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Example Question - coordinates

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Finding the General Coordinates of a Point in a Triangle

Let the coordinates of A be $(x_1, y_1, z_1)$ and B be $(x_2, y_2, z_2)$, and let C be $(x_3, y_3, z_3)$. The centroid G of triangle ABC has coordinates given by: $G \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3} \right)$ Given that G is at (1, 1, 1), we have: $1 = \frac{x_1+x_2+x_3}{3}$ $1 = \frac{y_1+y_2+y_3}{3}$ $1 = \frac{z_1+z_2+z_3}{3}$ Given the coordinates of A $(x_1, y_1, z_1) = (3, -5, 7)$ and B $(x_2, y_2, z_2) = (-1, 7, 6)$, we can substitute them into the equations: $1 = \frac{3 - 1 + x_3}{3}$ $1 = \frac{-5 + 7 + y_3}{3}$ $1 = \frac{7 + 6 + z_3}{3}$ Now solve for $x_3, y_3, z_3$: $1 = \frac{2 + x_3}{3}$ -> Multiply both sides by 3 -> $3 = 2 + x_3$ -> $x_3 = 1$ $1 = \frac{2 + y_3}{3}$ -> Multiply both sides by 3 -> $3 = 2 + y_3$ -> $y_3 = 1$ $1 = \frac{13 + z_3}{3}$ -> Multiply both sides by 3 -> $3 = 13 + z_3$ -> $z_3 = -10$ Thus, the coordinates of point C are $(x_3, y_3, z_3) = (1, 1, -10)$.

Calculation of Trigonometric Expression and Centroid Coordinates

\text{For the trigonometric expression:} \sin(105^\circ) + \cos(105^\circ) = \sin(60^\circ+45^\circ) + \cos(60^\circ+45^\circ)\ \text{Using the sine and cosine addition formulas:} \sin(A+B) = \sin{A}\cos{B} + \cos{A}\sin{B} \cos(A+B) = \cos{A}\cos{B} - \sin{A}\sin{B} \text{Substitute } A = 60^\circ \text{ and } B = 45^\circ: \sin(60^\circ)\cos(45^\circ) + \cos(60^\circ)\sin(45^\circ) + \cos(60^\circ)\cos(45^\circ) - \sin(60^\circ)\sin(45^\circ)\ \text{Using known values of sine and cosine for } 60^\circ \text{ and } 45^\circ: \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} + \frac{1}{2}\cdot\frac{\sqrt{2}}{2} + \frac{1}{2}\cdot\frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} \text{Simplify the expression:} \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} + \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} \text{The terms with } \sqrt{6} \text{ cancel out, so we are left with:} \frac{\sqrt{2}}{4} + \frac{\sqrt{2}}{4} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2} \text{For the centroid coordinates:} \text{Let the coordinates of } C \text{ be } (x_C, y_C). \text{ The centroid coordinates (G) can be found by:} G_x = \frac{x_A + x_B + x_C}{3}, G_y = \frac{y_A + y_B + y_C}{3} \text{Substitute the given values with } G (1,1,1): 1 = \frac{3 + (-1) + x_C}{3}, 1 = \frac{-5 + (-7) + y_C}{3} \text{Solve for } x_C \text{ and } y_C: x_C = 3 - (3 + (-1)) = 3 - 2 = 1 y_C = 3 - (-5 -7) = 3 + 12 = 15 \text{So, the coordinates of point C are } (1, 15). \text{For the trigonometric identity:} \cot{(\theta)} + \tan{(\theta)} = 2\csc{(\theta)} \text{Using the identities } \cot{(\theta)} = \frac{1}{\tan{(\theta)}} \text{ and } \csc{(\theta)} = \frac{1}{\sin{(\theta)}}, \text{ we get:} \frac{1}{\tan{(\theta)}} + \tan{(\theta)} = \frac{2}{\sin{(\theta)}} \text{Multiply by } \tan{(\theta)}\sin{(\theta)} \text{ to clear the denominators:} \sin{(\theta)} + \tan^2{(\theta)}\sin{(\theta)} = 2\tan{(\theta)} \text{Use the identity } \sin^2{(\theta)} + \cos^2{(\theta)} = 1 \text{ to express } \tan{(\theta)} \text{ in terms of } \sin{(\theta)}: \tan{(\theta)} = \frac{\sin{(\theta)}}{\cos{(\theta)}}, \text{ so } \sin{(\theta)} + \frac{\sin^3{(\theta)}}{\cos^2{(\theta)}} = 2\frac{\sin{(\theta)}}{\cos{(\theta)}} \text{Multiply by } \cos^2{(\theta)} \text{ to clear the denominators:} \sin{(\theta)}\cos^2{(\theta)} + \sin^3{(\theta)} = 2\sin{(\theta)}\cos{(\theta)} \text{Rearrange and factor out } \sin{(\theta)}: \sin{(\theta)}(\cos^2{(\theta)} - 2\cos{(\theta)} + \sin^2{(\theta)}) = 0 \text{Since } \sin{(\theta)} = 0 \text{ leads to no solution that satisfies the original equation, focus on:} \cos^2{(\theta)} - 2\cos{(\theta)} + \sin^2{(\theta)} = 0 \text{Using } \sin^2{(\theta)} = 1 - \cos^2{(\theta)}: \cos^2{(\theta)} - 2\cos{(\theta)} + (1 - \cos^2{(\theta)}) = 0 -2\cos{(\theta)} + 1 = 0 \cos{(\theta)} = \frac{1}{2} \text{Hence, the general values of } \theta \text{ can be } 60^\circ + n360^\circ \text{ or } 300^\circ + n360^\circ, \text{ where } n \text{ is an integer.}

