Let's solve each question step by step. --- ## Q6: The line segment joining points P(-3, 2) and Q(5, 7) is divided by the y-axis in the ratio Let the point where the line is divided by y-axis be \( R(0, y) \). Let the ratio be \( k:1 \). Using the section formula, \[ x = \frac{kx_2 + x_1}{k+1} \] Here, \( x = 0 \), \( x_1 = -3 \), \( x_2 = 5 \) \[ 0 = \frac{k \cdot 5 + (-3)}{k+1} \implies 0 = 5k - 3 \implies 5k = 3 \implies k = \frac{3}{5} \] So, ratio = 3:5 **Final Answer: (d) 3:5** --- ## Q7: The base BC of an equilateral ΔABC lies on the y-axis. The coordinates of C are (0, -3). If the origin is the midpoint of BC, what are the coordinates of A and B? Let \( B(0, y_1) \), \( C(0, -3) \), and origin \( (0, 0) \) is the midpoint of BC. Midpoint formula: \[ \left( \frac{0+0}{2}, \frac{y_1 + (-3)}{2} \right) = (0, 0) \implies \frac{y_1 - 3}{2} = 0 \implies y_1 = 3 \] So \( B(0, 3) \), \( C(0, -3) \). Now, for an equilateral triangle, A must be at a distance \( BC \) from both B and C, and not on the y-axis (since B and C are both on the y-axis). So, \( A(a, 0) \) or \( A(-a, 0) \). Distance BC = \( |3 - (-3)| = 6 \). Distance from A(a, 0) to B(0, 3) = \( \sqrt{a^2 + 9} \) Distance from A(a, 0) to C(0, -3) = \( \sqrt{a^2 + 9} \) Distance BC = 6 So A must be at \( (\pm 3\sqrt{3}, 0) \) **Final Answer: (a) (±3√3, 0), B(0, 3)** --- ## Q8: The distance of point A(5, 6) from the origin is \[ \text{Distance} = \sqrt{(5-0)^2 + (6-0)^2} = \sqrt{25 + 36} = \sqrt{61} \] **Final Answer: (b) √61 units** --- ## Q9: Vertices of parallelogram A(1,2), B(4, y), C(x, 6), D(3,5). Find (x, y). In a parallelogram, diagonals bisect each other. Let diagonals AC and BD. Midpoint of AC = Midpoint of BD Midpoint of AC = \( \left( \frac{1 + x}{2}, \frac{2 + 6}{2} \right) = \left( \frac{1 + x}{2}, 4 \right) \) Midpoint of BD = \( \left( \frac{4 + 3}{2}, \frac{y + 5}{2} \right) = \left( 3.5, \frac{y + 5}{2} \right) \) Equate: \[ \frac{1 + x}{2} = 3.5 \implies 1 + x = 7 \implies x = 6 \] \[ 4 = \frac{y + 5}{2} \implies y + 5 = 8 \implies y = 3 \] **Final Answer: (a) (6, 3)** --- ## Q10: Equation of perpendicular bisector of line joining (4,5) and B(-2,3) Midpoint: \( \left( \frac{4 + (-2)}{2}, \frac{5 + 3}{2} \right) = (1, 4) \) Slope of AB: \( m = \frac{5 - 3}{4 - (-2)} = \frac{2}{6} = \frac{1}{3} \) Slope of perpendicular bisector = negative reciprocal = -3 Equation: \( y - 4 = -3(x - 1) \) \[ y - 4 = -3x + 3 \implies 3x + y - 7 = 0 \] **Final Answer: (c) 3x + y - 7 = 0** --- ## Q11: If the distance of point (a, 4) from x-axis is double the distance from y-axis, then a = ? Distance from x-axis = |4| = 4 Distance from y-axis = |a| Given: 4 = 2|a| ⇒ |a| = 2 ⇒ a = 2 or -2 **Final Answer: (a) 2** --- ## Q12: If (x, y) is the midpoint of (3, 4) and (k, 7), and line 2x + 2y + 1 = 0 passes through the midpoint, find k. Midpoint: \( \left( \frac{3 + k}{2}, \frac{4 + 7}{2} \right) = \left( \frac{3 + k}{2}, \frac{11}{2} \right) \) Plug into the line: \[ 2 \left( \frac{3 + k}{2} \right) + 2 \left( \frac{11}{2} \right) + 1 = 0 \] \[ (3 + k) + 11 + 1 = 0 \implies k + 15 = 0 \implies k = -15 \] **Final Answer: (b) -15** --- **Summary of Answers:** - **Q6:** (d) 3:5 - **Q7:** (a) (±3√3, 0), B(0, 3) - **Q8:** (b) √61 units - **Q9:** (a) (6, 3) - **Q10:** (c) 3x + y - 7 = 0 - **Q11:** (a) 2 - **Q12:** (b) -15