Question

if A(1,2), B(1,4), C(3,2) are the vertices of a triangle. The slope of the line joining the mid-points of BA and BC is

Answer 1

Given,
A≡ (1, 2) , B≡(1,4) and C≡(3,2)
Let the midpoint of BA is P and midpoint of BC = Q
Midpoint of BA = [(1 + 1)/2 , (4 + 2)/2 ]
P= [(2/2) , (6/2) ]
P = (1, 3)

midpoint of BC = [(1 + 3)/2, (4 + 2)/2 ]
Q= [(4/2) , (6/2) ]
Q = (2, 3)

Slope of midpoint of BA and midpoint of BC = slope of points P(1,3) and Q(2,3) .
∵Slope = (y₂ - y₁)/(x₂ - x₁)
∴ slope of PQ = (3 - 3)/(2 - 1) = 0/1
∴slope of PQ = 0

Hence, slope of midpoint of BA and midpoint of BC = 0

Answer 2

Hello Dear.

Here is the answer---


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Let the co-ordinates of the point A(1,2) be (x₁, y₁)
-------------------------------------------B(1,4) be (x₂, y₂)
--------------------------------------------C(3,2) be (x₃,y₃)

For the Midpoint of BA,

 Using the Mid-Point Formula, 
  
          Midpoint = $[ \frac{x1 + x2}{2} , \frac{y1 + y2}{2} ]$

                        =  $[\frac{1 + 1}{2} , \frac{2 + 4}{2} ]$
                        = [1, 3]

  Let the Midpoint of AB (1,3) be (a,b)

For the Mid-Point of BC,

       Midpoint =  $[ \frac{x2 + x3}{2} , \frac{y2 + y3}{2} ]$
                      = $[ \frac{1 + 3}{2} , \frac{2 + 4}{2} ]$
                      =  [2, 3]

Let the Midpoint of BC be (x,y)

For the slope of the line joining the midpoints of line BA and BC,

Using the Formula,

 Slope(m)  = $\frac{y - b}{x - a}$
                 = $\frac{3 - 3}{2 - 1}$
                 = 0/1
                 = 0

Thus, the slope of the line joining the mid point of BC and Ba will be 0.


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Hope it helps.


Have a Marvelous Day.