Which theorem or postulate could be used to prove the congruence of the triangles?

A contractor is building a new subdivision on the outside of a city. He has started work on the first street and is planning for

ruslelena [56]

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 $2y = x + 9$

<h3>What is the equation of a straight line passing through two given points?</h3>

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:

$y - b = \dfrac{(d-b)}{(c-a)}(x-a)$

<h3>What is slope intercept form of equation of straight line?</h3>

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:

$y - (-6) = \dfrac{-2- (-6)}{3-(-5)} (x -(-5))\\
\\
y + 6 = \dfrac{1}{2}(x+5)\\
\\
y + 6 = \dfrac{x}{2} + \dfrac{5}{2}\\\\
y = \dfrac{x}{2} -\dfrac{7}{2}$

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:

$y = \dfrac{1}{2}x + c$

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,

$5 = \dfrac{1}{2} \times 1 + c\\\\
c = 5 - \dfrac{1}{2} = \dfrac{9}{2}$