step 1
we have the curve
![y=2x+5](https://img.qammunity.org/2023/formulas/mathematics/college/qz6sm2qok56u07bywkqdeu10nr1b8ijgr3.png)
Find out the derivative
![y^(\prime)=2\text{ ----slope of the curve}](https://img.qammunity.org/2023/formulas/mathematics/college/b19dd3ska0jclncijqj10g8o1jiw7kvw7j.png)
step 2
Find out the equation of the line perpendicular to the given curve that passes through the point (0,8)
Remember that
If two lines are perpendicular, then their slopes are negative reciprocal
so
The slope of the perpendicular line is m=-1/2
The equation of the line is given by
![y=mx+b](https://img.qammunity.org/2023/formulas/mathematics/high-school/smsb8cbft03lwblmi49nf2l6jby2ofxzws.png)
where
m=-1/2
point (0,8) ----> y-intercept
substitute
![y=-(1)/(2)x+8](https://img.qammunity.org/2023/formulas/mathematics/college/70q25aowac000lchrcyk6cz1wvfl6qsdmj.png)
step 3
Find out the intersection of both lines
![\begin{gathered} y=2x+5 \\ y=-(1)/(2)x+8 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/rj0ayp4hf70xwupm2ne6bet2gzdqiyrwqk.png)
Equate both equations
![\begin{gathered} 2x+5=-(1)/(2)x+8 \\ 2x+(1)/(2)x=8-5 \\ \\ (5)/(2)x=3 \\ \\ x=(6)/(5) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ajmkw8capv5onj3krt6akhjacc1gsr0099.png)
Find out the y-coordinate of the point
![\begin{gathered} y=2((6)/(5))+5 \\ \\ y=(12)/(5)+5 \\ \\ y=(37)/(5) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/s3xh3wnq2yqxi0988e97cdscorfszvmgfk.png)
The coordinates of the point are
![(x,y)=((6)/(5),(37)/(5))](https://img.qammunity.org/2023/formulas/mathematics/college/jz9z4cxztqrr873ayxypg1834vnn806c91.png)