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 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 k5, 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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