💡Okay, so I have this problem about using the oblique axonometric method to draw a perspective diagram of a triangle. The question is asking for the ratio of the area of the perspective triangle to the area of the original triangle. The options are A: √2/4, B: √3/4, C: √2/2, and D: √3/2.First, I need to recall what oblique axonometric projection is. From what I remember, it's a type of parallel projection where the object is rotated around one or more of its axes relative to the plane of projection. This method is often used in technical drawing to show three-dimensional objects in two dimensions while maintaining some of their original proportions.In oblique projection, typically, one face of the object is kept parallel to the projection plane, and the receding lines (those going into the depth) are drawn at an angle, usually 45 degrees, and often scaled down. This scaling is to make the drawing more readable and less distorted.Now, the problem mentions that when using this method, the shape obtained is still a triangle. That makes sense because if the original figure is a triangle, its projection should also be a triangle, just scaled or skewed.The key here is to figure out how the area changes. I think in oblique projection, the scaling factor affects the area. If one dimension is scaled by a certain factor, the area, which is two-dimensional, will be scaled by the square of that factor. But wait, in oblique projection, usually, only one axis is scaled, not both. So maybe the area scaling factor is just the scaling factor of that one axis.Let me think. In oblique axonometric projection, typically, the scaling factor for the depth (usually the z-axis) is less than 1. For example, in cavalier projection, the scaling factor is 1, which doesn't reduce the depth, but that can cause some distortion. In cabinet projection, the scaling factor is 0.5, which reduces the depth by half, making the projection less distorted.Wait, so if the scaling factor is 0.5, then the area would be scaled by 0.5 as well because only one dimension is scaled. Let me verify that.Suppose the original triangle has a base 'b' and height 'h'. The area is (1/2)*b*h. After projection, the base remains the same if it's parallel to the projection plane, but the height (or depth) is scaled by 0.5. So the projected height would be 0.5*h. Therefore, the area of the projected triangle would be (1/2)*b*(0.5*h) = (1/2)*(1/2)*b*h = (1/4)*b*h. So the ratio of the projected area to the original area is (1/4), which is 0.25.But wait, none of the options are 1/4. The options are all square roots over 4 or 2. Hmm. Maybe I'm missing something here.Alternatively, perhaps the scaling factor isn't 0.5 but something else. Maybe it's related to the angle used in the projection. If the angle is 45 degrees, the scaling factor might be cos(45°) or sin(45°), which is √2/2.Let me think again. If the projection is done at an angle θ, then the scaling factor for the depth would be cosθ. If θ is 45°, then cos45° is √2/2. So the scaling factor is √2/2, not 0.5.Therefore, the projected height would be (√2/2)*h. Then the area would be (1/2)*b*(√2/2)*h = (√2/4)*b*h. So the ratio of the projected area to the original area is √2/4.Looking back at the options, that's option A: √2/4.Wait, but earlier I thought the scaling factor was 0.5, but that might be specific to cabinet projection. Maybe in this case, the angle is 45°, so the scaling factor is √2/2, leading to the area ratio of √2/4.Alternatively, maybe the problem is referring to a different scaling factor. Let me check.In oblique projections, the scaling factor 'k' is often used, where k is the ratio of the projected depth to the actual depth. For cabinet projection, k is 0.5, and for cavalier, k is 1. But sometimes, other scaling factors are used depending on the desired effect.If the problem doesn't specify the angle or the scaling factor, maybe it's assuming a standard angle, like 45°, which would make the scaling factor √2/2.Alternatively, maybe the area scaling factor is derived from the projection distortion. In axonometric projection, the area scaling factor can be found using the formula involving the angles of the axes.But since it's oblique axonometric, which is a type of parallel projection, the area scaling factor is equal to the scaling factor of the receding axis. So if the receding axis is scaled by 'k', the area is scaled by 'k'.Wait, but area is two-dimensional, so if only one dimension is scaled, the area scales by 'k'. If two dimensions are scaled, it scales by 'k^2'. But in oblique projection, only one dimension (depth) is scaled, so the area scales by 'k'.But in the case of a triangle, if the base is along the projection plane, the base remains the same, and the height (depth) is scaled by 'k'. So the area of the triangle is (1/2)*base*height, so the projected area is (1/2)*base*(k*height). Therefore, the ratio is k.So if k is √2/2, then the ratio is √2/2. But that's option C. Wait, now I'm confused.Wait, maybe I need to consider the angle. If the projection is at 45°, then the scaling factor is cos(45°) = √2/2. So the depth is scaled by √2/2, so the area is scaled by √2/2.But earlier, I thought the area was scaled by (√2/2)^2 = 1/2, but that would be if both dimensions were scaled. But in this case, only one dimension is scaled, so the area scales by √2/2.Wait, let me think again. If I have a rectangle, and I scale one side by 'k', the area scales by 'k'. So for a triangle, it's similar. The base remains the same, the height is scaled by 'k', so the area is scaled by 'k'.But in the problem, it's a triangle, not a rectangle. So if the original triangle has base 'b' and height 'h', the area is (1/2)*b*h. After projection, the base is still 'b', but the height is scaled by 'k', so the projected area is (1/2)*b*(k*h) = k*(1/2)*b*h. Therefore, the ratio is k.So if k is √2/2, then the ratio is √2/2, which is option C.But earlier, I thought the scaling factor was √2/4. Hmm, maybe I made a mistake.Wait, let me think about the projection again. In oblique projection, the scaling factor for the depth is usually cos(theta), where theta is the angle between the projection direction and the projection plane. If theta is 45°, then cos(theta) is √2/2.But the area scaling factor is not just the scaling factor of the depth, but also considering the foreshortening. Wait, no, in oblique projection, the scaling factor is applied directly to the depth, so the area scales by that factor.Wait, maybe I need to consider the projection of the triangle. If the triangle is in 3D space, and we're projecting it onto the plane, the area scaling factor depends on the angle between the triangle's plane and the projection plane.But in oblique axonometric projection, the projection is parallel, so the scaling factor is uniform along the receding axis.Wait, maybe I should think about the determinant of the projection matrix. In parallel projection, the area scaling factor is equal to the determinant of the projection matrix, which for oblique projection is 1 * k, where k is the scaling factor for the depth.So if the scaling factor is √2/2, then the area scales by √2/2.But in the problem, the options are √2/4, √3/4, √2/2, and √3/2.So if the scaling factor is √2/2, then the ratio is √2/2, which is option C.But earlier, I thought the ratio was √2/4, which is option A. So which one is