Geometria Analitica Conamat Ejercicios Resueltos -

The article includes theory reminders, step-by-step solved problems, and practical tips. Analytic geometry combines algebra and geometry to study geometric figures using coordinates and equations. It is essential for understanding lines, circles, parabolas, ellipses, and hyperbolas.

: [ d = \sqrt(7 - 3)^2 + (5 - 2)^2 = \sqrt4^2 + 3^2 = \sqrt16 + 9 = \sqrt25 = 5 ]

: ( y = -3x + 11 ) 5. Equation of a Circle (Center and Radius) Standard form : [ (x - h)^2 + (y - k)^2 = r^2 ] Center ( C(h, k) ), radius ( r ). ✅ Solved Exercise 5 Find the equation of the circle with center ( C(3, -2) ) and radius ( r = 4 ). geometria analitica conamat ejercicios resueltos

: ( m = 2 ) 4. Equation of a Line (Point-Slope Form) Formula : [ y - y_1 = m(x - x_1) ] ✅ Solved Exercise 4 Find the line equation with slope ( m = -3 ) passing through ( (2, 5) ).

: ( d = 5 ) 2. Midpoint of a Segment Formula : [ M = \left( \fracx_1 + x_22, \fracy_1 + y_22 \right) ] ✅ Solved Exercise 2 Find the midpoint of ( P(-2, 4) ) and ( Q(6, -8) ). : [ d = \sqrt(7 - 3)^2 +

Below, you will find covering the most common topics, explained step by step. 1. Distance Between Two Points Formula : [ d = \sqrt(x_2 - x_1)^2 + (y_2 - y_1)^2 ] ✅ Solved Exercise 1 Find the distance between ( A(3, 2) ) and ( B(7, 5) ).

: [ M_x = \frac-2 + 62 = \frac42 = 2, \quad M_y = \frac4 + (-8)2 = \frac-42 = -2 ] : ( m = 2 ) 4

: ( (x - 3)^2 + (y + 2)^2 = 16 ) 6. Circle from General Form to Standard Form ✅ Solved Exercise 6 Convert ( x^2 + y^2 - 6x + 4y - 3 = 0 ) to standard form and find center and radius.

: Center ( (1, -2) ), ( a^2 = 25 \implies a = 5 ), ( b^2 = 9 \implies b = 3 ). Vertices: ( (1 \pm 5, -2) ) → ( (6, -2) ) and ( (-4, -2) ). ( c = \sqrta^2 - b^2 = \sqrt25 - 9 = 4 ). Foci: ( (1 \pm 4, -2) ) → ( (5, -2) ) and ( (-3, -2) ). 10. Hyperbola (Horizontal Transverse Axis) Equation : [ \frac(x - h)^2a^2 - \frac(y - k)^2b^2 = 1 ] Center ( (h, k) ), vertices ( (h \pm a, k) ), foci ( (h \pm c, k) ), ( c^2 = a^2 + b^2 ). ✅ Solved Exercise 10 Find center, vertices, foci of ( \frac(x - 2)^216 - \frac(y + 1)^29 = 1 ).

: ( M(2, -2) ) 3. Slope of a Line Formula : [ m = \fracy_2 - y_1x_2 - x_1 ] ✅ Solved Exercise 3 Find the slope through ( A(1, 3) ) and ( B(4, 9) ).

: [ y - 5 = -3(x - 2) \implies y - 5 = -3x + 6 \implies y = -3x + 11 ]