Example 1 :
Write the equation of the lines through the point (1, -1)
(i) parallel to x + 3y - 4 = 0
(ii) perpendicular to 3x + 4y = 6
Solution :
(i) Since the required line is parallel to the line x + 3y - 4 = 0 is, slopes of the required line and given line will be equal.
Slope of the line x + 3y − 4 = 0
= - Coefficient of x/coefficient of y = -1/3
Slope of the required line = -1/3.
Equation of the required line :
y - y_{1} = m (x - x_{1})
y - 1 = (-1/3) (x - 1)
3(y - 1) = -1(x - 1)
3y - 3 = -x + 1
x + 3y - 4 = 0
(ii) perpendicular to 3x + 4y = 6
Since the required line is perpendicular to the given line, product of their slopes will be equal to -1.
Slope of the line 3x + 4y = 6
= - Coefficient of x/coefficient of y = -3/4
Slope of the required line = 4/3
Equation of the required line :
y - y_{1} = m(x - x_{1})
y - 1 = (4/3)(x - 1)
3(y - 1) = 4(x - 1)
3y - 3 = 4x - 4
4x - 3y - 4 + 3 = 0
4x - 3y - 1 = 0
Example 2 :
If (−4, 7) is one vertex of a rhombus and if the equation of one diagonal is 5x − y + 7 = 0, then find the equation of another diagonal.
Solution :
In a rhombus, both diagonals will intersect each other at right angle.
So, the required diagonal will be perpendicular to the line 5x - y + 7 = 0 and passing through the point (-4, 7).
Slope of the line = Coefficient of x/Coefficient of y
= -5/(-1) = 5
Slope of required diagonal = -1/5.
Equation of other diagonal :
y - y_{1} = m(x - x_{1})
y - 7 = (-1/5)(x + 4)
5(y - 7) = -1(x + 4)
5y - 35 = -x - 4
x + 5y - 35 + 4 = 0
x + 5y - 31 = 0
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