How do I calculate the distance between these two lines?

Y=-2/3x - 1/2 and Y=-2/3 + 1/5 using the dist formula
|Ax + By +C| / root of a^2 + b^2

  d = (21√13)/130 ≈ 0.582435

Step-by-step explanation:

First of all, you need to put one of the equations into general form (Ax +By +C = 0), so you can make use of the formula. Multiplying the first equation by 6, we have ...

  6y = -4x -3

Adding the opposite of the right side, we have the general form equation ...

  4x +6y +3 = 0

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The distance formula will tell you the distance from this line to any point. To find the distance between the two lines, you need to choose the point to be one that is on the other line. It is probably convenient to use the y-intercept, (0, 1/5).

The formula is ...

  d = |4x +6y +3|/√(4²+6²)

The distance from the point (0, 1/5) is ...

  d = |4·0 +6(1/5) +3|/√52 = 4.2/(2√13)

  d = (21/130)√13 . . . . . with denominator rationalized

Alternate solution

Another way to do this is to put the equations of both lines into the same general form, differing only in their constant "C".

Multiplying both equations by 30, we get ...

  30y = -20x -15

  30y = -20x +6

So, the two general form equations are ...

  20x +30y +15 = 0

  20x +30y -6 = 0

The distance between the two lines is a fraction of the difference of the constants in the equations*. It will be ...

  |15 -(-6)|/√(20² +30²) = 21/√1300 = 21/(10√13) = (21√13)/130 . . . . as above

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* Any (x, y) pair that satisfies the first equation will make 20x+30y = -15. Using the second equation in the distance formula, you then have |-15-6|/√( ) = d. The number in the numerator is the difference of the two constants "C".

a. -0.3 > -0.33

triangle abc is reflected on the line x=-6.5 then reflected again on the x axis

step-by-step explanation: