Example Question - intersection points
Here are examples of questions we've helped users solve.
Graphing Linear Equations to Find Intersection Points
To graph each pair of lines from the provided sets A, B, and C, follow these steps for each set: **Set A:** Equations: 1. \(y = 2x + 1\) 2. \(y = -x + 7\) For each equation, find two points by choosing values for \(x\) and calculating the corresponding \(y\) value. Then plot these points on a graph and draw a line through them. Equation 1: Let \(x = 0\), then \(y = 2(0) + 1 = 1\). So, one point is (0,1). Let \(x = 1\), then \(y = 2(1) + 1 = 3\). Another point is (1,3). These two points will help you graph the first line. Equation 2: Let \(x = 0\), then \(y = -(0) + 7 = 7\). One point is (0,7). Let \(x = 1\), then \(y = -(1) + 7 = 6\). Another point is (1,6). These two points will help you graph the second line. **Set B:** Equations: 1. \(2y + 8 = x\) 2. \(4y - 2x = -16\) First, rearrange these equations into slope-intercept form (\(y = mx + b\)). Equation 1: \(x = 2y + 8\) can be written as \(y = \frac{1}{2}x - 4\) Equation 2: \(4y - 2x = -16\) can be simplified to \(2y - x = -8\), which can be written as \(y = \frac{1}{2}x + 4\) Now, find two points for each of these equations using the method shown in set A, and graph them. **Set C:** Equations: 1. \(6x - 2y = 20\) 2. \(y + 3x - 5 = 0\) First, rearrange these equations into slope-intercept form. Equation 1: \(6x - 2y = 20\) can be simplified to \(3x - y = 10\), which can be written as \(y = 3x - 10\) Equation 2: \(y + 3x - 5 = 0\) can be rearranged to \(y = -3x + 5\) Again, find two points for each equation and graph each line. Please note that I cannot directly graph them here, but this explanation should help you plot them on your own graph paper. Once graphed, you can identify the intersection point of each pair of lines, which is the solution to each set of simultaneous equations.
Calculating Minimal Area of Triangle
The image contains a problem written in Italian. Here's the translation and the solution: "Determine the points A and B of intersection of the line with equation 2x + 2y + 5 = 0 with the curve of equation xy = 1. Consider a point P of the curve belonging to the first quadrant and find for which position of P the area of triangle ABP is minimal." Let's solve this step-by-step. First, determine points A and B where the line intersects the curve: The line is given by: 2x + 2y + 5 = 0 We can solve for y to see the relationship more clearly: y = -(2x + 5)/2 Now consider the curve: xy = 1 Substitute the expression for y from the line into the curve: x * (-(2x + 5)/2) = 1 -(2x^2 + 5x) = 2 2x^2 + 5x - 2 = 0 To find the roots of this quadratic equation, we'll use the quadratic formula x = [-b ± sqrt(b^2 - 4ac)] / (2a), with a=2, b=5, and c=-2: The discriminant is ∆ = b^2 - 4ac = 5^2 - 4*2*(-2) = 25 + 16 = 41 The roots are: x_A = [-5 + sqrt(41)] / 4 x_B = [-5 - sqrt(41)] / 4 Since the curve is in the first quadrant, we will only consider the positive root for the x-coordinate of point A, as the negative root will result in a point in the third quadrant. So, point A has coordinates: x_A = [-5 + sqrt(41)] / 4 We can find the y-coordinate of A by substituting x_A back into the equation y = -(2x + 5)/2: y_A = -[(2 * [-5 + sqrt(41)] / 4) + 5] / 2 Now, for point B, since it's the intersection with the negative root: x_B = [-5 - sqrt(41)] / 4 y_B is found by substituting x_B into y's expression: y_B = -[(2 * [-5 - sqrt(41)] / 4) + 5] / 2 To determine for which position of P the area of triangle ABP is minimal, we must first find a general expression for the area of the triangle with vertices A, B, and a generic point P(x,1/x) which belongs to the curve xy=1 in the first quadrant. The area of a triangle given three points A(x1,y1), B(x2,y2), and P(x3,y3) can be found using the Shoelace formula: Area = 1/2 * |x1(y2-y3) + x2(y3-y1) + x3(y1-y2)| Substitute the coordinates of A, B, and P into this formula to get the area of triangle ABP as a function of x (the x-coordinate of P). Therefore the expression will involve x, and we would then need to differentiate this area function with respect to x and set the derivative equal to zero to find the value of x that minimizes the area. Solving the derivative equation would give us the x-coordinate of P that minimizes the area. We would also want to ensure that it is indeed a minimum by checking the second derivative or using a sign-chart argument. Due to the complexity of algebraic calculations involved, I'm summarizing the steps. You can perform these detailed calculations to find the exact expression for the area, its derivative, and solve for the points that minimize the area.
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