105k views
4 votes
Write an equation that passes through (5,-1) and is perpendicular to y=5/2x-5

1 Answer

2 votes

Answer:


y=-(2)/(5)x+1

Explanation:

The equation of a line is represented by the following equation:


\begin{gathered} y=mx+b \\ \text{where,} \\ m=\text{ slope} \\ b=y-\text{intercept} \end{gathered}

If the given line is perpendicular to the one we want to find, the slope of the missing line would be the negative reciprocal of the given slope:


\begin{gathered} m=-((2)/(5)) \\ m=-(2)/(5) \end{gathered}

Use the slope-point form of the line equation to determine the equation in slope-intercept form:


\begin{gathered} y-y_1=m(x-x_1) \\ y-(-1)=-(2)/(5)(x-5) \\ y+1=-(2)/(5)x+2 \\ y=-(2)/(5)x+2-1 \\ y=-(2)/(5)x+1 \end{gathered}

User Rasx
by
8.6k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories