Created by cian.buckley
about 11 years ago
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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 |
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