Analysis of Coordinates and Linear Equation of Points in Cartesian Plane

Given point Q(8, k) and point S(-6, 0), it is known that line PQ is parallel to the x-axis. (a) Since PQ is parallel to the x-axis, the y-coordinates of points P and Q are equal, which means the y-coordinate of point Q is k = 0 (the same as point S). $k = 0$ (b) To find the equation of line PS, we can calculate the slope (m) using the coordinates of points P and S: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 0}{-6 - 8}$ Since the change in y is 0 (the line is horizontal), the slope is: $m = 0$ The general equation of a line is $y = mx + b$. Since the slope m = 0, the equation simplifies to $y = b$ To find b, we use the fact that the line passes through point S(-6, 0): $0 = 0 \cdot (-6) + b$ $b = 0$ The equation of line PS is therefore $y = 0$.

Analysis of a Quadratic Function Graph

To determine the coordinates of point $Q$: Point $Q$ lies on the x-axis, which means the y-coordinate is 0. Set the function $f(x)=x^2+6x-5$ equal to 0 and solve for $x$: $x^2+6x-5=0$ $x=\frac{-6\pm\sqrt{6^2-4(1)(-5)}}{2(1)}$ $x=\frac{-6\pm\sqrt{36+20}}{2}$ $x=\frac{-6\pm\sqrt{56}}{2}$ $x=\frac{-6\pm 2\sqrt{14}}{2}$ $x=-3\pm\sqrt{14}$ Since $Q$ is to the right of the y-axis, we take the positive value: $Q=(x, y)$ $Q=(-3+\sqrt{14}, 0)$ To determine the maximum point $P$: The vertex of a parabola $y=ax^2+bx+c$ is given by the formula $x=-\frac{b}{2a}$. In this function $a=1$ and $b=6$, so: $x_P=-\frac{6}{2(1)}$ $x_P=-3$ Substitute $x_P$ into the function to find $y_P$: $y_P=(x_P)^2+6x_P-5$ $y_P=(-3)^2+6(-3)-5$ $y_P=9-18-5$ $y_P=-14$ So the maximum point $P$ is: $P=(x_P, y_P)$ $P=(-3, -14)$

Coordinates of Madrid on a Graph

El eje horizontal (x) representa los valores negativos a la izquierda del origen y los valores positivos a la derecha del origen. El eje vertical (y) representa los valores positivos por encima del origen y los valores negativos debajo del origen. Para encontrar la ubicación exacta de Madrid, buscamos el punto que representa a Madrid en la gráfica y luego determinamos sus coordenadas (x, y). Observando la gráfica, el punto para Madrid se encuentra en x = -3 y y = -4. Por lo tanto, las coordenadas ordenadas (x, y) para Madrid son (-3, -4).

