Question

1. find the midpoint, distance, and slope given (d(4,2)) and (a(6, - 5)).
2. write the equation of the line through the points in problem #1.

Explanation:

Step1: Find mid - point

The mid - point formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $M=(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Given $D(4,2)$ and $A(6,-5)$, we have $x_1 = 4,y_1=2,x_2 = 6,y_2=-5$. Then $M=(\frac{4 + 6}{2},\frac{2+( - 5)}{2})=(\frac{10}{2},\frac{2 - 5}{2})=(5,-\frac{3}{2})$.

Step2: Find distance

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Substitute $x_1 = 4,y_1=2,x_2 = 6,y_2=-5$ into the formula: $d=\sqrt{(6 - 4)^2+(-5 - 2)^2}=\sqrt{2^2+( - 7)^2}=\sqrt{4 + 49}=\sqrt{53}$.

Step3: Find slope

The slope formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$. Substitute $x_1 = 4,y_1=2,x_2 = 6,y_2=-5$ into the formula: $m=\frac{-5 - 2}{6 - 4}=\frac{-7}{2}=-\frac{7}{2}$.

Step4: Write equation of line

The point - slope form of a line is $y - y_1=m(x - x_1)$. We can use the point $D(4,2)$ and $m =-\frac{7}{2}$. Then $y - 2=-\frac{7}{2}(x - 4)$. Expand it: $y-2=-\frac{7}{2}x + 14$. Add 2 to both sides to get the slope - intercept form $y=-\frac{7}{2}x+16$.

Answer:

Mid - point: $(5,-\frac{3}{2})$
Distance: $\sqrt{53}$
Slope: $-\frac{7}{2}$
Equation of line: $y=-\frac{7}{2}x + 16$