# 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

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