Changing the way you learn

Question	Answer
\(y=mx+c\)...\[y-y_1=m(x-x_1)\]...\[ax+by+c=0\]	equation of a line
\[\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]	distance between two points
\[(\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2})\]	midpoint of a line segment
\[\frac{y_2-y_1}{x_2-x_1}\]	slope of a line
\[\frac{1}{2}.b.h\]	area of a triangle (1)
\[\frac{1}{2} \| x_1.y_2 - x_2.y_1 \|\]	area of a triangle (2) one point must be (0,0)
\[d=\frac{ \| ax_1+by_1+c \| }{\sqrt{a^2+b^2}}\]	perpendicular distance from a line to a point
\[d=\frac{ \| ax_1+by_1+c \| }{\sqrt{a^2+b^2}}\]	perpendicular distance from a line to a point
\[c=(\frac{hx_2+kx_1}{h+k} , \frac{hy_1+ky_2}{h+k})\]	internal division in a line segment
\[(\frac{x_1+x_2+x_3}{3}),(\frac{y_1+y_2+y_3}{3})\]	centroid of a triangle

Formulae

	Created by cian.buckley over 10 years ago