MA005: Calculus I

Practice 1: $Length = Dist( –7, –2 ) = | (–7) – (–2) | = | –5 | = 5$.

The midpoint is at $\dfrac{(–7) + (–2)}{2} =\dfrac{–9}{2} = – 4.5$.

Practice 2: $Dist(P,Q) = Dist(P,r)$ so $\sqrt{(x – 1)^2 + (y + 4)^2} = \sqrt{(x – 0)^2 + (y + 3)^2}$.

Squaring each side and simplifying, we eventually have $y = x – 4$ .

Practice 3: The point $P = ( x , y)$ is on the circle when it is 5 units from the center $C = ( –2, 6)$ so $Dist(P,C) = 5$. Then $Dist( (x,y) , (–2,6) ) = 5$ so

$\sqrt{(x + 2)^2 + (y – 6)^2} = 5$ or $(x + 2)^2 + (y – 6)^2 = 25$.

Practice 4: $∆x = 5 – (–3) = 8$, $∆y = –14 – 2 = –16$ , and $slope = \dfrac{∆y}{∆x} = \dfrac{–16}{8} = – 2$.

Practice 5: $slope =\dfrac{∆y}{∆x} = \dfrac{ (–3 + 5h) – (–3)}{(2 + h) – 2} = \dfrac{5h}{h} = 5$.

The midpoint is at $( \dfrac{ (2) + (2 + h)}{2} , \dfrac{(–3 + 5h) + (–3)}{2} ) = ( 2 +\dfrac{h}{2} , –3 + \dfrac{5h}{2})$.

Practice 6: $slope =\dfrac{∆y}{∆x} = \dfrac{(3a^2 + 5a) – (3x^2 + 5x)}{a – x}$

$= \dfrac{ 3(a^2 – x^2) + 5(a – x)}{a – x} = \dfrac{3(a + x)(a – x) + 5(a – x)}{a – x} = 3(a + x) + 5$.

Practice 7: Let $y_1 = mx_1 + b$ and $y_2 = mx_2 + b$ . Then

$slope =\dfrac{∆y}{∆x} = \dfrac{(mx_2 + b) – (mx_1 + b)}{x_2 – x_1} = \dfrac{m(x_2 – x_1)}{x_2 – x_1 } = m$.

Practice 8: The line $3x + 5y = 17$ has slope $\dfrac{–3}{5}$ so the slope of the parallel line is $m = \dfrac{–3}{5}$.

Using the form $y = \dfrac{–3}{5} x + b$ and the point ( –2, 3) on the line, we have

$3 = \dfrac{–3}{5} (–2) + b$ so $b = \dfrac{9}{5}$ and

$y = \dfrac{–3}{5} x + \dfrac{9}{5}$ or $5y + 3x = 9$ ..

Practice 9: The line $3y – 7x = 2$ has slope $\dfrac{7}{3}$ so the slope of the perpendicular line is $m= \dfrac{–3}{7}$.

Using the form $y = \dfrac{–3}{7} x + b$ and the point ( 2, –5) on the line, we have $–5 =\dfrac{–3}{7} (2) + b$ so $b = \dfrac{–29}{7}$ and $y =\dfrac{–3}{7} x + \dfrac{–29}{7}$ or $7y + 3x = –29$ .