To solve the problem, we firstly need to find the perpendicular distance between the given parallel lines and the perpendicular distance of the third line from one of these lines and then compute the ratio.
The equations of the lines
3x+4y+5=0 and
3x+4y−5=0 are parallel since their normal vectors are the same (the coefficients of
x and
y are identical in both equations). Next, we calculate the distance between these two parallel lines.
The general formula for the perpendicular distance
d between a point
(x0,y0) and a line
Ax+By+C=0 is given by:
d=‌To find the distance between two parallel lines
Ax+By+C1=0 and
Ax+By+C2=0, we substitute a point from one line in the equation of the other line (the point need not be specifically calculated as any point on a line satisfies its line's equation). For practical purposes, we just compute:
d=‌For the lines
3x+4y+5=0 and
3x+4y−5=0, we have:
‌A=3‌B=4‌C1=5‌C2=−5The distance
d between these lines can be calculated as follows:
‌d=‌‌d=‌=2Next, we find the distance of the line
3x+4y+2=0 from one of the previous lines, say,
3x+4y+5=0 :
Since the total distance between the two parallel lines is 2 , and the distance from one of the parallel lines to the line that divides these (i.e.,
3x+4y+2=0 ) is
d1=‌, the remaining distance
d2 to the other line will be:
d2=2−‌=‌−‌=‌The distance ratio in which the line
3x+4y+2=0 divides the total distance between the lines is
d1:d2=‌:‌ or simplifying,
3:7.