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Sagot :
Given:
The equation of a line is x-6y-5 = 0.
Another line passes perpendicular to the first line through the point (8, -8).
The objective is to find the equation of the second line in point slope form and general form.
Explanation:
The general equation of line is,
[tex]y=mx+c\text{ . . . . .(1)}[/tex]Here, m represents the slope of the line.
To find slope of first line:
The given equation of line 1 can be rewritten as,
[tex]\begin{gathered} x-6y-5=0 \\ -6y=-x+5 \\ y=\frac{-x+5}{-6} \\ y=\frac{-x}{-6}+\frac{5}{-6} \\ y=\frac{x}{6}-\frac{5}{6}\text{ . . . . .(2)} \end{gathered}[/tex]By comparing the equation (2) with the equation (1), the slope of the first line is,
[tex]m_1=\frac{1}{6}[/tex]To find slope of line 2:
The slope of line 2 which is perpendicular to line 1 can be calculated as,
[tex]\begin{gathered} m_1\times m_2=-1 \\ \frac{1}{6}\times m_2=-1 \\ m_2=-1\times6 \\ m_2=-6 \end{gathered}[/tex]To find equation of line in point slope form:
The general formula of point slope form is,
[tex]y-y_1=m(x-x_1)\text{ . . . . . (3)}[/tex]Consider the given coordinate as,
[tex](x_1,y_1)=(8,-8)[/tex]Substitute the value of coordinate and the slope in equation (3),
[tex]\begin{gathered} y-(-8)=-6(x-8) \\ y+8=-6(x-8)\text{ . . . .(4)} \end{gathered}[/tex]To find equation of line in general form:
The general formula of general form is,
[tex]ax+by+c=0[/tex]Then, equation (4) can be rewritten as,
[tex]\begin{gathered} y+8=-6x+48 \\ 6x+y+8-48=0 \\ 6x+y-40=0 \end{gathered}[/tex]Hence,
The equation of line in point slope form is y +8 = -6(x-8).
The equation of line in general form is 6x + y -40 = 0.
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