correct?Wait, maybe I need to consider that the height is scaled by √2/2, but the base remains the same, so the area is (base)*(height scaled)/2, so the ratio is (1)*(√2/2)/1 = √2/2.But wait, in the problem, it's a triangle on a plane, so maybe the triangle is in 3D space, and when projected, both the base and height are scaled? No, in oblique projection, only the depth is scaled, not the base.Wait, no, the base is along the projection plane, so it remains the same. The height is perpendicular to the base, but in 3D, it might have a component along the depth. So when projected, the height is scaled by the projection factor.Wait, maybe I need to think in terms of vectors. Let's say the original triangle has vectors along the x and y axes. In oblique projection, the x-axis remains the same, and the y-axis is projected at an angle, scaled by a factor.So if the original triangle has base along x and height along y, then after projection, the base remains the same, and the height is scaled by k, so the area is scaled by k.Therefore, if k is √2/2, the area ratio is √2/2, which is option C.But earlier, I thought the scaling factor was √2/4, but that might have been a mistake.Wait, maybe the problem is considering the projection of a triangle in 3D space, not just scaling one axis. So if the triangle is in a plane that's not parallel to the projection plane, the area scaling factor is the cosine of the angle between the planes.But in oblique axonometric projection, the projection is parallel, so the scaling factor is determined by the angle of the projection.Wait, maybe I need to use the formula for the area scaling factor in parallel projection, which is 1 / cos(theta), where theta is the angle between the projection direction and the normal to the projection plane.But I'm getting confused now.Alternatively, maybe I should look up the standard scaling factor for oblique axonometric projection.Wait, in oblique projection, the scaling factor for the depth is usually 0.5 or 1, but sometimes other values are used. If it's cabinet projection, it's 0.5, which would make the area ratio 0.5.But the options don't have 0.5, but they have √2/2, which is approximately 0.707, and √3/2, which is approximately 0.866.Wait, maybe the scaling factor is related to the angle of the projection. If the projection is at 45°, the scaling factor is cos(45°) = √2/2, so the area ratio is √2/2.Alternatively, if the projection is at 30°, the scaling factor would be cos(30°) = √3/2.But the problem doesn't specify the angle, so maybe it's assuming a standard angle, like 45°, leading to a scaling factor of √2/2.Therefore, the area ratio would be √2/2, which is option C.But wait, earlier I thought the area ratio was √2/4, but that was based on a different assumption.I think I need to clarify this.In oblique projection, the scaling factor 'k' is applied to the depth. The area scaling factor is 'k' because only one dimension is scaled. So if k = √2/2, then the area ratio is √2/2.But if k = 1/2, then the area ratio is 1/2.But the options don't have 1/2, but they have √2/4, which is approximately 0.3535, and √3/4, which is approximately 0.4330.Wait, maybe the scaling factor is applied differently. Maybe the projection is not just scaling the depth, but also involves some foreshortening.Wait, in axonometric projection, the scaling factors for the three axes are determined by the angles between the axes and the projection plane. In oblique axonometric, two axes are usually at 90°, and the third is oblique.But I'm not sure.Alternatively, maybe the area scaling factor is the product of the scaling factors of the two axes. If one axis is scaled by 1 and the other by √2/2, then the area scaling factor is 1 * √2/2 = √2/2.But if both axes are scaled, say, one by 1 and the other by √2/2, then the area is scaled by √2/2.Wait, but in the case of a triangle, if the base is along the x-axis and the height is along the y-axis, and the projection scales the y-axis by √2/2, then the area is scaled by √2/2.Therefore, the ratio is √2/2, which is option C.But earlier, I thought the ratio was √2/4, but that might have been a mistake.Wait, maybe I need to think about the projection matrix. In oblique projection, the projection matrix can be represented as:[1 0 0][0 1 0][α β γ]Where α, β, γ are the scaling factors for the projection direction. But I'm not sure.Alternatively, the area scaling factor can be found by the determinant of the projection matrix. If the projection is parallel, the determinant would be 1 for the in-plane dimensions, and the scaling factor for the depth.But I'm getting too deep into linear algebra, which might not be necessary.Let me try a different approach. Suppose the original triangle has area A. After projection, the area becomes A'. The ratio A'/A is what we need.In oblique projection, the area scaling factor is equal to the scaling factor of the depth, which is usually cos(theta), where theta is the angle between the projection direction and the projection plane.If theta is 45°, then cos(theta) is √2/2, so A' = A * √2/2.Therefore, the ratio is √2/2, which is option C.But wait, in the problem, it's mentioned that the shape obtained is still a triangle. So the projection doesn't distort the shape into a different polygon, which makes sense because the projection is parallel and the original is a triangle.Therefore, the area ratio is determined by the scaling factor of the depth, which is √2/2.So the answer should be option C: √2/2.But wait, in the initial thought process, I thought it was √2/4, but that was based on a different assumption.I think I need to confirm this with an example.Suppose we have a right triangle with legs of length 1 along the x and y axes. The area is 0.5.After oblique projection at 45°, the y-axis is scaled by √2/2, so the projected triangle has legs of 1 and √2/2. The area is (1 * √2/2)/2 = √2/4.Wait, so the area ratio is √2/4, which is option A.But now I'm confused because earlier I thought it was √2/2.Wait, let me recast this.Original area: 0.5.Projected area: (1 * (√2/2))/2 = √2/4.So the ratio is (√2/4)/0.5 = √2/2.Wait, no, the ratio is projected area over original area, which is (√2/4)/0.5 = √2/2.Wait, that's not correct. Let me compute it correctly.Original area: (1*1)/2 = 0.5.Projected area: (1*(√2/2))/2 = (√2/2)/2 = √2/4.So the ratio is (√2/4)/0.5 = (√2/4)/(1/2) = √2/2.Wait, so the ratio is √2/2, which is option C.But in the projected area, it's √2/4, which is half of the original area scaled by √2/2.Wait, I'm getting tangled up.Let me think again.Original area: 0.5.Projected area: (base * height)/2 = (1 * (√2/2))/2 = √2/4.So the ratio is √2/4 divided by 0.5, which is (√2/4)/(1/2) = √2/2.Therefore, the ratio is √2/2, which is option C.But wait, in the problem, the options are A: √2/4, B: √3/4, C: √2/2, D: √3/2.So the correct answer is C: √2/2.But earlier, I thought the projected area was √2/4, which is half of the original area scaled by √2/2.Wait, no, the projected area is √2/4, which is (√2/2)^2 * original area.Wait, no, original area is 0.5, projected area is √2/4, which is approximately 0.3535, which is less than 0.5.So the ratio is 0.3535 / 0.5 = 0.707, which is √2/2.Therefore, the ratio is √2/2, which is option C.But in the initial thought process, I thought it was √2/4, but that was a mistake because I didn't correctly compute the ratio.So the correct answer is C: √2/2.

