Triangle sum (3x+1)^0+(10x-11)^0+(3x-2)^0=180
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You can use those two given points of street 1 to form its equation, then can use the fact that parallel lines have same slope to find the equation of second street with the help of the point (1,5) which lies in second street.
The equation of the location of the second street in standard form is given as
Let the two given points be (a,b) and (c,d). Then the equation of straight line passing through these two points is given by:
y = mx + c is the slope intercept form of straight line where m is the slope and c is the y-intercept (where the straight line cut on y = c at y axis) of the given line.
<h3>How to find equation of second street's location in terms of equation of straight line?</h3>Firstly we will find the equation of straight line which represents street 1.
Since street 1 goes from point (-5,-6) and (3,-2), thus, its equation would be:
Thus, this above equation represents equation of location of street 1. The slope is 1/2 and y-intercept is -7/2.
Since the street 2 is parallel to street 1, thus we have its slope same as that of street 1. Writing the equation in slope intercept form we get:
Since street 2 passes through (1,5)( x= 1, y = 5), thus, this point must satisfy above equation which represents all points lying on street 2.
Thus,
Thus, the equation representing points on street 2 is given by:
Learn more about equation of straight line here: