Question 884621
Find the value of the derivative at the point.
The derivative is equal to the slope of the tangent line.
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{{{x^2+y^2-2x+4y-20=0}}}
{{{2xdx+2ydy-2dx+4dy=0}}}
{{{2ydy+4dy=2dx-2xdx}}}
{{{(2y+4)dy=(2-2x)dx}}}
{{{dy/dx=(2-2x)/(2y+4)=(1-x)/(y+2)}}}
At (5,1),
{{{m=dy/dx=(1-5)/(1+2)}}}
{{{m=-4/3}}}
Using the point slopt form of a line,
{{{y-1=-(4/3)(x-5)}}
{{{y=-(4/3)x+20/3+3/3}}}
{{{highlight(y=-(-4/3)x+23/3)}}}
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{{{drawing(300,300,-10,10,-10,10,circle(5,1,0.3),circle(1,-2,5),graph(300,300,-10,10,-10,10,-(4/3)x+23/3))}}}