💡First, I need to determine the elements of set ( B ) by solving the inequality ( x^2 - x leq 0 ).To solve ( x^2 - x leq 0 ), I can factor the quadratic expression:[x(x - 1) leq 0]This inequality holds when ( x ) is between 0 and 1, inclusive. Therefore, ( B = {x | 0 leq x leq 1} ).Next, I look at set ( A = {-1, 0, 1, 2} ) and identify which elements fall within the interval ( [0, 1] ).The elements of ( A ) that satisfy ( 0 leq x leq 1 ) are 0 and 1.Thus, the intersection of sets ( A ) and ( B ) is ( {0, 1} ).

💡Okay, so I have this problem about a sequence {a_n} where the sum of the first n terms, S_n, satisfies S_n = 2a_n - 2. I need to find the first three terms, a_1, a_2, a_3, and then guess the general formula for a_n. After that, I have to prove this formula using mathematical induction. Hmm, let's start with part (1).First, I know that S_n is the sum of the first n terms. So, S_1 should be just a_1. According to the given formula, S_n = 2a_n - 2. So, for n=1, that would be S_1 = 2a_1 - 2. But S_1 is also equal to a_1 because it's just the first term. So, setting them equal: a_1 = 2a_1 - 2. Let me solve this equation.Subtract 2a_1 from both sides: a_1 - 2a_1 = -2, which simplifies to -a_1 = -2. Multiplying both sides by -1 gives a_1 = 2. Okay, so a_1 is 2. That seems straightforward.Now, moving on to a_2. For n=2, S_2 = a_1 + a_2. According to the given formula, S_2 = 2a_2 - 2. So, substituting the known value of a_1, which is 2, we have 2 + a_2 = 2a_2 - 2. Let me solve for a_2.Subtract a_2 from both sides: 2 = a_2 - 2. Then, add 2 to both sides: 4 = a_2. So, a_2 is 4. That makes sense so far.Next, let's find a_3. For n=3, S_3 = a_1 + a_2 + a_3. Using the given formula, S_3 = 2a_3 - 2. Plugging in the known values of a_1 and a_2, we have 2 + 4 + a_3 = 2a_3 - 2. Simplifying the left side: 6 + a_3 = 2a_3 - 2.Subtract a_3 from both sides: 6 = a_3 - 2. Then, add 2 to both sides: 8 = a_3. So, a_3 is 8. Hmm, so the terms so far are 2, 4, 8. That looks like powers of 2. 2 is 2^1, 4 is 2^2, 8 is 2^3. So, maybe the general formula is a_n = 2^n. Let me check if that makes sense.If a_n = 2^n, then S_n, the sum of the first n terms, would be the sum from k=1 to n of 2^k. That's a geometric series with first term 2 and common ratio 2. The sum of the first n terms of a geometric series is S_n = a(r^n - 1)/(r - 1). Plugging in a=2 and r=2, we get S_n = 2(2^n - 1)/(2 - 1) = 2(2^n - 1)/1 = 2^{n+1} - 2.Wait, according to the given formula, S_n = 2a_n - 2. If a_n = 2^n, then 2a_n - 2 = 2*2^n - 2 = 2^{n+1} - 2, which is exactly what we got for S_n. So, that seems to check out. So, my conjecture is that a_n = 2^n.Alright, so part (1) is done. Now, part (2) is to prove this formula using mathematical induction. I remember that mathematical induction has two main steps: the base case and the inductive step.First, the base case. I need to show that the formula holds for n=1. Well, when n=1, a_1 should be 2^1 = 2, which we already found earlier. So, the base case holds.Next, the inductive step. I need to assume that the formula holds for some arbitrary positive integer k, meaning that a_k = 2^k. Then, I have to show that it must also hold for k+1, i.e., a_{k+1} = 2^{k+1}.So, let's assume that a_k = 2^k. Now, I need to find a_{k+1}. I know that S_{k+1} = S_k + a_{k+1}. According to the given formula, S_{k+1} = 2a_{k+1} - 2.But S_k is the sum of the first k terms, which we can express using the formula we have for S_n. Since we're assuming a_n = 2^n up to n=k, then S_k = 2^{k+1} - 2, as we saw earlier.So, substituting into the equation for S_{k+1}:S_{k+1} = S_k + a_{k+1} = (2^{k+1} - 2) + a_{k+1}But we also have S_{k+1} = 2a_{k+1} - 2. Therefore, setting these equal:(2^{k+1} - 2) + a_{k+1} = 2a_{k+1} - 2Let me solve for a_{k+1}. Subtract a_{k+1} from both sides:2^{k+1} - 2 = a_{k+1} - 2Then, add 2 to both sides:2^{k+1} = a_{k+1}So, a_{k+1} = 2^{k+1}, which is exactly what we wanted to show. Therefore, by the principle of mathematical induction, the formula a_n = 2^n holds for all positive integers n.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from S_{k+1} = S_k + a_{k+1}, and S_{k+1} is also equal to 2a_{k+1} - 2. Then, substituting S_k as 2^{k+1} - 2, which is correct because S_k for a geometric series is 2^{k+1} - 2. Then, adding a_{k+1} to that gives 2^{k+1} - 2 + a_{k+1}, which equals 2a_{k+1} - 2. Subtracting a_{k+1} from both sides gives 2^{k+1} - 2 = a_{k+1} - 2, then adding 2 gives 2^{k+1} = a_{k+1}. Yep, that seems correct.So, all in all, the induction step holds, and since the base case is true, the formula a_n = 2^n is valid for all n.I think I've got it. So, summarizing:1. Calculated a_1, a_2, a_3 as 2, 4, 8 respectively, leading to the conjecture that a_n = 2^n.2. Proved by induction that this formula holds for all positive integers n.I don't see any gaps or errors in my reasoning, so I feel confident about this solution.**Final Answer**The general formula for the sequence is boxed{a_n = 2^n}.

