How do you write the equation that is perpendicular to \[y=-\dfrac{1}{2}x+\dfrac{2}{3}\], passes through \[\left( 2,3 \right)\]?

Hint: In order to find the solution of the given question that is to write the equation that is perpendicular to \[y=-\dfrac{1}{2}x+\dfrac{2}{3}\], passes through \[\left( 2,3 \right)\] use the standard equation of point-slope form to write the equation line passing through given point where point slope formula is \[\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)\] where \[m\] is the slope of the line and \[\left( {{x}_{1}},{{y}_{1}} \right)\] is the point that passes through the line. And having the slope of the given equation of line. Compare the equation to the general equation of line and find the value of its slope. Use that value of slope and write slope of perpendicular. Use that value of slope to form a new equation.

Complete step by step answer:
Compare the given equation of line with general slope intercept form that is \[y=mx+c\]
We get the value of its slope as \[m=-\dfrac{1}{2}\].
Now we have to find the equation of the line perpendicular to the given line. Since we know for a line having slope m, the slope of line perpendicular to the line will be \[-\dfrac{1}{m}\]; then the slope for the perpendicular will be as follows:
\[\Rightarrow {m}'=-\dfrac{1}{m}\]
\[\Rightarrow {m}'=-\dfrac{1}{\left( -\dfrac{1}{2} \right)}\]
\[\Rightarrow {m}'=2\]
So, now we know the slope of the line we have to find is \[2\] and the point that passes through the line \[\left( 2,3 \right)\].
We know that the standard equation of point-slope formula \[\left( y-{{y}_{1}} \right)={m}'\left( x-{{x}_{1}} \right)\] where \[{m}'\] is the slope of the line and \[\left( {{x}_{1}},{{y}_{1}} \right)\] is the point that passes through the line.
Now, substituting the given values in the above formula we will have:
\[\Rightarrow \left( y-3 \right)=2\left( x-2 \right)\]
After solving the bracket, we get:
\[\Rightarrow y-3=2x-4\]
To solve the above equation further, add \[3\] both sides. So, we are left with:
\[\Rightarrow y-3+3=2x-4+3\]
After simplifying it further we get:
\[\Rightarrow y=2x-1\]

Therefore, the equation of line that is perpendicular to \[y=-\dfrac{1}{2}x+\dfrac{2}{3}\], passes through \[\left( 2,3 \right)\] is \[y=2x-1\].

Note: Students make mistakes by writing the slope of line perpendicular to a given line as same as the slope of line which is completely wrong and leads to the wrong answer. It’s important to remember that the slope of perpendicular will be negative if the reciprocal of the given slope of a line as perpendicular of a line is not in the same direction, it’s exactly making the right angle with the given line.