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Sagot :
Solution:
Given:
[tex]\begin{gathered} y-x=5 \\ \\ \text{Through the point (8,3)} \end{gathered}[/tex]The first line given is y - x = 5
[tex]\begin{gathered} Hence, \\ y=x+5 \end{gathered}[/tex]To get the slope of line 1, we compare the equation with the general equation of a straight line.
[tex]\begin{gathered} y=mx+b \\ \text{where m is the slope} \end{gathered}[/tex]Thus,
[tex]\begin{gathered} y=mx+b \\ y=x+5 \\ \\ \text{Comparing both equations,} \\ m=1 \end{gathered}[/tex]The slope of line 1 is 1.
Since line 2 is perpendicular to line 1, then their slopes are negative reciprocals of one another.
This means the product of their slopes is -1.
[tex]m_1m_2=-1[/tex]Hence, the slope of line 2 is gotten by;
[tex]\begin{gathered} m_1m_2=-1 \\ 1\times m_2=-1 \\ m_2=-\frac{1}{1} \\ m_2=-1 \\ \\ \text{The slope of line 2 is -1} \end{gathered}[/tex]Hence, the equation of the perpendicular line through the point (8,3) will be;
[tex]\begin{gathered} \frac{y-y_1}{x-x_1}=m \\ \text{where;} \\ x_1=8 \\ y_1=3 \\ m=-1 \\ \\ \frac{y-3}{x-8}=-1 \\ \text{Cross multiplying,} \\ y-3=-1(x-8) \\ y-3=-x+8 \\ y=-x+8+3 \\ y=-x+11 \\ y=11-x \end{gathered}[/tex]Therefore, the equation for the perpendicular line passing through the point (8,3) is;
[tex]y=11-x[/tex]
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