💡Okay, so I have this problem here: Given a complex number z equals i times (1 minus i), where i is the imaginary unit, and I need to find the absolute value of z, which is written as |z|. The options are A: 1, B: 1 plus i, C: 2, and D: square root of 2.Alright, let me start by recalling what a complex number is. A complex number is usually written as a + bi, where a is the real part and b is the imaginary part. The absolute value or modulus of a complex number a + bi is given by the square root of (a squared plus b squared). So, |z| equals sqrt(a² + b²). That's the formula I need to use here.But first, I need to find what z is. The problem gives z as i times (1 minus i). So, I need to multiply i by (1 minus i). Let me write that out:z = i * (1 - i)Hmm, okay, so I need to distribute the i across the parentheses. Let's do that step by step. Multiplying i by 1 gives me i, and multiplying i by (-i) gives me -i squared. So, z equals i minus i squared.Wait, i squared is equal to -1, right? Because i is the imaginary unit, and by definition, i squared is -1. So, substituting that in, z equals i minus (-1). Subtracting a negative is the same as adding the positive, so that becomes i plus 1.So, z equals 1 plus i. Now, I can rewrite z as 1 + i, which is in the standard form of a complex number, where the real part is 1 and the imaginary part is also 1.Now, to find the modulus |z|, I can use the formula I remembered earlier. So, |z| equals sqrt(a² + b²), where a is 1 and b is 1.Let me compute that. First, square the real part: 1 squared is 1. Then, square the imaginary part: 1 squared is also 1. Adding those together, 1 plus 1 equals 2. Now, take the square root of 2. So, |z| equals sqrt(2).Looking back at the options, option D is sqrt(2), which matches my result. So, I think the correct answer is D.Wait a second, let me double-check my steps to make sure I didn't make any mistakes. Starting from z = i*(1 - i). Distribute the i: i*1 is i, and i*(-i) is -i². Since i² is -1, that becomes -(-1), which is +1. So, z = i + 1, which is 1 + i. That seems correct.Calculating the modulus: sqrt(1² + 1²) = sqrt(1 + 1) = sqrt(2). Yep, that still checks out. So, I'm confident that the answer is D: sqrt(2).Just to make sure I didn't overlook anything, let me consider if there's another way to approach this problem. Maybe using the property that |ab| = |a||b| for complex numbers a and b. So, |z| = |i*(1 - i)| = |i| * |1 - i|.Calculating |i| is straightforward. Since i is 0 + 1i, its modulus is sqrt(0² + 1²) = 1.Now, |1 - i| is the modulus of the complex number 1 - i. Using the same formula, sqrt(1² + (-1)²) = sqrt(1 + 1) = sqrt(2).So, multiplying these together, |z| = |i| * |1 - i| = 1 * sqrt(2) = sqrt(2). That's the same result as before. So, this alternative method also confirms that the answer is D: sqrt(2).I think I've covered all the bases here. I expanded the expression for z, calculated its modulus directly, and then used the property of moduli to confirm the result. Both methods led me to the same answer, so I feel pretty certain that D is correct.Just to be thorough, let me glance over the other options to see if any of them could make sense or if I might have confused something. Option A is 1, which is the modulus of i, but z isn't just i; it's i times (1 - i), which we saw becomes 1 + i. So, 1 is too small.Option B is 1 + i, which is actually the complex number z itself, not its modulus. The modulus is a real number, not a complex one, so B can't be right.Option C is 2. If I had made a mistake in calculating the squares or the square root, maybe I could have gotten 2. For instance, if I had added 1 and 1 to get 2 and then not taken the square root, that would give me 2. But since the modulus requires taking the square root of the sum, 2 isn't the right answer.So, yeah, D is definitely the correct choice here. I think I've thought it through thoroughly and checked my work in multiple ways, so I'm confident in my answer.

