MH Board · Chapter

Straight Line

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equal slopes through a common point mean collinear

x = X + h

,

y = Y + k

;

substitute, don't subtract the wrong way

Locked

m = \tan\theta = \frac{y_2 - y_1}{x_2 - x_1}

Parallel and perpendicular linesParallel:

m_1 = m_2

; perpendicular:

m_1 m_2 = -1

(or one vertical, one horizontal)Locked

Point–slope and two-point forms $y - y_1 = m(x - x_1)$ ; through two points, $m$ from them firstLocked
Slope–intercept, intercept and normal forms $y = mx + c$ ; $\frac{x}{a} + \frac{y}{b} = 1$ ;
The general form ax + by + c = 0Slope $-\frac{a}{b}$ , $x$ -intercept $-\frac{c}{a}$ ,
Perpendiculars, medians and altitudesParallel: same slope; perpendicular: slope $-\frac{1}{m}$ . A median goes through a midpoint; an altitude is perpendicular to a sideLocked

Angle between two lines $\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$ gives the acute angleLocked
Distance of a point from a line $\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$
Distance between parallel linesMake the $x$ and $y$ coefficients equal first; then $\frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$
Lines through an intersection $L_1 + kL_2 = 0$ passes through the meeting point of $L_1 = 0$

x\cos\alpha + y\sin\alpha = p

Locked

y

-intercept

-\frac{c}{b}

Locked

and

L_2 = 0

;

find

k

from one more condition

Locked