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Sol 1 Using the slope-intercept form, $y=mx+5$ where $y=-2$ when $x=8$, so $-2=8m+5$ and therefore $m=-7/8$. Therefore the line has equation $y=-7/8x+5$.

Sol 2 The slope of the line is given by $m=\frac{4-2}{1-(-3)}=1/2$, so its equation is given by $y-4=1/2(x-1)$ or $y=1/2x+7/2$.

Sol 3 Since the two points have the same x-coordinate, the line passing through these points is vertical; so its equation is simply $x=5$.

Sol 4 Solving the equation $2x-5y=20$ for y gives $y=2/5x-4$, so its slope is given by $m_1=2/5$ and therefore a perpendicular line has slope given by $m_2=\frac{-1}{m_1}=-5/2$. Using the point-slope form, we get the equation $y-(-3)=-5/2(x-4)$; and simplifying gives $y=-5/2x+7$.

Sol 5 The perpendicular bisector will pass through the midpoint of the line segment, which is given by $P=((6+2)/2,(-3+5)/2)=(4,1)$. The slope of the line segment is given by $m_1=(-3-5)/(6-2)=-2$, so the slope of the perpendicular bisector will be $m_2=\frac{-1}{m_1}=1/2$. Therefore the perpendicular bisector has the equation $y-1=1/2(x-4)$ or $y=1/2x-1$.

Sol 6 The line through the two given points has slope $m=(3-(-1))/(6-4)=2$, so the line we want will also have slope $m=2$ since the two lines are parallel. Thus its equation is given by $y-(-4)=2(x-7)$ or $y=2x-18$.

Sol 7 The tangent line at the point $(4,-3)$ will be perpendicular to the line segment from the center $(0,0)$ to the point $(4,-3)$. Since this line segment has slope $m=(-3-0)/(4-0)=-3/4$, the tangent line has the slope $m_1=\frac{-1}{m}=4/3$. Since the tangent line passes through the point $(4,-3)$, its equation is $y-(-3)=4/3(x-4)$ or $y=4/3x-25/3$.

Sol 8 Multiplying the first equation by 2 gives $2x-4y=10$, and then subtracting the second equation from the first gives $(2x-4y)-(2x-5y)=10-11$ or $y=-1$. Then substituting back into the first equation gives $x-2(-1)=5$ so $x=3$. Therefore $(3,-1)$ is the point of intersection of the lines.

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Lawrence Marx 2002-07-12