pllzz ans.32nd no link pllzz

pllzz ans.32nd no link pllzz 31. Find the equation of the line through (0, -3) and perpendicular to -the points (—3, 2) and (9, 1). 32. The vertices of a triah1gle are A(IO, 4), B(4, —9) and C(—2, —1). Find the equation oi the altitude through A. [The perpendicular drawn from vertex ofa triangle to the opposite side is called altitude.] 33. A(2, —4), B (3, 3) and C (—1, 5) are the vertices of triangle ABC. Find the equation (t) the median of the triangle through A. (ii) the altitude of the triangle through B. 34. Find the equation of the right bisector of the line segment joining the points (l, 2) and (5, —6). 35. Points A and B have coordinates (7, —3) and (l, 9) respectively. Find (i) the slope of AB. (ii) the equation of the perpendicular bisector of the line segment AB. (iii) the value of p if (—2, p) lies on it. (2008) 36. The points B(l, 3) and D(6, 8) are two opposite vertices of a square ABCD. Find the equation of the diagonal AC. I-dine AC is the right bisector of BD

Dear student

Since AD ⊥ BC \Rightarrow Slope of AD \times Slope of BC = - 1 \Rightarrow m \times (\frac{- 1 + 9}{- 2 - 4}) = - 1 \Rightarrow m \times (\frac{8}{- 6}) = - 1 \Rightarrow m = \frac{3}{4} Since AD passes through A (10, 4) and has slope m = \frac{3}{4} . So its equation is y - 4 = \frac{3}{4} (x - 10) \Rightarrow 4 y - 16 = 3 x - 30 \Rightarrow 3 x - 4 y - 14 = 0

Regards