Example 1 :
Find the equation of the straight line passing through point of intersection of the lines
2x + y – 3 = 0
5x + y – 6 = 0
and parallel to the line joining the points (1, 2) and (2, 1).
Solution :
First let us
find the point of intersection of two lines
2x + y – 3 = 0 ----(1)
5x + y – 6 = 0 ----(2)
(1) - (2) :
2x + y – 3 - (5x + y – 6) = 0
2x - 5x + y - y - 3 + 6 = 0
-3x + 3 = 0
x = 1
By applying the value of x in (1), we get
2(1) + y – 3 = 0
2 + y – 3 = 0
-1 + y = 0
y = 1
The point of intersection of two lines (1, 1)
Slope of the line joining two points (1, 2) and (2, 1)
m = (y_{2 }- y_{1})/(x_{2 }- x_{1})
m = (1 - 2)/(2 - 1)
m = -1
Equation of required line :
(y - y_{1}) = m(x - x_{1})
(y - 1) = -1(x – 1)
y - 1 = -x + 1
x + y – 1 – 1 = 0
x + y – 2 = 0
So, equation of the required line is x + y - 2 = 0.
Example 2 :
Find the equation of the straight line passing through the point of intersection of the
5x – 6y = 1
3x + 2y + 5 = 0
and is perpendicular to the straight line
3x – 5y + 11 = 0
Solution :
First let us find the point of intersection of two lines
5x - 6y = 1 ----(1)
3x + 2y = -5 ----(2)
(1) + 3(2) ==>
5x - 6y + 9x + 6y = 1 - 15
14x = -14
x = -1
By applying the value of x in (1), we get
5(-1) – 6y = 1
-5 – 6y = 1
-6y = 1 + 5
y = -1
The point of intersection of two lines (-1, -1).
Slope of the line 3x – 5y + 11 = 0
m = -3/(-5)
m = 3/5
Equation of required line :
y - y_{1} = m(x - x_{1})
y – (-1) = (3/5)(x – (-1))
y + 1 = (3/5)(x + 1)
5(y + 1) = 3(x + 1)
5y + 5 = 3x + 3
3x – 5y – 2 = 0
So, equation of the required line is 3x – 5y – 2 = 0.
Example 3 :
Find the equation of the straight line joining the point of intersection of the lines
3x – y + 9 = 0
x + 2y = 4
and the point of intersection of the lines
2x + y – 4 = 0
x – 2y + 3 = 0
Solution :
Point of intersection of the lines
3x – y = -9 ----(1)
x + 2y = 4 ----(2)
2(1) + (2) ==>
6x – 2y + x + 2y = -18 + 4
7x = -14
x = -2
By applying the value of x in (1), we get
3(-2) – y = -9
-6 – y = -9
-y = -9 + 6
y = 3
So, point of intersection of first two lines is (-2, 3).
Point of intersection of the lines
2x + y = 4 ----(3)
x - 2y = -3 ----(4)
2(1) + (2) ==>
4x + 2y + (x - 2y) = 8 - 3
5x = 5
x = 1
Substitute x = 1 in (3).
2(1) + y = 4
2 + y = 4
y = 2
Point of intersection of first two lines is (2, 1).
Equation of the required line will pass through the points (-2, 3) and (2, 1)
(y - y_{1})/(y_{2 }- y_{1}) = (x - x_{1})/(x_{2 }- x_{1})
(y - 3)/(1 - 3) = (x - (-2))/(2 - (-2))
(y - 3)/(-2) = (x + 2)/4
4(y - 3) = -2(x + 2)
4y – 12 = -2x – 4
2x + 4y – 12 + 4 = 0
2x + 4y – 8 = 0
x + 2y - 4 = 0
So, the equation of the required line is x + 2y – 4 = 0.
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Oct 04, 22 11:07 PM
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