EQUATION OF THE LINE PASSING THROUGH THE POINT

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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