Any first degree equation in two variables x and y of the form ax + by +c = 0 …(1) where a, b, c are real numbers and at least one of a, b is non-zero is called “Straight line” in xy plane.

**Straight line**

Any first degree equation in two variables *x* and *y* of the
form *ax* + *by* +*c* = 0 …(1) where *a*, *b*,
*c* are real numbers and at least one of *a*, *b* is non-zero is
called “Straight line” in *xy* plane.

The *X* axis and *Y* axis together are called coordinate
axes. The *x* coordinate of every point on *OY* (*Y* axis) is 0.
Therefore equation
of *OY *(*Y* axis) is *x *= 0* *(fig 5.27)

The *y* coordinate of every point on *OX* (*X*
axis) is 0. Therefore the equation of *OX*
(*X* axis) is *y* = 0 (fig 5.28)

Let *AB* be a straight line parallel to *X* axis, which is at a distance ‘*b*’. Then *y*
coordinate of every point on ‘*AB*’ is ‘*b*’. (fig 5.29)

Therefore, the equation of *AB* is *y* = *b*

**Note**

·
If *b* > 0 , then the line *y=b* lies above the *X*
axis

·
If *b* < 0 , then the line *y=b* lies below the *X*
axis

·
If *b* = 0 , then the line *y=b* is the *X* axis
itself.

Let *CD* be a straight line parallel to *Y* axis, which is at a distance ‘c’. Then *x* coordinate of every point on *CD* is ‘*c*’. The
equation of *CD* is *x* = *c*. (fig 5.30)

**Note**

·
If *c*>0, then the line* x=c *lies right to the
side of the* Y *axis

·
If* c *<* *0, then the line* x=c *lies left to
the side of the* Y *axis

·
If *c*=0, then the line* x=c *is the* Y *axis
itself.

Find the
equation of a straight line passing through (5,7) and is (i) parallel
to *X* axis
(ii) parallel to *Y* axis.

(i) The equation of any straight line parallel to *X* axis is
*y=b*.

Since it passes through (5,7), *b* = 7.

Therefore, the required equation of the line is *y=*7.

(ii) The equation of any straight line parallel to *Y* axis
is *x=c*

Since it passes through (5,7), *c* = 5

Therefore, the required equation of the line is *x=*5.

Every straight line that is not vertical will cut the *Y*
axis at a single point. The coordinate of this point is called *y*
intercept of the line.

A line with slope *m* and *y* intercept *c* can be expressed
through the equation *y*=*mx*+*c *We call this equation as the slope-intercept form of the equation of a
line.

**Note**

·
If a line with slope *m*, *m*≠0 makes *x*
intercept *d*, then the equation of the straight line is *y *=* m *(*x*–*d*).

·
*y *=* mx *represent equation of a straight line with slope*
m *and passing through the* *origin.

Find the equation of a straight line whose

(i) Slope is 5 and *y* intercept is -9

(ii) Inclination is 45° and *y* intercept is 11

(i) Given, Slope** **=

Therefore, equation of a straight line is *y* = *mx* +*c*

*y *=* *5*x *−* *9* *gives* *5*x *−* y
*−* *9* *=* *0

(ii) Given, *θ* = 45° , *y* intercept, *c* = 11

Slope *m* = tan *θ *= tan 45° = 1

Therefore, equation of a straight line is of the form *y* = *mx*
+*c*

Hence we get, *y*= *x* + 11 gives *x* − *y*
+ 11 = 0

Calculate the slope and** ***y*** **intercept of the straight line** **8*x*** **−** **7*y*** **+** **6** **=** **0

Equation of the given straight line is** **8

7*y* = 8*x* + 6 (bringing it to the form *y*
= *mx* +*c* )

* y* = 8/7 *x* + 6/7 …(1)

Comparing (1) with *y* = *mx* +*c*

Slope *m* = 8/7 and *y* intercept *c* = 6/7

The graph relates temperatures** ***y*** **(in Fahrenheit degree) to temperatures** ***x*** **(in Celsius degree) (a)
Find the slope and *y* intercept (b) Write an equation of the line What is
the mean temperature of the earth in Fahrenheit degree if its mean temperature
is 25° Celsius?

(a) From the figure, slope** **=

The line crosses the *Y* axis at (0, 32) So

the slope is 9/5 and *y* intercept is 32.

