Given:
A line passes through the points A (1,3) and B (7,1).
The objective is,
a) To find gradient of AB.
b) To find gradient of a line perpendicular to AB.
c) To find the equation line passing throught (4,2) and perpendicular to AB.
Explanation:
a)
Consider the given coordinates of line AB as,
[tex]\begin{gathered} (x_1,y_1)=(1,3) \\ (x_2,y_2)=(7,1) \end{gathered}[/tex]The general formula to find the gradient of line AB is,
[tex]m_{AB}=\frac{y_2-y_1}{x_2-x_1}[/tex]To find gradient of AB:
Substitute the given coordinates in the above gradient formula.
[tex]\begin{gathered} m_{AB}=\frac{1-3}{7-1} \\ =\frac{-2}{6} \\ =-3 \end{gathered}[/tex]Hence, the gradient of line AB is -3.
b)
To find gradient of perendicular line:
Consider the perpendicular line as CD.
The product of gradients of two perpendicular lines will be -1.
So, the gradient of the perpendicular line CD can be calculated as,
[tex]\begin{gathered} m_{AB}\times m_{CD}=-1 \\ m_{CD}=\frac{-1}{m_{AB}} \\ m_{CD}=-\frac{1}{-3} \\ m_{CD}=\frac{1}{3} \end{gathered}[/tex]Hence, the gradient of a line perndicular to AB is (1/3).
c)
To find equation of line perpendicular to AB:
Since, the perpendicular line passes through the point,
[tex](x_1,y_1)=(4,2)[/tex]From part (b) the gradient of this perpendicular line is (1/3),
Then, the equation can be calculated using point slope fomula as,
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ y-2=\frac{1}{3}(x-4) \\ y-2=\frac{x}{3}-\frac{4}{3} \\ y=\frac{x}{3}-\frac{4}{3}+2 \\ y=\frac{x}{3}+0.666666\ldots.. \\ y=\frac{x}{3}+0.67 \end{gathered}[/tex]Hence, the equation of perpendicular line is y = (x/3) + 0.67.
Answers:
a) Gradient AB : (-3)
b) Gradient CD : (1/3)
c) Equation of perpendiular line is y = (x/3) + 0.67.
