Straight Lines Class 11

Ex9.1, 4 teachoo.com
Draw a quadrilateral in the Cartesian plane, whose vertices are
(- 4, 5), (0, 7), (5, - 5) and (— 4, -2). Also, find its area
D (-4, -2)
Let points be c(5,-5)
A (-4, 5), B (0, 7), C (5, -5), D (-4, -2)
B (0, 7)
Area of quadrilateral ABCD A (-4, 5)
= Area of A ABD + Area of ABCD
Finding Area of A ABD & Area of ABCD separately

Ex 9.1, 2 teachoo.com
The base of an equilateral triangle with side 2a lies along they y-axis
such that the mid point of the base is at the origin. Find vertices of
the triangle.
Y
Let ABC be an equilateral triangle (0, a) B
2a
In equilateral triangle all sides are equal Ne
x’ Xx
So, AB=BC =AC=2a 2a A
2a
Given that {0, -a) C
y’
the base of equilateral triangle i.e. BC lies along the y-axis
such that mid point of BC is O(0, 0)
So, OB =OC=a
-. Coordinates of B = (0, a)
& Coordinates of C = (0, —a)

Ex 9.1, 3 teachoo.com
Find the distance between P (x,, y,) and Q (x, y,) when:
(i) PQ is parallel to the y-axis.
yraxis P(X1, Va)
Since PQ is parallel to y-axis
X= Xp. .
x-axis
Q(x, Yo)
Distance PQ = ./(x2 — x4)? + (v2 — y1)?
=f (%2 — X2)? + (y2 -— 1?
=V¥02— 917) +0
= (¥2 — ¥1)?
=+(y2 —¥)
= l¥2 — yal

Find the distance between P (x1, y1) and Q (x2, y2) when:
(ii) PQ is parallel to the x-axis.

Ex9.1, 4 teachoo.com
Find a point on the x-axis, which is equidistant from the points (7, 6)
and (3, 4).
Let the given points be
A(7, 6) & B (3, 4)
Let C be a point on the x-axis
Coordinates of C = C(x, 0) (As it is on the x-axis, y = 0)
Given that point C is
equidistant from the points A & B

Ex 9.1, 5 teackoo.com
Find the slope of a line, which passes through the origin, and the
mid-point of the line segment joining the points P (0, —4) & B (8, 0).
We need to find
Slope of a line passing through origin © (0, 0)
& mid point of P (0, -4) & B (8, 0)
o_o
P(0,-4) M_ B(8, 0)
Let M be the mid point of P (0, -4) & B (8, 0)
We know that
+ +
mid point of (x,, y,) and (x2, y2) = (Ft 2)
, : 0+8 —4+0
Mid point M of (0, -4) & (8, 0) = CH)

Ex 9.1, 6 teachoo.com
Without using the Pythagoras theorem, show that the points (4, 4),
(3, 5) and (-1, -1) are the vertices of a right angled triangle.

Let the 3 points of triangle be

A (4, 4), B (3, 5), C (-1,-1)

Lets calculate slope of AB, BC and AC

If product of slope is -1

It means lines are perpendicular

and it is a right angle triangle

Ex 9.1, 7 teackoo.com
Find the slope of the line, which makes an angle of 30° with the
positive direction of y-axis measured anticlockwise.
Y
30°
« 120°
x 42) xX
The line makes angle of 30° from positive y axis y
Hence it makes angle of (90° + 30°) from positive x axis
i.e. 120° from positive x axis
Hence 6 = 120°
Slope = m = tan 9
= tan 120°

Ex 9,1, 8 teachoo.com
Without using distance formula, show that points (—2, —1), (4, 0),
(3, 3) and (-3, 2) are vertices of a parallelogram.
D (-3, 2) € (3,3)
Let the given points be
A (-2,-1) , B (4, 0), C (3, 3), D (-3, 2)
A(-2,-1) B (4, 0}
We have to prove if ABCD is a parallelogram
ABCD is a parallelogram if both pairs of opposite sides are parallel
ie. AB || CD & AD || BC
So, we have to prove
Slope of AB = Slope of CD
Slope of AD = Slope of BC

Ex9,1,9 teachoo.com
Find the angle between the x-axis and the line joining the points
(3, -1) and (4, -2).
First we find slope of line joining the points (3, -1) and (4, -2).
We know that slope of line passing through (x,, y,) and (x), y.) is
¥2 7591
m=——
x2 4
Here x, = 3, y, =-1
&x% =4, y. =-2
Slope of line joining (3, -1) and (4, -2) is
—2-(-1)
m =———
4-3

Ex 9.1, 10 teachoo.com
The slope of a line is double of the slope of another line. If tangent
of the angle between them is 3? find the slopes of the lines.
Let m, & m, be the slopes of two lines
. m,—™m,
We know that angles between two lines are tan 6 = |—__—
1+m,m,
1
Here tan 0 = 3
&m, = 2m,
Putting values
m., — ™m™.
tan 6 = ||
1+ mm,
1 _| 2m,-m,
3 Jl +m,(2m,)

