Question

Point A has coordinates (5, -6) and point B has coordinates (7, 2).a)Find the coordinates of the midpoint of A and B.b)Find the gradient of the straight line AB.C)Find the equation of the straight line AB, giving the answer in the formy= mx + C.d)Find an equation of the perpendicular bisector of AB, giving the answer in theform ax+ by +C = 0, where a, b and care integers.

Answer 1

We have the coordinates of point A (5, -6) and point B has coordinates (7, 2).

Part A.

To find the coordinates of the midpoints

We will use the relationship

$\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}$

so that our midpoint will be

$\begin{gathered} \frac{5+7}{2},\frac{-6+2}{2} \\ \frac{12}{2},\frac{-4}{2} \\ \Rightarrow(6,-2) \end{gathered}$

Part B

To find the gradient

$\frac{y_2-y_1}{x_2-x_1}=\frac{2-(-6)}{7-5}=\frac{8}{2}=4$

Gradient = 4.

Part C

The equation of the straight line AB, can be obtained as follow

$\begin{gathered} \frac{y-y_1}{x-x_1}=m \\ \frac{y-(-6)}{x-5}=4 \\ \\ \frac{y+6}{x-5}=4 \end{gathered}$

Cross multiply

$\begin{gathered} y+6=4(x-5) \\ y+6=4x-20 \\ y=4x-20-6 \\ y=4x-26 \end{gathered}$

The equation of the straight line is: y= 4x -26

Part D

To get the equation of the perpendicular bisector, we will find the slope that is perpendicular to the line AB

since the slope of AB is 4

$\begin{gathered} \text{the slope perpendicular to the line will be} \\ m=-\frac{1}{4} \end{gathered}$

The midpoint is (6, -2)

Then the equation of the perpendicular bisector can be obtained as follow

$\begin{gathered} \frac{y-(-2)}{x-6}=-\frac{1}{4} \\ \\ \frac{y+2}{x-6}=-\frac{1}{4} \end{gathered}$

Cross multiply

$\begin{gathered} 4y+8=-x+6 \\ 4y+x+8-6=0 \\ 4y+x+2=0 \\ x+4y+2=0 \end{gathered}$

The equation of the perpendicular bisector is

x+4y+2=0