💡Alright, so I have this problem about a parabola and a line intersecting it. Let me try to figure it out step by step. First, the problem says: A line ( l ) passes through the focus ( F ) of the parabola ( C: y^2 = 4x ) and has a slope of ( k ). This line intersects the parabola at points ( A ) and ( B ), and the distance between ( A ) and ( B ) is 8. Part (1) asks for the equation of line ( l ). Okay, so I need to find ( k ) such that the distance between ( A ) and ( B ) is 8. Let me recall some properties of parabolas. The standard form ( y^2 = 4ax ) has its focus at ( (a, 0) ). In this case, ( 4a = 4 ), so ( a = 1 ). Therefore, the focus ( F ) is at ( (1, 0) ). So, the line ( l ) passes through ( (1, 0) ) and has a slope ( k ). The equation of such a line can be written in point-slope form: ( y = k(x - 1) ). Now, this line intersects the parabola ( y^2 = 4x ). To find the points of intersection, I can substitute ( y = k(x - 1) ) into the parabola's equation. Substituting, we get:[ [k(x - 1)]^2 = 4x ]Expanding the left side:[ k^2(x^2 - 2x + 1) = 4x ][ k^2x^2 - 2k^2x + k^2 = 4x ]Bring all terms to one side:[ k^2x^2 - (2k^2 + 4)x + k^2 = 0 ]So, this is a quadratic in ( x ). Let me denote this as:[ k^2x^2 - (2k^2 + 4)x + k^2 = 0 ]Let’s denote the roots of this quadratic as ( x_1 ) and ( x_2 ), which correspond to the x-coordinates of points ( A ) and ( B ). From quadratic theory, we know that:- The sum of the roots ( x_1 + x_2 = frac{2k^2 + 4}{k^2} )- The product of the roots ( x_1x_2 = frac{k^2}{k^2} = 1 )So, ( x_1x_2 = 1 ). Interesting, that might come in handy later.Now, the distance between points ( A ) and ( B ) is given as 8. Let me recall that the distance between two points ( (x_1, y_1) ) and ( (x_2, y_2) ) is:[ sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = 8 ]But since both points lie on the parabola ( y^2 = 4x ), we can express ( y_1 ) and ( y_2 ) in terms of ( x_1 ) and ( x_2 ). Specifically, ( y_1 = k(x_1 - 1) ) and ( y_2 = k(x_2 - 1) ).Let me compute ( (y_2 - y_1) ):[ y_2 - y_1 = k(x_2 - 1) - k(x_1 - 1) = k(x_2 - x_1) ]So, the distance squared is:[ (x_2 - x_1)^2 + (k(x_2 - x_1))^2 = 64 ]Factor out ( (x_2 - x_1)^2 ):[ (x_2 - x_1)^2(1 + k^2) = 64 ]Therefore:[ (x_2 - x_1)^2 = frac{64}{1 + k^2} ]Taking square roots:[ |x_2 - x_1| = frac{8}{sqrt{1 + k^2}} ]But I also know that for a quadratic equation ( ax^2 + bx + c = 0 ), the difference of the roots is given by ( sqrt{(x_1 + x_2)^2 - 4x_1x_2} ). So, let's compute that.We have:[ (x_1 + x_2)^2 - 4x_1x_2 = left( frac{2k^2 + 4}{k^2} right)^2 - 4(1) ]Simplify:[ left( frac{2k^2 + 4}{k^2} right)^2 - 4 = frac{(2k^2 + 4)^2}{k^4} - 4 ]Expand the numerator:[ (2k^2 + 4)^2 = 4k^4 + 16k^2 + 16 ]So,[ frac{4k^4 + 16k^2 + 16}{k^4} - 4 = frac{4k^4 + 16k^2 + 16 - 4k^4}{k^4} = frac{16k^2 + 16}{k^4} = frac{16(k^2 + 1)}{k^4} ]Therefore, ( |x_2 - x_1| = sqrt{frac{16(k^2 + 1)}{k^4}} = frac{4sqrt{k^2 + 1}}{k^2} )But earlier, we had:[ |x_2 - x_1| = frac{8}{sqrt{1 + k^2}} ]So, equate the two expressions:[ frac{4sqrt{k^2 + 1}}{k^2} = frac{8}{sqrt{1 + k^2}} ]Multiply both sides by ( k^2 sqrt{1 + k^2} ):[ 4(k^2 + 1) = 8k^2 ]Simplify:[ 4k^2 + 4 = 8k^2 ][ 4 = 4k^2 ][ k^2 = 1 ]So, ( k = pm 1 )Therefore, the slope ( k ) is either 1 or -1. Thus, the equations of line ( l ) are:- For ( k = 1 ): ( y = 1(x - 1) ) which simplifies to ( y = x - 1 )- For ( k = -1 ): ( y = -1(x - 1) ) which simplifies to ( y = -x + 1 )So, part (1) is solved. The equations are ( y = x - 1 ) and ( y = -x + 1 ).Moving on to part (2). It says: If point ( A ) is symmetric to point ( D ) with respect to the ( x )-axis, prove that line ( BD ) always passes through a fixed point and find the coordinates of that point.Hmm, okay. So, point ( D ) is the reflection of ( A ) over the ( x )-axis. So, if ( A ) is ( (x_1, y_1) ), then ( D ) is ( (x_1, -y_1) ).We need to show that line ( BD ) passes through a fixed point regardless of the position of ( A ) and ( B ), i.e., regardless of the slope ( k ). But wait, in part (1), we found specific slopes ( k = pm 1 ). So, does this mean that for these specific lines, the line ( BD ) always passes through a fixed point?Wait, actually, in part (1), we found that for the distance ( AB = 8 ), the slope ( k ) must be ( pm 1 ). So, perhaps in part (2), we are to consider these specific lines, i.e., when ( k = 1 ) or ( k = -1 ), and show that line ( BD ) passes through a fixed point.Alternatively, maybe part (2) is a general statement regardless of ( k ), but given that in part (1) we found specific ( k ), perhaps it's under the condition that ( AB = 8 ). I need to clarify.But let's proceed. Let me denote:- ( A = (x_1, y_1) )- ( B = (x_2, y_2) )- ( D = (x_1, -y_1) )We need to find the equation of line ( BD ) and show that it passes through a fixed point.First, let me find the coordinates of ( B ) and ( D ):- ( B = (x_2, y_2) )- ( D = (x_1, -y_1) )So, line ( BD ) connects ( (x_2, y_2) ) and ( (x_1, -y_1) ).Let me compute the slope of line ( BD ):Slope ( m = frac{-y_1 - y_2}{x_1 - x_2} )So, the equation of line ( BD ) can be written as:( y - y_2 = m(x - x_2) )Plugging in ( m ):( y - y_2 = frac{-y_1 - y_2}{x_1 - x_2}(x - x_2) )Alternatively, we can write it in parametric form or find the intercept.Alternatively, perhaps using the two-point formula.But maybe a better approach is to find the equation of line ( BD ) and see if it passes through a fixed point regardless of ( x_1, x_2, y_1, y_2 ).But perhaps we can use the properties of the parabola and the given conditions.From part (1), we know that ( x_1x_2 = 1 ) because the product of the roots is 1. Also, from the quadratic equation, ( x_1 + x_2 = frac{2k^2 + 4}{k^2} ). But in part (1), ( k^2 = 1 ), so ( x_1 + x_2 = frac{2(1) + 4}{1} = 6 ). So, ( x_1 + x_2 = 6 ).Also, since ( y_1 = k(x_1 - 1) ) and ( y_2 = k(x_2 - 1) ), and ( k = pm 1 ), so ( y_1 = pm(x_1 - 1) ) and ( y_2 = pm(x_2 - 1) ).But since ( k ) can be positive or negative, perhaps we need to consider both cases. But maybe the fixed point is the same regardless.Alternatively, perhaps we can find a relationship between ( y_1 ) and ( y_2 ).From the equation ( y_1^2 = 4x_1 ) and ( y_2^2 = 4x_2 ), so ( y_1^2 y_2^2 = 16x_1x_2 = 16(1) = 16 ). Therefore, ( y_1 y_2 = pm 4 ). But since ( y_1 = k(x_1 - 1) ) and ( y_2 = k(x_2 - 1) ), multiplying them gives ( y_1 y_2 = k^2(x_1 - 1)(x_2 - 1) ). Since ( k^2 = 1 ), ( y_1 y_2 = (x_1 - 1)(x_2 - 1) ).But we also have ( x_1 + x_2 = 6 ) and ( x_1x_2 = 1 ). So, let me compute ( (x_1 - 1)(x_2 - 1) ):( (x_1 - 1)(x_2 - 1) = x_1x_2 - x_1 - x_2 + 1 = 1 - 6 + 1 = -4 )Therefore, ( y_1 y_2 = -4 ). So, ( y_1 y_2 = -4 ).So, that's a useful relationship.Now, going back to the slope of line ( BD ):( m = frac{-y_1 - y_2}{x_1 - x_2} )But ( y_1 y_2 = -4 ), so perhaps we can express ( -y_1 - y_2 ) in terms of something.Alternatively, let me express ( y_2 ) in terms of ( y_1 ): from ( y_1 y_2 = -4 ), we have ( y_2 = -frac{4}{y_1} ).So, substituting into the slope:( m = frac{-y_1 - (-frac{4}{y_1})}{x_1 - x_2} = frac{-y_1 + frac{4}{y_1}}{x_1 - x_2} )But ( x_1 - x_2 = -(x_2 - x_1) ). From earlier, ( |x_2 - x_1| = frac{8}{sqrt{1 + k^2}} = frac{8}{sqrt{2}} = 4sqrt{2} ). Wait, but in part (1), we found ( k^2 = 1 ), so ( k = pm 1 ). So, ( sqrt{1 + k^2} = sqrt{2} ). Therefore, ( |x_2 - x_1| = frac{8}{sqrt{2}} = 4sqrt{2} ). So, ( x_2 - x_1 = pm 4sqrt{2} ). But depending on which root is larger, but since ( x_1 ) and ( x_2 ) are roots, their order depends on the slope.But perhaps instead of getting bogged down here, let me think differently.Given that ( x_1 + x_2 = 6 ) and ( x_1x_2 = 1 ), we can write ( x_2 = frac{1}{x_1} ). Because ( x_1x_2 = 1 ), so ( x_2 = 1/x_1 ).Similarly, ( y_2 = -4/y_1 ), as we saw earlier.So, let me express everything in terms of ( x_1 ) and ( y_1 ).So, point ( B ) is ( (1/x_1, -4/y_1) ), and point ( D ) is ( (x_1, -y_1) ).So, line ( BD ) connects ( (1/x_1, -4/y_1) ) and ( (x_1, -y_1) ).Let me compute the slope ( m ):( m = frac{-y_1 - (-4/y_1)}{x_1 - 1/x_1} = frac{-y_1 + 4/y_1}{x_1 - 1/x_1} )Simplify numerator and denominator:Numerator: ( -y_1 + 4/y_1 = (-y_1^2 + 4)/y_1 )Denominator: ( x_1 - 1/x_1 = (x_1^2 - 1)/x_1 )So, slope ( m = frac{(-y_1^2 + 4)/y_1}{(x_1^2 - 1)/x_1} = frac{(-y_1^2 + 4)x_1}{y_1(x_1^2 - 1)} )But from the parabola equation, ( y_1^2 = 4x_1 ), so substitute:( m = frac{(-4x_1 + 4)x_1}{y_1(x_1^2 - 1)} = frac{(-4x_1^2 + 4x_1)}{y_1(x_1^2 - 1)} )Factor numerator:( -4x_1(x_1 - 1) )Denominator:( y_1(x_1 - 1)(x_1 + 1) )So, we have:( m = frac{-4x_1(x_1 - 1)}{y_1(x_1 - 1)(x_1 + 1)} )Cancel out ( (x_1 - 1) ) terms (assuming ( x_1 neq 1 ), which it isn't because ( x_1x_2 = 1 ) and ( x_1 + x_2 = 6 ), so ( x_1 ) can't be 1):( m = frac{-4x_1}{y_1(x_1 + 1)} )But ( y_1 = k(x_1 - 1) ), and ( k = pm 1 ). So, ( y_1 = pm(x_1 - 1) ). Let me substitute:( m = frac{-4x_1}{pm(x_1 - 1)(x_1 + 1)} = frac{-4x_1}{pm(x_1^2 - 1)} )But ( x_1^2 - 1 = (x_1 - 1)(x_1 + 1) ), which is in the denominator.Now, let me write the equation of line ( BD ). Using point ( D ) as ( (x_1, -y_1) ):( y - (-y_1) = m(x - x_1) )( y + y_1 = m(x - x_1) )Substitute ( m ):( y + y_1 = frac{-4x_1}{pm(x_1^2 - 1)}(x - x_1) )But this seems a bit messy. Maybe there's a better approach.Alternatively, let me use the two-point form of the line. The line through ( B(1/x_1, -4/y_1) ) and ( D(x_1, -y_1) ).The equation can be written as:( frac{y - (-4/y_1)}{x - 1/x_1} = frac{-y_1 - (-4/y_1)}{x_1 - 1/x_1} )Simplify numerator and denominator:Numerator: ( y + 4/y_1 )Denominator: ( x - 1/x_1 )Slope: as before, ( m = frac{-y_1 + 4/y_1}{x_1 - 1/x_1} )But perhaps instead of going this route, let me consider parametric equations or find a relationship that eliminates ( x_1 ) and ( y_1 ).Alternatively, let me consider that line ( BD ) passes through a fixed point ( (h, k) ). So, for any ( x_1 ), the point ( (h, k) ) must satisfy the equation of line ( BD ).So, let me write the equation of line ( BD ) in terms of ( x_1 ) and ( y_1 ), and then see if there exists ( (h, k) ) that satisfies it for all ( x_1 ).From earlier, the equation is:( y + y_1 = frac{-4x_1}{y_1(x_1 + 1)}(x - x_1) )Let me rearrange this:Multiply both sides by ( y_1(x_1 + 1) ):( y_1(x_1 + 1)(y + y_1) = -4x_1(x - x_1) )Expand left side:( y_1(x_1 + 1)y + y_1^2(x_1 + 1) = -4x_1x + 4x_1^2 )But ( y_1^2 = 4x_1 ), so substitute:( y_1(x_1 + 1)y + 4x_1(x_1 + 1) = -4x_1x + 4x_1^2 )Simplify the right side:( -4x_1x + 4x_1^2 )Left side:( y_1(x_1 + 1)y + 4x_1^2 + 4x_1 )So, bringing all terms to one side:( y_1(x_1 + 1)y + 4x_1^2 + 4x_1 + 4x_1x - 4x_1^2 = 0 )Simplify:( y_1(x_1 + 1)y + 4x_1 + 4x_1x = 0 )Factor out ( 4x_1 ):( y_1(x_1 + 1)y + 4x_1(1 + x) = 0 )Hmm, this is getting complicated. Maybe I need a different approach.Let me consider specific cases. Since ( k = pm 1 ), let's take ( k = 1 ) first.Case 1: ( k = 1 )So, the line is ( y = x - 1 ). Let's find points ( A ) and ( B ).From part (1), the quadratic equation is ( x^2 - 6x + 1 = 0 ). Wait, no, let me recompute.Wait, in part (1), we had ( k^2 = 1 ), so the quadratic was ( x^2 - 6x + 1 = 0 ). So, roots are:( x = [6 pm sqrt(36 - 4)]/2 = [6 pm sqrt(32)]/2 = [6 pm 4sqrt(2)]/2 = 3 pm 2sqrt(2) )So, ( x_1 = 3 + 2sqrt{2} ), ( x_2 = 3 - 2sqrt{2} )Then, ( y_1 = x_1 - 1 = 2 + 2sqrt{2} ), ( y_2 = x_2 - 1 = 2 - 2sqrt{2} )So, point ( A = (3 + 2sqrt{2}, 2 + 2sqrt{2}) ), point ( B = (3 - 2sqrt{2}, 2 - 2sqrt{2}) )Point ( D ) is the reflection of ( A ) over the x-axis: ( D = (3 + 2sqrt{2}, -2 - 2sqrt{2}) )Now, find the equation of line ( BD ).Points ( B(3 - 2sqrt{2}, 2 - 2sqrt{2}) ) and ( D(3 + 2sqrt{2}, -2 - 2sqrt{2}) )Compute the slope ( m ):( m = frac{(-2 - 2sqrt{2}) - (2 - 2sqrt{2})}{(3 + 2sqrt{2}) - (3 - 2sqrt{2})} = frac{-2 - 2sqrt{2} - 2 + 2sqrt{2}}{4sqrt{2}} = frac{-4}{4sqrt{2}} = frac{-1}{sqrt{2}} )So, slope ( m = -1/sqrt{2} )Using point ( B ):( y - (2 - 2sqrt{2}) = -1/sqrt{2}(x - (3 - 2sqrt{2})) )Let me simplify this equation.Multiply both sides by ( sqrt{2} ):( sqrt{2}(y - 2 + 2sqrt{2}) = - (x - 3 + 2sqrt{2}) )Expand:( sqrt{2}y - 2sqrt{2} + 4 = -x + 3 - 2sqrt{2} )Bring all terms to left:( sqrt{2}y - 2sqrt{2} + 4 + x - 3 + 2sqrt{2} = 0 )Simplify:( x + sqrt{2}y + ( -2sqrt{2} + 2sqrt{2} ) + (4 - 3) = 0 )( x + sqrt{2}y + 1 = 0 )So, the equation is ( x + sqrt{2}y + 1 = 0 )Now, does this line pass through a fixed point? Let me see if it passes through ( (-1, 0) ):Plug ( x = -1 ), ( y = 0 ):( -1 + 0 + 1 = 0 ). Yes, it satisfies.So, for ( k = 1 ), line ( BD ) passes through ( (-1, 0) ).Case 2: ( k = -1 )So, the line is ( y = -x + 1 ). Let's find points ( A ) and ( B ).The quadratic equation is the same as in part (1): ( x^2 - 6x + 1 = 0 ). So, roots are ( x = 3 pm 2sqrt{2} )But since ( k = -1 ), ( y = -x + 1 ), so:For ( x_1 = 3 + 2sqrt{2} ), ( y_1 = - (3 + 2sqrt{2}) + 1 = -2 - 2sqrt{2} )For ( x_2 = 3 - 2sqrt{2} ), ( y_2 = - (3 - 2sqrt{2}) + 1 = -2 + 2sqrt{2} )So, point ( A = (3 + 2sqrt{2}, -2 - 2sqrt{2}) ), point ( B = (3 - 2sqrt{2}, -2 + 2sqrt{2}) )Point ( D ) is the reflection of ( A ) over the x-axis: ( D = (3 + 2sqrt{2}, 2 + 2sqrt{2}) )Now, find the equation of line ( BD ).Points ( B(3 - 2sqrt{2}, -2 + 2sqrt{2}) ) and ( D(3 + 2sqrt{2}, 2 + 