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A line is perpendicular to y=-4x-6 and passes through the point (0,-3). Identify the equation of this line. Can somebody help me with this

User Savion
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1 Answer

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12 votes

Two lines are perpendicular if their slopes satisfy the following equation:


m_2=-(1)/(m_1)

In this case, we have the line y = -4x - 6 which slope is m = -4, so the line perpendicular to this has a slope equal to:


\begin{gathered} m_1=-4 \\ m_2=-(1)/(m_1) \\ m_2=-(1)/((-4))=(1)/(4) \end{gathered}

Also, we know the line we are looking for passes through the point (0, -3), so we can found the equation:


\begin{gathered} \text{The general equation of a line is:} \\ y=mx+b \\ \text{Where m is the slope and b is the y-intercept value.} \\ In\text{ this case the slope is 1/4 and also we know that the point (0, -3) satisfy the equation:} \\ m=(1)/(4) \\ y=(1)/(4)x+b \\ We\text{ use the point (0, -3):} \\ -3=(1)/(4)\cdot0+b \\ b=-3 \end{gathered}

Above we found the value of the y-intercept value as -3, so the equation of the line is:


y=(1)/(4)x-3

User Joon
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