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# General Form of a Straight Line

The linear equation (first degree polynomial in two variables x and y) ax + by +c = 0 (where a, b and c are real numbers such that at least one of a, b is non-zero) always represents a straight line. This is the general form of a straight line.

General Form of a Straight Line

The linear equation (first degree polynomial in two variables x and y) ax + by +c = 0 (where a, b and c are real numbers such that at least one of a, b is non-zero) always represents a straight line. This is the general form of a straight line.

Now, let us find out the equations of a straight line in the following cases

(i) parallel to ax + by + c = 0

(ii) perpendicular to  ax + by + c = 0

(iii) The point of intersection of two intersecting straight lines

## 1. Equation of a line parallel to the lineax+by+c=0

The equation of all lines parallel to the line ax + by +c = 0 can be put in the form ax + by + k = 0 for different values of k

## 2. Equation of a line perpendicular to the lineax+by+c=0

The equation of all lines perpendicular to the line ax + by +c = 0 can be written as bx ay + k = 0 for different values of k.

## 3. Slope of a straight line

The general form of the equation of a straight line is ax + by +c = 0 . (at least one of a, b is non-zero)

coefficient of x = a , coefficient of y = b , constant term = c.

The above equation can be rewritten as by = −axc

Gives comparing (1) with the form y = mx +l

We get, slope m = − a/b

m = −coefficient of x / coefficient of y

y intercept l = − c/b

y intercept = −constant term / coefficient of y ### Example 5.30

Find the slope of the straight line 6x + 8y + 7 = 0 .

### Solution

Given 6x + 8y + 7 = 0

slope Therefore, the slope of the straight line is – 3/4.

Example 5.31

Find the slope of the line which is

(i) parallel to 3x − 7y = 11

(ii) perpendicular to 2x − 3y + 8 = 0

Solution

(i) Given straight line is 3x − 7y = 11

gives 3x - 7y –11  = 0

Slope m= − 3/-7 = 3/7

Since parallel lines have same slopes, slope of any line parallel to

3x − 7y = 11  is  3/7

(ii) Given straight line is 2x − 3y + 8 = 0

Slope m = −2/−3  = 23

Since product of slopes is −1 for perpendicular lines, slope of any line perpendicular to 2x − 3y + 8 = 0 is ### Example 5.32

Show that the straight lines 2x + 3y − 8 = 0  and  4x + 6y + 18 = 0 are parallel.

### Solution

Slope of the straight line 2x + 3y − 8 = 0 is

m1 = − coefficient of x / coefficient of y

m1 = − 2/3

Slope of the straight line 4x + 6y + 18 = 0 is

m2 = −4/6 = −2/3

Here, m1   = m2

That is, slopes are equal. Hence, the two straight lines are parallel.

Example 5.33

Show that the straight lines x 2y + 3 = 0 and 6x + 3y + 8 = 0 are perpendicular.

Solution

Slope of the straight line x − 2y + 3 = 0 is

m1 = −1/−2 = 1/2

Slope of the straight line 6x + 3y + 8 = 0 is

m2 = −6/3 = −2

Now, m1 × m2  = 1/2 × (−2) = −1

Hence, the two straight lines are perpendicular.

Example 5.34

Find the equation of a straight line which is parallel to the line 3x 7y = 12 and passing through the point (6,4).

Solution

Equation of the straight line, parallel to 3x − 7y −12 = 0 is 3x − 7y + k = 0 Since it passes through the point (6,4)

3(6) − 7(4) + k= 0

k = 28 −18 = 10

Therefore, equation of the required straight line is 3x − 7y + 10 = 0.

Example 5.35

Find the equation of a straight line perpendicular to the line and passing through the point (7, –1).

Solution

The equation can be written as 4x − 3y − 21 = 0 .

Equation of a straight line perpendicular to 4x − 3y − 21 = 0 is 3x + 4y + k = 0

Since it is passes through the point (7, –1),

21 − 4 + k = 0 we get, k = −17

Therefore, equation of the required straight line is 3x + 4y −17 = 0 .

### Example 5.36

Find the equation of a straight line parallel to Y axis and passing through the point of intersection of the lines 4x + 5y = 13 and x − 8y + 9 = 0 .

### Solution

Given lines 4x + 5y −13= 0   ...(1)

x 8y + 9 = 0                        ...(2)

To find the point of intersection, solve equation (1) and (2) The equation of the line is x = 59/37 gives 37x − 59 = 0

### Example 5.37

The line joining the points A(0, 5) and B(4, 1) is a tangent to a circle whose centre C is at the point (4, 4) find

(i) the equation of the line AB.

(ii) the equation of the line through C which is perpendicular to the line AB.

(iii) the coordinates of the point of contact of tangent line AB with the circle.

### Solution

(i) Equation of line AB, A(0, 5) and B(4, 1) (ii) The equation of a line which is perpendicular to the line AB : x + y − 5 = 0 is xy + k = 0

Since it is passing through the point (4,4), we have 4 – 4 + k = 0 gives k = 0

The equation of a line which is perpendicular to AB and through C is xy = 0...(2)

(iii) The coordinate of the point of contact P of the tangent line AB with the circle is point of intersection of lines.

x + y 5 = 0  and x y = 0

solving, we get x = 5/2 and y = 5/2

Therefore, the coordinate of the point of contact is P(5/2,5/2).

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