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

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*

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*.

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* = −*ax* –*c*

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

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

Given 6*x* + 8*y* + 7 = 0

slope *m *=

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

**Example 5.31**

Find the slope of the line which is

(i) parallel to 3*x* − 7*y* = 11

(ii) perpendicular to 2*x* − 3*y* + 8 = 0

*Solution*

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

gives 3*x* - 7*y* –11 = 0

Slope *m*= − 3/-7 = 3/7

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

3*x* − 7*y* =
11 is 3/7

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

Slope *m* = −2/−3 = 23

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

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

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

*m*_{1}* *= − coefficient of *x
/* coefficient of *y*

*m*_{1 }= − 2/3

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

*m*_{2} = −4/6 = −2/3

Here, *m*_{1}* *=* m*_{2}

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

**Example 5.33**

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

*Solution*

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

*m*_{1}* *=* *−1/−2 =* *1/2

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

*m*_{2}* *=* *−6/3* *= −2

Now, *m*_{1} × *m*_{2} = 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** **3*x*** **−** **7*y*** **=** **12** **and passing through the
point (6,4).

*Solution*

Equation of the straight line, parallel to 3*x* − 7*y*
−12 = 0 is 3*x* − 7*y* + *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 3*x* − 7*y*
+ 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 4

Equation of a straight line perpendicular to 4*x* − 3*y*
− 21 = 0 is 3*x* + 4*y* + *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 3*x* + 4*y*
−17 = 0 .

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

Given lines 4*x* + 5*y* −13= 0 ...(1)

*x *−* *8*y *+* *9 = 0
...(2)

To find the point of intersection, solve equation (1) and (2)

The equation of the line is *x*
= 59/37 gives 37*x* − 59 = 0

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.

(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 *x* − *y* + *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 *x* − *y* = 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).

Tags : Equations, Example Solved Problem | Coordinate Geometry , 10th Mathematics : UNIT 5 : Coordinate Geometry

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10th Mathematics : UNIT 5 : Coordinate Geometry : General Form of a Straight Line | Equations, Example Solved Problem | Coordinate Geometry

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