Equation of the Line Passing Through the Intersection of the Lines :
Here we are going to see, how to find the equation of the line passing through the intersection of the lines
Question 1 :
Find the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x − 4y = 1 and parallel to the line 13x + 5y +12 = 0
Solution :
First, let us find the point of intersection of the given lines 7x + 3y = 10, 5x − 4y = 1.
7x + 3y = 10 ----(1)
5x − 4y = 1 ----(2)
4(1) + 3(2)
28x + 12y = 40
15x - 12y = 3
------------------
43x = 43
x = 1
By applying the value of x in (1), we get
7(1) + 3y = 10
7 + 3y = 10
3y = 10 - 7
3y = 3
y = 1
So, the point of intersection of the given lines is (1, 1)
The required line is passing through the point (1, 1) and parallel to the line 13x + 5y + 12 = 0
Slope of the required line = -13/5
(y - y_{1}) = m(x - x_{1})
(y - 1) = (-13/5)(x - 1)
5(y - 1) = -13(x - 1)
5y - 5 = -13x + 13
13x + 5y - 5 - 13 = 0
13x + 5y - 18 = 0
Question 2 :
Find the equation of a straight line through the intersection of lines 5x −6y = 2, 3x + 2y = 10 and perpendicular to the line 4x −7y +13 = 0
Solution :
5x −6y = 2 -----(1)
3x + 2y = 10 -----(2)
(1) ==> 5x −6y = 2
3(2) ==> 9x + 6y = 30
------------------
14x = 32
x = 32/14 = 16/7
By applying the value of x in (1), we get
5(16/7) - 6y = 2
(80/7) - 6y = 2
6y = (80/7) - 2
6y = (80 - 14)/7
y = 11/7
The point of intersection of the given lines is (16/7, 11/7)
The required line is perpendicular to the line 4x −7y +13 = 0
slope of the required line = -4/(-7) = 4/7
(y - y_{1}) = (-1/m)(x - x_{1})
(y - (11/7)) = (-7/4)(x - (16/7))
(7y - 11)/7 = (-7/4)(7x - 16)/7)
4(7y - 11) = -7(7x - 16)
28y - 44 = -49x + 112
49x + 28y - 44 - 112 = 0
49x + 28y - 156 = 0
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