Geometry problem: find the distance between the parallel lines y=-3/2x+1 and 2y+3x=10

d = 1.847

Step-by-step explanation:

To avoid any ambiguity, please enclose the slope -3/2 inside parentheses:

y = (-3/2)x + 1.  Next, solve the 2nd equation for y:  2y = - 3x + 10 => y = (-3/2)x + 5.  Notice how these two lines have the same slope (-3/2)?  Thus, the two given lines are parallel.

It's important to realize that the distance between these two lines is a line segment with perpendicular to both lines.  Thus, the segment representing this distance has the form y = (2/3)x + c.  Let's say that this segment passes through the y-intercept of y = (-3/2)x + 1; it thus passes through (0,1), and thus has the equation y = (2/3)x + 1.  

Find the point at which this y = (2/3)x + 1 intersects the line y = (3/2)x + 5.  Note that the xresultant point

Setting these two equations = to one another results in (2/3)x + 1 = (-3/2)x + 5.

Multiplying both sides by the LCD (6) will eliminate the fractions:

4x + 6 = -9x + 30.  Combining like terms:  13x = 24, and x = 24/13 = 1.846.

Substitute this x-value into   y = (2/3)x + 1 to find the y-coordinate of the point of intersection of y = (2/3)x + 1 and y = (-3/2)x + 5:

When x = 1.846, y = (2/3)(1.846) + 1 = 1.231.

Thus, the desired distance is that between (0,1) and (1.846, 1.231), and is:

d = √ [ (1.846 - 0)^2 + (1.231 - 1)^2 ] = √ [ (1.846)^2 + (0.053)^2 ]

This simplifies to                              d = √ [ 3.408 + 0.003 ] = √3.411, or

                                                          d = 1.847   (answer)