Finding Point of Intersection of Two Lines

To find the coordinates of the point of intersection of the two lines, we first need the equations of both lines. The equation of the first line is given as $ 2x - 3y + 1 = 0 $. The second line passes through the points S(7,-6) and T(-3,2). We can find the slope of this line (m) using the slope formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] \[ m = \frac{2 - (-6)}{-3 - 7} \] \[ m = \frac{8}{-10} \] \[ m = -\frac{4}{5} \] Next, we will use the point-slope form to write the equation for the second line using point S(7, -6) or T(-3, 2). I'll choose S(7, -6) for this example: \[ y - y_1 = m(x - x_1) \] \[ y - (-6) = -\frac{4}{5}(x - 7) \] \[ y + 6 = -\frac{4}{5}x + \frac{4}{5} \cdot 7 \] \[ y + 6 = -\frac{4}{5}x + \frac{28}{5} \] \[ y = -\frac{4}{5}x + \frac{28}{5} - \frac{30}{5} \] \[ y = -\frac{4}{5}x - \frac{2}{5} \] To find the intersection, we solve the two equations together. Let's write them in the standard form: Equation of line 1: $ 2x - 3y + 1 = 0 $ Equation of line 2: $ \frac{-4}{5}x - y - \frac{2}{5} = 0 $ Now multiply equation 2 by 5 to remove fractions: Equation of line 2: $ -4x - 5y - 2 = 0 $ To solve the system, we can use either substitution or elimination. Let's use elimination: Multiply the first equation by 4 to align coefficients for x: \[ 8x - 12y + 4 = 0 \] \[ -4x - 5y - 2 = 0 \] Add these equations together: \[ 4x - 17y + 2 = 0 \] \[ 2 \rightarrow 8x - 17y = -2 \] Now, we will solve for $ x $ using equation 1 and substitution into equation 2: \[ 2x - 3y + 1 = 0 \rightarrow 2x = 3y - 1 \] \[ x = \frac{3}{2}y - \frac{1}{2} \] Substitute $ x $ in equation $ 8x - 17y = -2 $: \[ 8(\frac{3}{2}y - \frac{1}{2}) - 17y = -2 \] \[ 12y - 4 - 17y = -2 \] \[ -5y = 2 \] \[ y = \frac{-2}{5} \] Now substitute $ y $ back into one of the original equations to find $ x $: \[ 2x - 3\left(\frac{-2}{5}\right) + 1 = 0 \] \[ 2x + \frac{6}{5} + 1 = 0 \] \[ 2x + \frac{6}{5} + \frac{5}{5} = 0 \] \[ 2x = -\frac{6}{5} - \frac{5}{5} \] \[ 2x = -\frac{11}{5} \] \[ x = -\frac{11}{10} \] Therefore, the coordinates of the point of intersection of the two lines are $x = -\frac{11}{10}$ and $y = \frac{-2}{5}$, or more clearly written as $(-1.1, -0.4)$.

