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Sagot :
Answer:
To find the equation of the line passes through (-5,4) and perpendicular to the line y=3/4 x +4
The slope of the line y=3x/4 +4 is 3/4
Let m be the slope of the required equation of the line,
Since these two lines are perpendicular to each other.
we know that,
Product of the slopes of the perpendicular lines is -1
That is,
[tex]\frac{3}{4}\times m=-1[/tex][tex]m=-\frac{4}{3}[/tex]Slope of the required line is -4/3
Passes through the point (-5,4)
The equation of the line passes through the point (x1,y1) and m as its slope is
[tex](y-y1)=m(x-x1)[/tex]Substitute x1=-5, y1=4 and m=-4/3
we get,
[tex](y-4)=-\frac{4}{3}(x+5)[/tex][tex]y-4=-\frac{4}{3}x-\frac{20}{3}[/tex][tex]y=-\frac{4}{3}x-\frac{20}{3}+4[/tex][tex]y=-\frac{4}{3}x+\frac{(-20)+12}{3}[/tex][tex]y=-\frac{4}{3}x-\frac{8}{3}[/tex]The required slope-intercept form of the equation of the line is y=-4/3 x-8/3.
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