Given:
![\begin{gathered} y=4x+1 \\ \text{ point }(2,3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/c2ghbysd2v1si70rqk0pt8l5gddibftsce.png)
To find:
Draw a graph of a line that is perpendicular to the given line and passing through a given point.
Step-by-step explanation:
As we know that relation between two slopes of perpendicular slopes of lines:
![m_1.m_2=-1](https://img.qammunity.org/2023/formulas/mathematics/college/x6v30agut3gkhd448n962b6416fe234pjb.png)
Slope of given line y = 4x + 1 is:
![m_2=4](https://img.qammunity.org/2023/formulas/mathematics/college/4qfpv1lfsaanoupzilqebcw2d0tnngkr04.png)
So, the slope of line perpendicular to given line is:
![m_2=-(1)/(4)](https://img.qammunity.org/2023/formulas/mathematics/college/xapqp3nrmnkvmaa1al5mrwlnrwqnvbukvd.png)
Also, so line equation that is perpendicular to given line is:
![y=-(1)/(4)x+c...........(i)](https://img.qammunity.org/2023/formulas/mathematics/college/i0mo858klmc1n0jv2kathsbbrrxjzvhz41.png)
Also, the required line is passing thorugh given point (2, 3), i.e.,
![\begin{gathered} 3=-(1)/(4)(2)+c \\ c=3+(1)/(2) \\ c=(7)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ubmdopa6vf4esqopbactplvvq17yepg93e.png)
So, line equation that is perpendicular to given line is:
![y=-(1)/(4)x+(7)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/w7fkw1stoqlfynq9hj2hvix5e1otsdvs61.png)
The required graph of line is: