Section Formula Worksheet solution5





In this page section formula worksheet solution5 we are going to see solution for each questions with detailed explanation.

(12) In what ratio is the line joining the points (-5 ,1) and (2 ,3) divided by y-axis? Also, find the point of intersection.

Let L : m be the ratio of the line segment joining the points (-5 , 1) and (2 ,3) and let p(0,y) be the point on the y axis

Section formula internally = (Lx2 + mx1)/(L + m) , (Ly2 + my1)/(L + m)

                            (0 , y)   = [L(2) + m(-5)]/(L + m) , [L(3) + m(1)]/(L + m)

                            (0 , y)   = [2L - 5 m]/(L + m) , [3L + m]/(L + m)

                                    [2L - 5 m]/(L + m) = 0

                                           2 L - 5 m = 0

                                                     2 L = 5 m

                                                          L/m = 5/2

                                           L : m = 5 : 2              

To find the required point we have to apply this ratio in the formula

                            (0 , y)   = [2(5) – 5(2)]/(5 + 2) , [3(5) + 2]/(5 + 2)

                            (0 , y)   = [10 – 10]/7 , [15 + 2]/7

                            (0 , y)   = (0 , 17/7)

Therefore the required point is (0,17/7)


(13) Find the length of the medians of the triangle whose vertices are (1 , -1) (0 ,4) and (-5 ,3)

Solution:   

Let A (1, -1) B (0, 4) and C (-5, 3) are the points vertices of the triangle

Let D, E and F are the midpoints of the sides AB, BC and CA respectively

Midpoint of AB = (x₁+x₂)/2 , (y₁+y₂)/2

                        = (1+0)/2 , (-1+4)/2

                   = D  (1/2,3/2)

Midpoint of BC = (x₁+x₂)/2 , (y₁+y₂)/2

                    = (0+(-5))/2 , (4+3)/2

                    = (0-5/2 , 7/2)

                    = E  (-5/2,7/2)

Midpoint of CA = (x₁+x₂)/2 , (y₁+y₂)/2

                     = (-5+1)/2 , (3+(-1))/2

                     = (-4/2 , 2/2)

                     = F  (-2,1)

Length of the median AD = √(x₂-x₁)² + (y₂-y₁)²

                                         = √(1+5/2)² + (-1-7/2)²

                                         = √(7/2)² + (-9/2)²

                                         = √(49/4)+(81/4)

                                         = √(49+81)/4

                                         = √130/4

                                         = √130/2

Length of the median BE = √(x₂-x₁)² + (y₂-y₁)²

                                         = √(-2-0)² + (1-4)²

                                         = √(-2)² + (-3)²

                                         = √4+9

                                         = √13

Length of the median CF = √(x₂-x₁)² + (y₂-y₁)²

                                         = √((1/2)+5)² + ((3/2)-3)²

                                         = √(11/2)² + (-3/2)²

                                         = √(121/4)+(9/4)

                                         = √130/4

                                         = √130/2

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