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Write an equation of the line passing through the point (4, 3) that is perpendicular to the line Y + 3 = -9/11 (x + 4)

Sagot :

We have to find the equation of a line that pass through the point (4,3) and is perpendicular to the line:

[tex]y+3=-\frac{9}{11}(x+4)[/tex]

This line, that is perpendicular, has a slope m=-9/11, as it is written in slope-point form.

The line we are looking for will have a slope that is the negative reciprocal of m=-9/11, so it will be:

[tex]m_2=-\frac{1}{m_1}=-\frac{1}{-\frac{9}{11}}=\frac{11}{9}[/tex]

Our line has a slope of m=11/9.

We can use the known point (4,3) to write the equation as:

[tex]y-3=\frac{11}{9}(x-4)[/tex]

If we want the equation in slope intercept form, we rearrange:

[tex]\begin{gathered} y=\frac{11}{9}(x+4)+3 \\ y=\frac{11}{9}x+\frac{44}{9}+3 \\ y=\frac{11}{9}x+\frac{44+27}{9} \\ y=\frac{11}{9}x+\frac{71}{9} \end{gathered}[/tex]

Answer: y-3=11/9*(x-4)

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