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 a points (5,‐4)and(‐7,6)
Answers
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!