Question

1. jason lives exactly halfway between the primary and secondary schools in his neighbourhood. the intervals between the grid lines represent 1 km.

a) how far apart are the schools?
b) determine the coordinates of jasons home.
c) what other locations are equidistant from the two schools? explain your reasoning.
d) determine an equation that represents all locations that are equidistant from the two schools.

Explanation:

Step1: Recall distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Given $P(4,3)$ and $S(14,8)$, where $x_1 = 4,y_1=3,x_2 = 14,y_2 = 8$.

Step2: Substitute values into formula

$d=\sqrt{(14 - 4)^2+(8 - 3)^2}=\sqrt{10^2+5^2}=\sqrt{100 + 25}=\sqrt{125}=5\sqrt{5}\approx 11.18$ km. Since each grid - line represents 1 km, the distance is approximately 11.18 km.

Step3: Find mid - point for equidistant locations (part b)

The mid - point formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $M=(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Substituting $x_1 = 4,y_1=3,x_2 = 14,y_2 = 8$, we get $M=(\frac{4 + 14}{2},\frac{3+8}{2})=(9,5.5)$.

Step4: Equation for equidistant points (part d)

Let a point be $(x,y)$. The distance from $(x,y)$ to $P(4,3)$ is $d_1=\sqrt{(x - 4)^2+(y - 3)^2}$, and the distance from $(x,y)$ to $S(14,8)$ is $d_2=\sqrt{(x - 14)^2+(y - 8)^2}$. Since $d_1=d_2$, we have $(x - 4)^2+(y - 3)^2=(x - 14)^2+(y - 8)^2$. Expanding: $x^{2}-8x + 16+y^{2}-6y+9=x^{2}-28x + 196+y^{2}-16y + 64$. Simplifying gives $20x+10y=235$, or $4x + 2y=47$.

Answer:

a) The two schools are approximately $5\sqrt{5}\approx11.18$ km apart.
b) The coordinates of the point equidistant from the two schools (mid - point) are $(9,5.5)$.
c) The locus of points equidistant from the two schools is a perpendicular bisector of the line segment joining the two schools. The equation of the set of points equidistant from the two schools is $4x + 2y=47$.
d) The equation $4x + 2y=47$ represents all locations that are equidistant from the two schools. The reason is that for any point $(x,y)$ on this line, the distance from $(x,y)$ to $P(4,3)$ is equal to the distance from $(x,y)$ to $S(14,8)$ based on the distance formula.