How do you write the equation in slope intercept form perpendicular to the line \[2x-3y=12\] and passes through the point \[\left( 2,6 \right)\] ?

Hint: This problem is of topic conic section and of sub topic “Straight Lines”. The general representation of a straight line is \[y=mx+c\] where ‘m’ is the slope of the line and ‘c’ is the y-intercept. We also need to remember one very important thing regarding perpendicular lines. Say a line \[ax+by=c\] is the equation of a straight line, then the equation of the line perpendicular to this line is given by \[bx-ay=k\]. Here the value of the constant \[k\] can easily be found out by putting another point on this line and equating it.

Complete step-by-step solution:
Now we start off with the solution to the given problem by writing that,
The equation of the given line is \[2x-3y=12\] . The equation of the line perpendicular to this given line is given by \[3x+2y=k\] . Now on this line we put the point \[\left( 2,6 \right)\] to find out the value of the constant ‘k’. We write,
\[\begin{align}
  & \Rightarrow 3\left( 2 \right)+2\left( 6 \right)=k \\
 & \Rightarrow 6+12=k \\
\end{align}\]
Thus, from this we get the value of the constant as \[\Rightarrow k=18\] . We put this value of the constant to the equation of the line that we have formed. Thus \[3x+2y=18\] is our required equation of the line. Writing this in the form of slope and intercept form we get the required answer as, \[y=-\dfrac{3}{2}x+9\]


Note: The problems regarding conic sections particularly of straight line can be solved easily once we understand the chapter in depth. We need to remember all the formulae and relations of conic sections to solve problems of coordinate geometry efficiently. We also need to keep in mind the transformation of one line to another line which is perpendicular to the given line and passes through a given point.