# Thread: Slope-intercept form of a line.

1. ## Slope-intercept form of a line.

How do I solve the following problems. For the first one, I know that 2/3x is the slope but what is the slope intercept form and how do I figure it out

1) Write the slope-intercept form of the equation of the line described:
through (-2,3), perpendicular to y=2/3x-3

This one below needs to be in matrix form, ie:[ ]. For this I am having trouble figuring out how to place the numbers in the matrix and do the inverse to solve it since there are not an equal amount of numbers in each equation
2) Solve the system
r+s+6t=23
3r-2s-3t=30
-r+t=-8

What the first question, find the gradient using the relationship $\displaystyle m_T\times m_P=-1$

Where 'm' are the gradients to the tangent and perpendicular.

$\displaystyle \left[ \begin{array}{c} r \\ s \\ t \end{array} \right] = \left[ \begin{array}{ccc} 1 & 1 & 6 \\ 3 & -2 & -3 \\ -1 & 0 & 1 \end{array} \right]^{-1} \left[ \begin{array}{c} 23 \\ 30 \\ 8 \end{array} \right]$
The line perpendicular to y= (2/3)x- 3 has slope -3/2. The equation of the line through point (a, b) with slope m is $\displaystyle y= m(x-a)+ b$.