COORDINATE GEOMETRY · HL

Distance & Slope

The distance and slope formulae, with parallel and perpendicular lines.

Section 1 of 5

Distance

$|AB| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

(i) $A(2,3)$ $B(5,4)$ — find $|AB|$

$(2,3) \to x_1,y_1 \qquad (5,4) \to x_2,y_2$

$\sqrt{(5-2)^2 + (4-3)^2}$

$\sqrt{3^2 + 1^2} = \sqrt{10}$

(ii) $P(3,5)$ $Q(-4,5)$

$(3,5) \to x_1,y_1 \qquad (-4,5) \to x_2,y_2$

$\sqrt{(-4-3)^2 + (5-5)^2}$

$\sqrt{49} = 7$

(iii) $(3,-5)$ $(-1,4)$

$\sqrt{(-1-3)^2 + (4+5)^2}$

$\sqrt{97}$

(iv) $(-4,3)$ $(-7,-1)$

$\sqrt{(-7+4)^2 + (-1-3)^2}$

$= 5$

(v) Distance from $(3,1)$ to $(5,8)$

$(3,1) \to x_1,y_1 \qquad (5,8) \to x_2,y_2$

$\sqrt{(5-3)^2 + (8-1)^2} = \sqrt{53}$

Section 2 of 5

Slope

Rate of change $= \dfrac{dy}{dx}$

$m = \dfrac{\text{rise}}{\text{run}}$

Remember

1.Run $=$ left to right.

2.$m = \dfrac{y_2-y_1}{x_2-x_1}$

(i) $A(1,2)$ and $B(3,5)$ — find slope

$m = \dfrac{3}{2}$

(ii) Slope of $(3,4)$ to $(5,9)$

$(3,4) \to x_1,y_1 \qquad (5,9) \to x_2,y_2$

$m = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{9-4}{5-3}$

$= \dfrac{5}{2}$

Section 3 of 5

Midpoint, distance & slope together

For each pair find (i) Midpoint (ii) Distance (iii) Slope.

(a) $P(3,-1)$ $Q(2,5)$

$(3,-1) \to x_1,y_1 \qquad (2,5) \to x_2,y_2$

Midpoint: $\left(\dfrac{3+2}{2},\ \dfrac{-1+5}{2}\right) = \left(\dfrac{5}{2},\ 2\right)$

Distance: $\sqrt{(2-3)^2 + (5+1)^2} = \sqrt{37}$

Slope: $m = \dfrac{5+1}{2-3} = \dfrac{6}{-1} = -6$

(b) $(-3,4)$ $(5,-2)$

$(-3,4) \to x_1,y_1 \qquad (5,-2) \to x_2,y_2$

Midpoint: $\left(\dfrac{-3+5}{2},\ \dfrac{4+(-2)}{2}\right) = (1,\ 1)$

Distance: $\sqrt{(5+3)^2 + (-2-4)^2} = 10$

Slope: $m = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{-2-4}{5--3} = -\dfrac{3}{4}$

(c) $A(3,5)$ $B(2,7)$

$(3,5) \to x_1,y_1 \qquad (2,7) \to x_2,y_2$

Midpoint: $\left(\dfrac{3+2}{2},\ \dfrac{5+7}{2}\right) = \left(\dfrac{5}{2},\ 6\right)$

Distance: $\sqrt{(2-3)^2 + (7-5)^2} = \sqrt{5}$

Slope: $m = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{7-5}{2-3} = -2$

(d) $(3,-5)$ $(-2,7)$

$(3,-5) \to x_1,y_1 \qquad (-2,7) \to x_2,y_2$

Midpoint: $\left(\dfrac{3+(-2)}{2},\ \dfrac{-5+7}{2}\right) = \left(\dfrac{1}{2},\ 1\right)$

Distance: $\sqrt{(-2-3)^2 + (7+5)^2} = 13$

Slope: $m = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{7+5}{-2-3} = -\dfrac{12}{5}$

Section 4 of 5

Parallel & perpendicular

Parallel

1.$\parallel$ means same slope.

Perpendicular

1.Right angle $= 90^{\circ} =$ tangent $\perp$.

2.Take first slope and invert and change sign.

Section 5 of 5

Slope table

Given a slope, write its parallel slope (same) and perpendicular slope (invert, change sign).

Slope	Parallel	Perpendicular
$\dfrac{2}{3}$	$\dfrac{2}{3}$	$-\dfrac{3}{2}$
$\dfrac{4}{7}$	$\dfrac{4}{7}$	$-\dfrac{7}{4}$
$-\dfrac{5}{9}$	$-\dfrac{5}{9}$	$\dfrac{9}{5}$
$-\dfrac{1}{6}$	$-\dfrac{1}{6}$	$\dfrac{6}{1} = 6$

SUM

The lot in one box

Distance & slope toolkit

1.Distance $= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

2.Slope $m = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{\text{rise}}{\text{run}}$, run = left to right.

3.Parallel $\Rightarrow$ same slope.

4.Perpendicular $\Rightarrow$ invert and change sign.

End of lesson

Distance & Slope — HL · Mathslive.ie