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write an equation in slope intercept form for a line that passes through the given point and is perpendicular to the graph of the equation.(3,-2); y = × + 4

User MrJD
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We know that two lines are perpendicular if and only if their slopes fullfil:


m_1m_2=-1

Then we need to find the slope of the equation given:


y=x+4

we notice that this equation is written in the slope intercept form:


y=mx+b

comparing this equations we notice that:


m_1=1

Now, plugging this value in the condition and solving for the second slope we have:


\begin{gathered} 1m_2=-1 \\ m_2=-(1)/(1) \\ m_2=-1 \end{gathered}

Now that we have the slope of the line we are looking form we plug its value, and the point in the equation:


y-y_1=m(x-x_1)

then:


\begin{gathered} y-(-2)=-1(x-3) \\ y+2=-x+3 \end{gathered}

finally we write the equation in the slope intercept form given above:


\begin{gathered} y+2=-x+3 \\ y=-x+3-2 \\ y=-x+1 \end{gathered}

Therefore the equation we are looking for is:


y=-x+1

User Chris Stryker
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