2sqrt{2}) )Compute the slope ( m ):( m = frac{(2 + 2sqrt{2}) - (-2 + 2sqrt{2})}{(3 + 2sqrt{2}) - (3 - 2sqrt{2})} = frac{2 + 2sqrt{2} + 2 - 2sqrt{2}}{4sqrt{2}} = frac{4}{4sqrt{2}} = frac{1}{sqrt{2}} )So, slope ( m = 1/sqrt{2} )Using point ( B ):( y - (-2 + 2sqrt{2}) = 1/sqrt{2}(x - (3 - 2sqrt{2})) )Simplify:Multiply both sides by ( sqrt{2} ):( sqrt{2}(y + 2 - 2sqrt{2}) = x - 3 + 2sqrt{2} )Expand:( sqrt{2}y + 2sqrt{2} - 4 = x - 3 + 2sqrt{2} )Bring all terms to left:( -x + sqrt{2}y + 2sqrt{2} - 4 + 3 - 2sqrt{2} = 0 )Simplify:( -x + sqrt{2}y -1 = 0 )( x - sqrt{2}y + 1 = 0 )Wait, that's different from the previous case. But let me check if this line passes through ( (-1, 0) ):Plug ( x = -1 ), ( y = 0 ):( -1 - 0 + 1 = 0 ). Yes, it satisfies.So, for ( k = -1 ), line ( BD ) also passes through ( (-1, 0) ).Therefore, regardless of whether ( k = 1 ) or ( k = -1 ), line ( BD ) passes through the fixed point ( (-1, 0) ).Hence, the fixed point is ( (-1, 0) ).**Final Answer**(1) The equations of line ( l ) are (boxed{y = x - 1}) and (boxed{y = -x + 1}).(2) Line ( BD ) always passes through the fixed point (boxed{(-1, 0)}).

💡Okay, so I have this inequality to solve: (3x - 1)/(2 - x) ≥ 1. Hmm, I need to find the values of x that satisfy this inequality. Let me think about how to approach this.First, I remember that when dealing with inequalities involving fractions, I have to be careful about the denominator because it can't be zero, and the sign of the denominator affects the direction of the inequality when multiplying both sides. So, let's note that 2 - x ≠ 0, which means x ≠ 2. That's important because x = 2 is excluded from the solution set.Now, to solve (3x - 1)/(2 - x) ≥ 1, I think I should bring all terms to one side to make it easier to handle. So, I'll subtract 1 from both sides:(3x - 1)/(2 - x) - 1 ≥ 0.Now, I need to combine these terms into a single fraction. To do that, I'll express 1 as (2 - x)/(2 - x) so that both terms have the same denominator:(3x - 1)/(2 - x) - (2 - x)/(2 - x) ≥ 0.Now, subtract the numerators:[(3x - 1) - (2 - x)] / (2 - x) ≥ 0.Let me simplify the numerator:3x - 1 - 2 + x = 4x - 3.So, the inequality becomes:(4x - 3)/(2 - x) ≥ 0.Hmm, this looks a bit simpler. Now, I need to find the values of x where this fraction is greater than or equal to zero. To do that, I can find the critical points where the numerator or denominator is zero because those points divide the number line into intervals where the expression may change its sign.The numerator, 4x - 3, equals zero when x = 3/4.The denominator, 2 - x, equals zero when x = 2.So, the critical points are x = 3/4 and x = 2. These points divide the real number line into three intervals:1. x < 3/42. 3/4 < x < 23. x > 2Now, I need to test each interval to see where the expression (4x - 3)/(2 - x) is positive or negative.Let's start with the first interval: x < 3/4.Pick a test point, say x = 0.Plug into the numerator: 4(0) - 3 = -3 (negative).Plug into the denominator: 2 - 0 = 2 (positive).So, the fraction is negative/positive = negative. Since we're looking for where the expression is ≥ 0, this interval doesn't satisfy the inequality.Next, the second interval: 3/4 < x < 2.Let's pick x = 1.Numerator: 4(1) - 3 = 1 (positive).Denominator: 2 - 1 = 1 (positive).So, the fraction is positive/positive = positive. This interval satisfies the inequality.Finally, the third interval: x > 2.Let's pick x = 3.Numerator: 4(3) - 3 = 9 (positive).Denominator: 2 - 3 = -1 (negative).So, the fraction is positive/negative = negative. This interval doesn't satisfy the inequality.Now, we also need to check the critical points themselves.At x = 3/4:Numerator: 4*(3/4) - 3 = 3 - 3 = 0.So, the expression equals zero, which satisfies the inequality (since it's ≥ 0). So, x = 3/4 is included.At x = 2:Denominator becomes zero, which makes the expression undefined. So, x = 2 is excluded.Therefore, combining all this information, the solution set is:3/4 ≤ x < 2.Looking back at the options:A: {x | 3/4 ≤ x ≤ 2} – This includes x = 2, which is not allowed.B: {x | 3/4 ≤ x < 2} – This is correct because it includes x = 3/4 and excludes x = 2.C: {x | x > 2} or {x ≤ 3/4} – This is incorrect because the solution is between 3/4 and 2, not outside.D: {x | x ≥ 3/4} – This is incorrect because it includes x > 2, which doesn't satisfy the inequality.So, the correct answer is B.