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Sagot :
Given:
The equation of a straight line is,
[tex]y=7x-7[/tex]The objective is to find,
a) The equation of perpendicular line passes throught the point (-8,5).
b) The equation of parallel line passes throught the point (-8,5).
Explanation:
The general equation of straight line is,
[tex]y=mx+c[/tex]Here, m represents the slope of the straight line and c represents the y intercept.
a)
For perpendicular lines, the prouct of slope of two lines will be (-1).
By comparing the general equation and the given equation the slope value will be,
[tex]m_1=7[/tex]Now, the slope value of perpendicular line can be calculated as,
[tex]\begin{gathered} m_1\times m_2=-1 \\ 7\times m_2=-1 \\ m_2=-\frac{1}{7} \end{gathered}[/tex]Since, the perpendicular line passes through the point (-8,5), the equation of line can be calculated using point slope formula.
[tex]\begin{gathered} y-y_1=m_2(x-x_1)_{} \\ y-5=-\frac{1}{7}(x-(-8)) \\ y-5=-\frac{1}{7}(x+8) \\ y-5=-\frac{x}{7}-\frac{8}{7} \\ y=-\frac{x}{7}-\frac{8}{7}+5 \\ y=-\frac{x}{7}-\frac{8}{7}+\frac{35}{7} \\ y=-\frac{x}{7}+\frac{27}{7} \end{gathered}[/tex]Hence, the equation of perpendicular line is obtained.
b)
For paralle lines the slope value will be equal for both lines.
[tex]m_1=m_3=7[/tex]Since, the parallal line passes through the point (-8,5), the equation of line can be calculated using point slope formula.
[tex]\begin{gathered} y-y_1=m_3(x-x_1) \\ y-5=7(x-(-8)) \\ y-5=7(x+8) \\ y-5=7x+56 \\ y=7x+56+5 \\ y=7x+61 \end{gathered}[/tex]Hence, the equation of parallel line is obtained.
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