(b) Use the slope and *y* intercept to write an equation

The equation is *y* = 9/5 *x* + 32

(c) In Celsius, the mean temperature of the earth is 25°. To find
the mean temperature in Fahrenheit, we find the value of *y* when *x*
= 25

*y* = 9/5 *x* + 32

*y* = 9/5 (25)+32

*y *=* *77

Therefore, the mean temperature of the earth is 77° F.

**Note**

The formula for converting Celsius to Fahrenheit is given by *F*
= 9/5 *C* + 32. which is the linear equation representing a stragiht
line derived in the example.

Here we will find the equation of a straight line passing through
a given point *A*(*x*_{1} , *y*_{1} ) and having
the slope *m*.

Let *P* (*x*, *y*) be any point other than *A*
on the given line. Slope of the line joining *A*(*x*_{1} , *y*_{1}
) and *P* (*x*, *y*) is given by

Therefore, the equation of the required line is *y* − *y*_{1} = *m* (*x* − *x*_{1}
) (Point slope form)

Find the
equation of a line
passing through the point (3, - 4) and having slope -5/7

Given, (*x*_{1} ,*y*_{1} ) = (3,
−4) and *m* = −5/7

The equation of the point-slope form of the straight line is *y*
− *y*_{1} = *m* (*x* − *x*_{1})

we write it as *y* + 4 = −
5/7 (*x*-3)

gives us 5*x* + 7*y* + 13 = 0

Find the equation of a line passing through the point** ***A*(1, 4)** **and** **perpendicular to the
line joining points (2, 5) and (4, 7) .

Let the given points be *A*(1, 4) , *B*(2, 5) and *C*(4,
7).

Slope of line *BC* = (7–5)/(4-2) = 2/2 = 1

Let *m* be the slope of the required line.

Since the required line is perpendicular to *BC*,

*m *×1 = −1

*m *= −1

The required line also pass through the point *A*(1,4).

The equation of the required straight line is *y* − *y* _{1}
= *m* (*x* − *x*_{1} )

*y*-* *4 =−1(*x* −1)

*y*-* *4= −*x* + 1

we get, *x* + *y* – 5 = 0

Let *A*(*x*_{1} , *y*_{1} )
and *B* (*x*_{2} , *y*_{2} ) be two given
distinct points. Slope of the straight line passing through these points
is given by *m = *

From the equation of the straight line in point slope form, we get

(is the equation of the line in two-point form)

Find the equation of a straight line passing through** **(5,** **-** **3)** **and** **(7,** **-** **4)** **.

The equation of a straight line passing through the two points** **(

Substituting the points we get,

Gives 6*y* + 2 = − *x* + 5

Therefore, *x *+* *2*y *+* *1 = 0

Two buildings of** **different heights are located at opposite sides
of each other. If a heavy rod is attached joining the terrace of the buildings
from (6, 10) to (14, 12) , find the equation of the rod joining the buildings ?

Let* **A*(6, 10)** **,

The equation of the rod is the equation of the straight line
passing through *A*(6,10) and *B*(14,12)

Therefore, *x *−* *4*y *+* *34 = 0

Hence, equation of the rod is *x* − 4*y* + 34 = 0

We will find the equation of a line whose intercepts are *a *and
*b *on the coordinate axes respectively.

Let *PQ* be a line meeting *X* axis at *A* and *Y
*axis at *B* . Let *OA*=*a, OB*=*b*. Then the
coordinates of *A *and *B* are (*a* , 0) and (0,*b*)
respectively. Therefore, the equation of the line joining *A* and *B*
is

Find the equation of a line which passes through** **(5,7)** **and makes intercepts** **on the axes equal in
magnitude but opposite in sign.

Let the *x* intercept be ‘*a*’ and *y* intercept be
‘*–a*’.

The equation of the line in intercept form is

Therefore, *x *−* y *=* a *...(1)

Since (1) passes through(5,7)

Therefore, 5 - 7 = *a *gives *a* = −2

Thus the required equation of the straight line is *x* − *y*
= −2 ; or *x* − *y* + 2 = 0

**Example 5.26**

Find the intercepts made by the line** **4*x*** **−** **9*y*** **+** **36** **=** **0** **on the coordinate axes.

