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Write an equation that passes through (4,-3) and is perpendicular to y=1/2x-3

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Given:

The equation of a line is y = (1/2)x-3.

Another line passes perpendicular to the first line through the point,


(x_1,y_1)=(4,-3)

The objective is to find the equation of the second line.

Step-by-step explanation:

In general, the product of the slope of perpendicular lines will be -1.

Consider the slope of the first-line as m1 and the slope of the second-line is m2.

To find m1:

From the given equation, the slope of the first line will be,


m_1=(1)/(2)

To find m2:

Then, the slope of the second line can be calculated as,


\begin{gathered} m_1* m_2=-1 \\ m_2=-(1)/(m_1) \\ m_2=-(1)/((1)/(2)) \\ m_2=-2 \end{gathered}

To find the equation of the second line:

The equation of a line using the slope and the point can be calculated as,


y-y_1=m_2(x-x_1)

On plugging the obtained values in the above equation.


\begin{gathered} y-(-3)=-2(x-4) \\ y+3=-2x+8 \\ y=-2x+8-3 \\ y=-2x+5 \end{gathered}

Hence, the equation of the line is y=-2x+5.

User Connor Peet
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