# What is the distance between the point of intersection of 6x+6y+8 = 0 and 6x+y=0 and the point (4,5).

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### 2 Answers

We first find the point of intersection of the given lines 6x+6y+8 = 0 and 6x+y=0

6x + y =0 => y = -6x

substitute in 6x+6y+8 = 0 => 6x + 6*(-6x) +8=0

=> 6x – 36 x + 8 =0

=> -30x = -8

=> x = 8/30 = 4/15

y = -6*x = -6*(4/15) = -8/5

The lines intercept at (4/15, -8/5)

Now the distance between (4, 5) and (4/15, -8/5) is sqrt [(4 - 4/15) ^2+ (5+ 8/5) ^2] = sqrt [12937/225]

**Therefore the required distance between the intersection of the lines and (4, 5) is sqrt [12937/225].**

The point of intersection of the lines 6x+6y+8 = 0 and 6x+y = 0 is got by solving the two equations.

(6x+6y+8) -(6x+y) = 0

5y +8 = o

5y = -8

y = -8/5. Substituting y = -8/5 in 6x+y = 0, we get 6x -8/5 = 0. So x= (8/5)/6 = 8/30 =4/15

So the point P of interesection = P (8/30.-8/5)

Now to find the disyance between (8/30 , -8/6) and (4,5) is

sqrt{ (4-4/15)^2 + (5- (-8/5))^2}

= sqrt{ 56/15)^2 +(33/5)^2}

= 7.58 nealy.

Therefore the distance between the point of intersection of the lines 6x+6y+8 = 0 and 6x+y = 0 and the point (4,5) is 7.58 nearly.