Question 1179087
<br>
The video referenced by the other tutor shows a perfectly good way of solving the problem:
(1) Using the slopes of the given lines and y-intercept of the first line, find the equation of the line perpendicular to that first line containing that y-intercept;
(2) use the equation of that perpendicular line and the equation of the second given line to find the point of intersection; and
(3) use the distance formula with that point of intersection and the y-intercept of the first line to get the distance between the two lines.<br>
There is a lot of good math in that solution process; you should understand it and know how to use it.<br>
But there are much easier and faster ways to answer the problem.  Below are two of them.<br>
First alternative: Make a sketch with a right triangle<br>
Sketch a graph of the two lines with slope 2 and y-intercepts -1 and +9;
Draw the perpendicular segment from (0,9) to the other line;
Look at the right triangle whose sides are that perpendicular segment, part of the second line, and part of the y-axis.  Since the slopes of the two lines are 2, we can call the lengths of the two legs of that right triangle x and 2x; and the hypotenuse is 10.  Then from the Pythagorean Theorem,<br>
{{{x^2+(2x)^2=10^2}}}
{{{x^2+4x^2=100}}}
{{{5x^2=100}}}
{{{x^2=20}}}
{{{x = 2*sqrt(5)}}}<br>
ANSWER: The distance between the two lines (the length of the perpendicular segment between the two lines) is x = 2*sqrt(5)<br>
Second alternative: Use the point-to-line distance formula<br>
A very useful formula to know is for finding the distance from a given point to a given line:<br>
Given the equation of a line in the form Ax+By+C=0 and a point (p,q), the distance from the point to the line is<br>
{{{abs((Ap+Bq+C)/(sqrt(A^2+B^2)))}}}<br>
Use the equation of the first line in the required form (2x-y-1=0) and the y-intercept of the second line (0,9) as the fixed point and plug the numbers into the formula:<br>
{{{abs((2(0)-1(9)-1)/(sqrt(2^2+1^2))) = abs((-10)/sqrt(5)) = 2*sqrt(5)}}}<br>