[tex]\begin{gathered} POINT\text{ (x,y)=(2,6)} \\ PERPENDICULAR\text{ TO y=-}\frac{1}{4}x+3 \\ The\text{ searched line must be perpendicular to the last line equation, } \\ slope\text{ m}\Rightarrow perpendicular\text{ slope }-\frac{1}{m} \\ I\text{n this case} \\ m=-\frac{1}{4}\text{ }\Rightarrow\text{ perpendicular slope is -}\frac{1}{-\frac{1}{4}} \\ or\text{ }\frac{1}{\frac{1}{4}}=\frac{\frac{1}{1}}{\frac{1}{4}}=\frac{1\cdot4}{1\cdot1}=4 \\ \text{hence, the line equation is in the form} \\ y=4x+b \end{gathered}[/tex][tex]\begin{gathered} \text{Now, the only thing to compute is the y-intercept "b". To find b, we must use the point given:} \\ y=4x+b\Rightarrow6=4(2)+b \\ 6=8+b \\ 6-8=b \\ -2=b \\ \text{hence, b=-2.} \\ \text{FINALLY, the searched equations is given by} \\ y=4x-2 \end{gathered}[/tex]
