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 - y1 = m (x - x1)
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 - y1 = m(x - x1)
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 - y1 = m(x - x1)
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
Example 3 :
The line with equation 2x − 3y + 5 = 0 is perpendicular to the line with equation 3x + ky − 1 = 0. Find the value of the constant k.
Solution :
The given lines are perpendicular to each other, then product of their slopes will be equal to -1.
Slope of the line 2x - 3y + 5 = 0 :
3y = 2x + 5
y = (2/3)x + (5/3)
Slope (m1) = 2/3
Slope of the line 3x + ky - 1 = 0 :
ky = -3x + 1
y = (-3/k)x + (1/k)
Slope (m2) = -3/k
Product of the slopes m1 x m2 = -1
(2/3) x (-3/k) = -1
-2/k = -1
k = 2
So, the value of k is 2.
Example 4 :
The straight line l1 passes through the points with coordinates (−3, 7) and (1, −5).
a) Find an equation of the line l1 in the form ax + by + c = 0, where a, b and c are integers. The line l2 is perpendicular to l1 and passes through the point with coordinates (4, 6).
b) Find, in the form k√5, the distance from the origin of the point where l1 and l2 intersect.
Solution :
(−3, 7) and (1, −5)
Slope of the line joining the points (−3, 7) and (1, −5)
= (y2 - y1)/(x2 - x1)
= (-5 - 7)/(1 + 3)
= -12/4
= -3
Equation of line l1 :
(y - y1) = m(x - x1)
y - 7 = -3(x + 3)
y = -3x - 9 + 7
3x + y = -2 -----(1)
Slope of line l2 = -1/(-3) ==> 1/3
Equation of line l2 :
(y - y1) = -1/m(x - x1)
Perpendicular line passes through (4, 6).
y - 6 = 1/3(x - 4)
3(y - 6) = 1(x - 4)
3y - 18 = x - 4
3y = x - 4 + 18
x - 3y = -14 -----(2)
Point of intersection of lines l1 and l2.
(1) x 3 ==>
9x + 3y = -6
x - 3y = -14
-------------
10x = -20
x = -2
Applying x = -2 in (1)
3(-2) + y = -2
-6 + y = -2
y = -2 + 6
y = 4
So, the point of intersections of these two lines is (-2, 4).
Distance from origin = √[(x2 - x1)2 + (y2 - y1)2]
= √[(0 + 2)2 + (0 - 4)2]
= √(4 + 16)
= √20
= 2 √5
Comparing with k√5, the value of k is 5.
Example 5 :
If AB is defined by the endpoints A(4, 2) and B(8, 6), write an equation of the line that is the perpendicular bisector of AB.
Solution :
Midpoint of line segment AB = (x1 + x2)/2, (y1 + y2)/2
= (4 + 8)/2, (2 + 6)/2
= 12/2, 8/2
= (6, 4)
Slope of the line AB = (y2 - y1) / (x2 - x1)
= (6 - 2) / (8 - 4)
= 4/4
= 1
Slope of the perpendicular bisector to AB
= -1/1
= -1
Equation of perpendicular bisector :
(y - y1) = -1/m(x - x1)
(y - 4) = -1(x - 6)
y - 4 = -x + 6
y = -x + 6 + 4
y = -x + 10
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