*Solution*

Equation of the given line is** **4

we write it as 4*x* - 9*y *= −36 (bringing it to the
normal form)

Dividing by -36 we get, ...(1)

Comparing (1) with intercept form, we get *x* intercept *a* = − 9 ; *y*
intercept *b* = 4

**Example 5.27**

A mobile phone is put to use when the battery power is 100%. The
percent**
**of battery power ‘*y*’
(in decimal) remaining after using the mobile phone for *x* hours is
assumed as *y* = −0. 25*x* + 1

(i) Draw a graph of the equation.

(ii) Find the number of hours elapsed if the battery power is 40%.

(iii) How much time does it take so that the battery has no power?

*Solution*

(i)

(ii) To find the time when the battery power is 40%, we have to
take *y* = 0. 40

0.40 = −0. 25*x* + 1
gives 0. 25*x* = 0. 60

we get, *x *=* *0.60/0.25* *=* *2.4* *hours.

(iii) If the battery power is 0 then *y* = 0

Therefore, 0 = −0.25*x* + 1 gives 0.25*x* = 1 hence *x*
= 4 hours.

Thus, after 4
hours, the battery of the mobile phone will have no
power.

A line makes positive intercepts on coordinate axes whose sum is 7
and it**
**passes through (-3, 8) .
Find its equation.

If *a* and *b* are the intercepts then *a* + *b*
= 7 or *b* = 7 −*a*

By intercept form *x/a* + *y/b* = 1 ...(1)

We have *x/a *+* y*/(7−*a) *=* *1

As this line pass through the point (-3, 8) , we have

−3/*a* + 8/(7 – *a)* = 1 gives –3(7–*a*) + 8*a* = *a*(7–*a*)

− 21 + 3*a* + 8*a*
= 7*a* −*a*^{2}

So, *a *^{2}* *+* *4*a *−* *21*
*=* *0

Solving this equation (*a* − 3)(*a* + 7) = 0

*a *=* *3* *or* a *= −7

Since *a* is positive, we have *a* = 3 and *b* = 7–*a*
= 7–3 =4.

Hence *x/*3 + *y/*4 = 1

Therefore, 4*x* + 3*y* −12 = 0 is the required equation.

A circular garden is bounded** **by East Avenue and Cross Road. Cross Road
intersects North Street at *D* and East Avenue at *E*. *AD* is
tangential to the circular garden at *A*(3, 10).* *Using the figure.

(a) Find the equation of

(i) East Avenue.

(ii) North Street

(iii) Cross Road

(b) Where does the Cross Road intersect the

(i) East Avenue ?

(ii) North Street ?

(a) (i) East Avenue is the straight line joining* **C*(0, 2)** **and

(ii) Since the point *D* lie vertically above *C*(0, 2)
. The *x* coordinate of *D* is 0.

Since any point on North Street has *x* coordinate value 0.

The equation of North Street is *x* = 0

(iii) To find equation of Cross Road.

Center of circular garden *M* is at (7, 7), *A* is (3,
10) We

first find slope of *MA*, which we call *m*_{1}

Thus *m*_{1} = (10 −7) / (3-7) = −3/4.

Since the Cross Road is perpendicular to MA, if *m*_{2}
is the slope of the Cross
Road then, *m*_{1}*m*_{2}
= −1 gives −3/4 *m*_{2} = −1 so *m*_{2} =
4/3.

Now, the cross road has slope 4/3 and it passes through the point *A*(
3, 10).

The equation of the Cross Road is *y* − 10 = 3/4 (*x* −
3)

3*y* − 30 = 4*x* −12

Hence, 4*x* − 3*y* + 18 = 0

(b) (i) If *D* is (0, *k*) then *D* is a point on
the Cross Road.

Therefore, substituting *x* = 0 , *y* = *k* in the
equation of Cross Road,

we get, 0 − 3*k* + 18 = 0

Value of *k *=* *6

Therefore, *D *is* *(0, 6).

(ii) To find *E*, let *E* be (*q* , 2)

Put *y* = 2 in the equation of the Cross Road,

we get, 4*q* − 6 + 18 = 0

4*q* = −12 gives *q* = –3

Therefore, The point *E* is (-3, 2)

Thus the Cross Road meets the North Street at *D*(0, 6) and
East Avenue at E (-3, 2) .

Tags : Equations, Example Solved Problem | Coordinate Geometry Equations, Example Solved Problem | Coordinate Geometry

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