Question

Find the equation of a straight line through the point of intersection of the lines 8x+3y=18, 4x+5y=9 and bisecting the line segment joining the points (5,-4) and (-7, 6).
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Answer 1

8x + 3y = 18 ----(1)

4x + 5y = 9  ----(2)

On solving (1) * 5 - (2) * 3, we get

40x + 15y = 90

12x + 15y  =  27  

--------------------

28x  =  63

x  =  63/28

By applying the value of x in (1), we get  

8(63/28) + 3y  =  18

3y  =  18 - (126/7)

3y  =  (126-126)/7

y  =  0  

Point of intersection of the given lines is (63/28, 0).  

Given points are (5,–4) and (–7,6)

Midpoint  =  (5 - 7)/2, (-4 + 6)/2

  =  -2/2, 2/2

 =  (-1, 1)

Equation of the line passing through the points (-1, 1) and (63/28, 0)

(y - y1)/(y2 - y1)  =  (x - x1)/(x2 - x1)

(y - 1)/(0 - 1)  =  (x + 1)/((63/28) + 1)

(y - 1)/(- 1)  =  (x + 1)/(91/28)

91(y - 1)  =  -28(x + 1)

91y - 91  =  -28x - 28

28x + 91y - 91 + 28  =  0

28x + 91y - 63  =  0

Dividing the entire equation by 7, we get

4x + 13y - 9  =  0

Hope this helps!