answer:1. **Let ( AB ) be the given segment on one of the parallel lines. Let ( C ) and ( D ) be arbitrary points on the other given parallel line.** 2. **Construct points ( D_{1}, D_{2}, ldots, D_{n} ) such that all segments ( D_{i} D_{i+1} ) are equal to the segment ( CD ). According to the given problem, this is possible.** - This step is crucial because it ensures that the distances between consecutive points ( D_{i} ) are equal. This construction can be achieved by using the division of the second parallel line into ( n ) equal segments through the method of proportional division. 3. **Let ( P ) be the point of intersection of lines ( AC ) and ( BD_{n} ).** - Here, we use the fact that ( AC ) and ( BD_{n} ) will intersect at some point due to the known construction methods in projective geometry. 4. **Let ( B_{1}, B_{2}, ldots, B_{n-1} ) be the points of intersection of line ( AB ) with lines ( PD_{1}, PD_{2}, ldots, PD_{n-1} ), respectively.** - These intersection points ( B_{1}, B_{2}, ldots, B_{n-1} ) are determined by drawing lines through ( P ) to each ( D_{i} ). 5. **It is clear that points ( B_{1}, B_{2}, ldots, B_{n-1} ) divide the segment ( AB ) into ( n ) equal parts.** - This conclusion follows from the properties of parallel lines and proportional division. Specifically, since ( PD_{i} ) are parallel to each other and by the intercept theorem, the segment ( AB ) will be divided into ( n ) equal parts by the constructed lines. Therefore, the segment ( AB ) is now divided into ( n ) equal parts by the points ( B_1, B_2, ldots, B_{n-1} ). ( boxed{} )

answer:Solution: (1) The equation of line l is y-2= dfrac{1}{2} (x-4), which simplifies to y= dfrac{1}{2} x, Substituting into the ellipse equation x^2+4y^2=36, we get x=±3sqrt{2}, y=±dfrac{3sqrt{2}}{2}, Therefore, |AB|= sqrt{(6sqrt{2})^2+(3sqrt{2})^2}=3sqrt{10}; (2) Given the coordinates of P, we have dfrac{16}{36}+ dfrac{4}{9} < 1, indicating that P is inside the ellipse. Let A(x_1, y_1) and B(x_2, y_2), then dfrac{x_1^2}{36}+ dfrac{y_1^2}{9}=1, and dfrac{x_2^2}{36}+ dfrac{y_2^2}{9}=1, Using the midpoint formula, we get x_1+x_2=8, y_1+y_2=4, From dfrac{(x_1−x_2)(x_1+x_2)}{36}+ dfrac{(y_1−y_2)(y_1+y_2)}{9}=0, Substituting the midpoint values, we find that the slope of AB, {k}_{AB}= dfrac{y_1−y_2}{x_1−x_2}=- dfrac{1}{2}, Therefore, the equation of the desired line is y-2=- dfrac{1}{2} (x-4), which simplifies to boxed{x+2y-8=0}.

answer:Let's denote Roja's speed as ( R ) km/hr. Since Pooja is moving at 3 km/hr and Roja is moving in the opposite direction, their relative speed when moving away from each other is the sum of their individual speeds. Therefore, the relative speed is ( R + 3 ) km/hr. After 4 hours, the distance between them is 40 km. We can use the formula for distance, which is: [ text{Distance} = text{Speed} times text{Time} ] Given that the time is 4 hours and the distance is 40 km, we can set up the equation: [ 40 = (R + 3) times 4 ] Now, we can solve for ( R ): [ 40 = 4R + 12 ] Subtract 12 from both sides: [ 40 - 12 = 4R ] [ 28 = 4R ] Divide both sides by 4: [ R = frac{28}{4} ] [ R = 7 ] So, Roja's speed is boxed{7} km/hr.

answer:Let's denote the number of years A lent money to B as "n" years. The interest A received from B can be calculated using the simple interest formula: Interest = Principal × Rate × Time For B: Interest from B = 5000 × 15/100 × n Interest from B = 750n For C: Interest from C = 3000 × 15/100 × 4 Interest from C = 3000 × 0.15 × 4 Interest from C = 450 × 4 Interest from C = 1800 According to the problem, the total interest received from both B and C is Rs. 3300. So, we can write the equation: 750n + 1800 = 3300 Now, let's solve for n: 750n = 3300 - 1800 750n = 1500 Divide both sides by 750 to find n: n = 1500 / 750 n = 2 Therefore, A lent money to B for boxed{2} years.