CAS View Supported Geometry Commands

CAS View Supported Geometry Commands

From GeoGebra 4.9.170.0 on, the CAS View supports exact versions of the following Geometry Commands.

Exact Calculations

Command	Evaluate	Numeric or Input, Rounding 2 Decimal Places
Angle[(1,0),(0,0),(1,2)]	arctan \left( 2 \right)	Numeric : 1.11 Input : 63.43° or 1.11 rad according Angle Unit selected
AngleBisector[(0,1),(0,0),(1,0)]	y = x	Numeric : y = x Input : - 0.71 x +0.71 y = 0
Circumference[x^2+y^2=1/sqrt(π)]	2 \; \sqrt{\pi \; \sqrt{\pi}}	4.72
Distance[(0,0), x + y = 1] Simplify[Distance[(0,0), x+y=1]]	\frac{1}{\sqrt{2}} \frac{\sqrt{2}}{2}	0.71
Distance[(0,0),x+2y=4] Simplify[Distance[(0,0),x+2y=4]]	\frac{4}{\sqrt{5}} 4 \cdot \frac{\sqrt{5}}{5}	1.79
Distance[(0,4),y=x^2] Simplify[Distance[(0,4),y=x^2]]	\sqrt{ \left( \frac{7}{2} - 4 \right)^{2} + \left( -\frac{1}{2} \; \sqrt{14} \right)^{2}} \frac{\sqrt{15}}{2}	1.94
Distance[(0.5,0.5),x^2+y^2=1] Simplify[ Distance[(0.5,0.5),x^2+y^2=1]]	\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} \; \sqrt{ \left( -\sqrt{2} + 1 \right) \; \left( -\sqrt{2} + 1 \right) \; \sqrt{2} \; \sqrt{2}} \frac{-\sqrt{2} + 2}{2}	0.29
Ellipse[(2,1),(5,2),(5,1)]	28 \; x^{2} - 24 \; x \; y - 160 \; x + 60 \; y^{2} - 96 \; y + 256 = 0	Numeric : 28 \; x^{2} - 24 \; x \; y - 160 \; x + 60 \; y^{2} - 96 \; y + 256 = 0 Input : 7 \; x^{2} - 6 \; x \; y + 15 \; y^{2} - 40 \; x + - 24 \; y = - 64
Ellipse[(2,1),(5,2),(6,1)]	32 \; x^{2} \; \sqrt{2} + 36 \; x^{2} - 224 \; x \; \sqrt{2} - 24 \; x \; y - 216 \; x \; ... \; ... + 32 \; \sqrt{2} \; y^{2} - 96 \; \sqrt{2} \; y + 256 \; \sqrt{2} + 68 \; y^{2} - 120 \; y + 196 = 0	Numeric : 81.25 \; x^{2} - 24 \; x \; y - 532.78 \; x + 113.25 \; y^{2} - 255.76 \; y + 558.04 = 0 Input : 81.25 \; x^{2} - 24 \; x \; y - 532.78 \; x + 113.25 \; y^{2} - 255.76 \; y = - 558.04
Radius[x^2+y^2=1/sqrt(π)]	\frac{\sqrt{\pi \; \sqrt{\pi}}}{\pi}	0.75

Symbolic Computations

Command	Evaluate	Numeric
Circle[(a,b),r]	(y - b)² + (x - a)² = r²
Distance[(a,b),(c,d)]	\sqrt{ \left( b - d \right)^{2} + \left( a - c \right)^{2}}	\sqrt{a^{2} - 2 \; a \; c + b^{2} - 2 \; b \; d + c^{2} + d^{2}}
Distance[(a,b),p x + q y = r]
Line[(a,b),(c,d)]	y = \frac{x}{a - c} \; \left( b - d \right) + \frac{1}{a - c} \; \left( a \; d - b \; c \right)	y = \frac{a \; d - b \; c + b \; x - d \; x}{a - c}
Line[(a,b),y=p x+q]	y = p x - a p + b	y = -a p + b + p x
MidPoint[(a,b),(c,d)]	\left( \frac{a + c}{2}, \frac{b + d}{2} \right)	\left( 0.5 \; a + 0.5 \; c, 0.5 \; b + 0.5 \; d \right)
PerpendicularBisector[(a,b),(c,d)]	y = \frac{-a + c}{b - d} \; x + \frac{a^{2} + b^{2} - c^{2} - d^{2}}{2 \; b - 2 \; d}	y = \frac{a^{2} - 2 \; a \; x + b^{2} - c^{2} + 2 \; c \; x - d^{2}}{2 \; b - 2 \; d}