Analyzing a Quadrilateral Using Coordinates and Slopes

It seems that you're requesting help with a math problem involving the coordinates of a quadrilateral. Unfortunately, the image is not clear enough for me to provide a definitive answer, but I can still guide you on how to solve this type of problem. We have a quadrilateral ABCD with points A(2,4), B(4,2), C(-1,-3), and D(3,-1). To answer the questions, you need to: a. Find the slope of each side of quadrilateral ABCD: - The slope of a line through points (x1, y1) and (x2, y2) is given by (y2 - y1) / (x2 - x1). For AB: slope = (2 - 4) / (4 - 2) = -2 / 2 = -1 For BC: slope = (-3 - 2) / (-1 - 4) = -5 / -5 = 1 For CD: slope = (-1 - (-3)) / (3 - (-1)) = 2 / 4 = 0.5 For DA: slope = (4 - (-1)) / (2 - 3) = 5 / -1 = -5 b. What type of quadrilateral is formed by the fences? Justify your answer using the slope and/or distance formulas: - To determine the type of quadrilateral, we need to check for parallel sides (equal slopes) and equal lengths (using the distance formula). The distance formula for points (x1, y1) and (x2, y2) is given by √[(x2 - x1)^2 + (y2 - y1)^2]. Find the distance of each side: AB = √[(4 - 2)^2 + (2 - 4)^2] = √[2^2 + (-2)^2] = √(4 + 4) = √8 BC = √[(-1 - 4)^2 + (-3 - 2)^2] = √[(-5)^2 + (-5)^2] = √(25 + 25) = √50 CD = √[(3 - (-1))^2 + (-1 - (-3))^2] = √[4^2 + 2^2] = √(16 + 4) = √20 DA = √[(2 - 3)^2 + (4 - (-1))^2] = √[(-1)^2 + 5^2] = √(1 + 25) = √26 The sides AB and CD are not parallel (their slopes are not equal), and sides BC and DA are also not parallel. Also, none of the side lengths are equal. Without plotting the points, we can't be entirely certain what specific type of quadrilateral it is, but we can confirm that it is not a parallelogram, rectangle, square, or rhombus since none of the sides are equal in length or parallel. It's most likely a general quadrilateral with no special properties regarding its sides or angles. Keep in mind that the calculations for the distances of sides AB, BC, CD, and DA should be recalculated with proper care for precision and to ensure accuracy in the answer. If the points were plotted on a graph, the shape could also be visually identified and confirmed.

Determining Points with Vertical Distance

The question asks to select the coordinates for two points that have a vertical distance between them of 2 units. Let's evaluate the provided options: A) (2, 1) - To have a vertical distance of 2 units, we would look for a point that is either 2 units above (i.e., (2, 3)) or 2 units below (i.e., (2, -1)) this point. However, no such points are listed in the options. B) (1, 0) - Following the same logic as above, we need a point that is (1, 2) or (1, -2). No points with x = 1 and y = 2 or -2 are provided. C) (1, 0) - This option is repeated in the list; therefore, it does not provide a pair of points by itself, so it doesn't satisfy the condition. D) (1, 2) - One way to have a vertical distance of 2 units from this point is to have the point (1, 0), which is also listed in the options. E) (2, 1) - This is the same as option A and doesn't provide a pair with a vertical distance of 2 units. Based on this evaluation, the points that have a vertical distance between them of 2 units are (1,2) and (1,0), which is option B matched with option D.

Finding Turning Point and Line of Symmetry of Quadratic Function

The question presented in the image is asking for the coordinates of the turning point (vertex) and the line of symmetry on the graph of the quadratic function $ y = 2x^2 + 2x - 1 $. To find the turning point, we first find the derivative of the function (to find the x-coordinate where the slope of the tangent is zero) and then substitute this x-coordinate back into the original function to get the y-coordinate. The derivative of $ y = 2x^2 + 2x - 1 $ with respect to $ x $ is: \[ \frac{dy}{dx} = 4x + 2 \] Setting the derivative equal to zero to find the x-coordinate of the turning point: \[ 4x + 2 = 0 \] \[ 4x = -2 \] \[ x = -\frac{1}{2} \] Substitute $ x = -\frac{1}{2} $ into the original equation to get the y-coordinate: \[ y = 2(-\frac{1}{2})^2 + 2(-\frac{1}{2}) - 1 \] \[ y = 2(\frac{1}{4}) - 1 - 1 \] \[ y = \frac{1}{2} - 2 \] \[ y = -\frac{3}{2} \] So the turning point's coordinates are $ (-\frac{1}{2}, -\frac{3}{2}) $. The line of symmetry for a parabola given by a quadratic function is always a vertical line that passes through the x-coordinate of the vertex. Thus, the line of symmetry for this graph is $ x = -\frac{1}{2} $.