Respuesta :
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\
slope=\cfrac{a}{{{ b}}}\qquad negative\implies -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\
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\textit{so if this equation has a slope of -1, a perpendicular line will have}
\\\\\\
\cfrac{-1}{1}\qquad negative\implies \cfrac{1}{1}\qquad reciprocal\implies \cfrac{1}{1}\implies 1[/tex]
so.. what's the equation of a line whose slope is 1 and passes through -2,-2?
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ -2}}\quad ,&{{ -2}})\quad \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies 1 \\\\\\ % point-slope intercept y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad \begin{array}{llll} \textit{plug in the values for } \begin{cases} y_1=-2\\ x_1=-2\\ m=1 \end{cases}\\ \textit{and solve for "y"} \end{array}\\ \left. \qquad \right. \uparrow\\ \textit{point-slope form}[/tex]
so.. what's the equation of a line whose slope is 1 and passes through -2,-2?
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ -2}}\quad ,&{{ -2}})\quad \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies 1 \\\\\\ % point-slope intercept y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad \begin{array}{llll} \textit{plug in the values for } \begin{cases} y_1=-2\\ x_1=-2\\ m=1 \end{cases}\\ \textit{and solve for "y"} \end{array}\\ \left. \qquad \right. \uparrow\\ \textit{point-slope form}[/tex]
Answer:
the equation of the line is equal to
[tex]y=x[/tex]
Step-by-step explanation:
Step 1
Find the slope of the line that is perpendicular to the given line
we know that
If two lines are perpendicular
then
the product of their slopes is equal to minus one
so
[tex]m1*m2=-1[/tex]
we have the given line
[tex]y-4=-(x-6)[/tex]
this is the equation of the line into point-slope form
the slope is equal to
[tex]m1=-1[/tex]
[tex]m1*m2=-1[/tex]
Find m2
[tex]m2=-1/m1[/tex]
[tex]m2=-1/(-1)=1[/tex]
The slope of the line perpendicular to the given line is equal to [tex]1[/tex]
Step 2
Find the equation of the line in slope-intercept form
we know that
the equation of the line in slope-intercept form is equal to
[tex]y=mx+b[/tex]
where
m is the slope
b is the y-intercept
in this problem we have
[tex]m=1[/tex]
point [tex](-2,-2)[/tex]
substitute and find the value of b
[tex]-2=(1)(-2)+b[/tex]
[tex]-2=-2+b[/tex]
[tex]b=0[/tex]
the equation of the line is equal to
[tex]y=x[/tex]
