hi,
im not too sure how to work this out:
equation of a straight line through A(-1,3) perpendicular to y=1/3x-1
i require some help on this one step at a time please.
thanks!
1. Two lines with the slopes $\displaystyle m_1$ and $\displaystyle m_2$ are perpendicular if the slopes satisfy the equation
$\displaystyle m_1 \cdot m_2 = -1$
So the slope of the perpendicular line is (-3).
2. Use the point-slope-formula of a straight line to determine the equation of the line in question:
$\displaystyle y-y_1 = m(x-x_1)$
where $\displaystyle x_1, y_1$ are the coordinates of the point which is located on the line:
$\displaystyle y-3=(-3)(x-(-1))~\implies~y=-3x$
Hi andyboy179.
The slope of the line perpendicular to y=1/3 x -1 is -3because the product of the slopes must be -1. Use the point slope formula to find the new line
y-3/x+1 = -3 finish it
bjh