requirements given.

*So there are two pair of parallel lines under the given condtions for u and v.*

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