Question

find a general form of an equation for the perpendicular bisector of the segment ab.
7.) a(3, -1), b(-2, 6)

Explanation:

Step1: Find the mid - point of segment AB

The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. For $A(3,-1)$ and $B(-2,6)$, the mid - point $M$ is $(\frac{3+( - 2)}{2},\frac{-1 + 6}{2})=(\frac{1}{2},\frac{5}{2})$.

Step2: Find the slope of segment AB

The slope formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2-y_1}{x_2-x_1}$. For $A(3,-1)$ and $B(-2,6)$, the slope $m_{AB}=\frac{6-( - 1)}{-2 - 3}=\frac{7}{-5}=-\frac{7}{5}$.

Step3: Find the slope of the perpendicular bisector

If two lines are perpendicular, the product of their slopes is $- 1$. Let the slope of the perpendicular bisector be $m$. Then $m\times(-\frac{7}{5})=-1$, so $m = \frac{5}{7}$.

Step4: Use the point - slope form to find the equation of the line

The point - slope form of a line is $y - y_0=m(x - x_0)$, where $(x_0,y_0)$ is a point on the line and $m$ is the slope. Using the point $M(\frac{1}{2},\frac{5}{2})$ and $m=\frac{5}{7}$, we have $y-\frac{5}{2}=\frac{5}{7}(x - \frac{1}{2})$.

Step5: Convert to general form

Multiply through by 14 to clear the fractions: $14y-35 = 10(x-\frac{1}{2})$. Expand the right - hand side: $14y-35 = 10x - 5$. Rearrange to get $10x-14y+30 = 0$, or $5x-7y + 15=0$.

Answer:

$5x-7y + 15 = 0$