So, we know some coordinates of the places:
- Home: (-3, -12)
- School: (-15, 6)
- Grocery: (24, 6)
- Gym: (0, y_gym)
First, we want to determine the line that represents a stree that connects School and Home, that is, the points (-15, 6) and (-3, -12).
Let's represent a line in the standard form:
[tex]y=a+bx[/tex]
The slope, "b", can be calculated using the points we know:
[tex]b=\frac{y_2-y_1}{x_2-x_1}=\frac{6-(-12)}{-15-(-3)}=\frac{6+12}{-15+3}=\frac{18}{-12}=-\frac{3}{2}[/tex]
Then, we can use the slope-point form and solve for y:
[tex]\begin{gathered} y-y_1=b(x-x_1) \\ y+12=-\frac{3}{2}(x+3) \\ y=-\frac{3}{2}x-\frac{9}{2}-12 \\ y=-\frac{3}{2}x-\frac{33}{2} \end{gathered}[/tex]
That is the equation of the First Street.
The Main Street passes through Home and Grocery, that is, points (-3 -12) and (24, 6).
Using the same method, we firts get the slope:
[tex]b=\frac{y_2-y_1}{x_2-x_1}=\frac{6-(-12)}{24-(-3)}=\frac{18}{27}=\frac{2}{3}[/tex]
And use the slope-point form:
[tex]\begin{gathered} y-y_1=b(x-x_1)_{} \\ y+12=\frac{2}{3}(x+3) \\ y=\frac{2}{3}x+2-12 \\ y=\frac{2}{3}x-10 \end{gathered}[/tex]
Lastly, we know that Ridge Road is parallel to Main Street and it passes through School. If they are parallel, they have the same slope, so the slope of Ridge Road is:
[tex]b=\frac{2}{3}[/tex]
Since we know School is in thir Road, we can use the slope-point to get the equation for Ridge Road:
[tex]\begin{gathered} y-y_1=b(x-x_1) \\ y-6=\frac{2}{3}(x+15) \\ y=\frac{2}{3}x+10+6 \\ y=\frac{2}{3}x+16 \end{gathered}[/tex]
And, since we know that the x-coordinate of the Gym is x = 0, we can put this into the equation of the Ridged Road to get its y-coordinate:
[tex]\begin{gathered} y=\frac{2}{3}\cdot0+16 \\ y=16 \end{gathered}[/tex]
The picture we get is the following:
So, the final answers are:
First Street:
[tex]y=-\frac{3}{2}x-\frac{33}{2}[/tex]
Main Street:
[tex]y=\frac{2}{3}x-10[/tex]
Ridge Road:
[tex]y=\frac{2}{3}x+16[/tex]
Coordinates of the Gym:
[tex](0,16)[/tex]