Ex 9.1, 11 teachoo.com
A line passes through (x,, y,) and (h, k) . If slope of the line is m,
show that k —y, =m (h—x,).
We know that slope of line passing through (x,, y,) and (x,, y,) is
m= y27-91
X_g—-Xy
So, Slope of line passing through (x,, y,) and (h, k) is
Here xy = Xx , V1 = V1
x =h,y, =k
Putting values
-K-y
m= h-x,
m(h—x,)=k—-y,
k-y, =m (h-x,)
Hence proved

teachoo.com

Ex10.1, 8
Find the value of x for which the points (x, —1), (2, 1) and (4, 5) are
collinear.
Let the three points be
A (x, -1), B (2, 1), C (4, 5)
If 3 points are collinear
Slope of AB = Slope of BC
We know that slope of a line through the points (x,, y,)(X>, ya)is
m= 220%

x, 7 xy

Ex 10.1, 13 teachoo.com
b
If three point (h, 0}, (a, b) & (0, k} lie on a line, show that + ko 1.
Let points be A (h, 0), B (a, 5), C (0, k)
Given that
A,B &Clie ona line
Hence the 3 points are collinear
-. Slope of AB = Slope of BC
We know that
Slope of a line through the points (x,, y,}, (X>, Ya) is
y27 41
m= ——
X27 X41

Ex 10.1, 14 teachoo.com
Consider the given population and year graph. Find the slope of the
line AB and using it, find what will be the population in the year 2010?
Y

The two points given are | 192

s
A (1985, 92), oo B

4 (1995, 97)
B (1995, 97) § A

zB 92

& (1985, 92)

87
Let plot a third point on
>
line C (2010, y) fe) 1985 1990 1995 2000 2005 2010
Years

Points A,B, C lie on the line
So, A, B & C are collinear
- Slope of AB = Slope of BC

Ex 9.2, 1 teachoo.com
Write the equations for the x and y-axes.
Y
At x-axis,
Value of y is always 0
Hence equation for x-axis is x oO x
y=0
Y
Similarly
At y-axis,
Value of x is always 0
Hence equation for y-axis is
x=0

Ex 9.2, 2 teachoo.com
Find the equation of the line which passes through the point (—4, 3)
+ 1
with slope 3°
We know that
equation of line passing through point (xp yo) with slope m is
Y= Yo = M(X— Xp)
Given
slope (m) =5
& point (Xp, Yo) = (-4, 3)
Putting values

Ex 9.2, 3 teachoo.com
Find the equation of the line which passes though (0, 0) with slope m.
We know that
equation of line passing through point (x, yo) with slope m is

¥— Yo = M(X— Xp)
Here Slope =m
Point (Xo, Yo) = (0, 0)
Hence x, = 0, yp = 0
Putting the values

(y- 0) = m(x-0)

y=mx

which is required equation

Ex 9.2, 4 teachoo.com
Find the equation of the line which passes though (2, 2v3) and is
inclined with the x-axis at an angle of 75°.
We know that
equation of line passing through point (xp, yp) with slope m is

¥ — Vo = M(x — Xp)
Here Point (x, yo) = (2, 2V3)
Hence x, = 2, Yo = 2v3
And slope = m = tan 6
Given 8 = 75°

2m = tan(75°)

Ex 9.2, 5 teackoo.com
Find the equation of the line which intersects the x-axis at a distance
of 3 units to the left of origin with slope —2.
Let line AB intersect the x-axis at a distance 3 units to the left of origin
At x-axis, y is always 0 y
A
-. Line AB cuts x-axis at P (—3, 0)
& slope of line AB is -2 x P(-3,0,, | © x
ie.m=—-2
YN
We know that
equation of line passing through (Xo, Yo) with slope m is
(¥— Yo) = m(x— Xo)

Ex 9.2, 6 teackoo.com
Find the equation of the line which intersects the y-axis at a
distance of 2 units above the origin and makes an angle of 30° with
the positive direction of the x-axis.
Y
Line AB intersects the y-axis 2 units above origin B
At y-axis, x will always 0, P(0,2)
-. Line AB cuts y-axis at P (0,2) Kw
x’ Oo x
A Y’
Also, line AB makes an angle of 30° with the x-axis
-. Slope = tan 6
m = tan 30°
=
"vB

Ex 9.2, 7 teachoo.com
Find the equation of the line which passes through the points (-1, 1)
and (2, -4).
We know that equation of line through two points (x,, y,) & (x, y) is
Y2— V4

yr = pam, 1)
Since equation of the line through the points (-1, 1) and (2, -4).
Here,

xX,=-1y,=1

& xX, =2, y. =-4

Ex 9.2, 8 teachoo.com
The vertices of APQR are P (2, 1), Q (—2, 3) and R (4, 5).
Find equation of the median through the vertex R.
R (4, 5}
Vertices are P (2, 1), Q (-2, 3), and R (4, 5).
We need to find equation of median
. . P S Q
.e.£
i.e. Equation of RS (2, 1) (-2, 3)
Since RS is median, S is the mid point of PQ
We know that mid point of a line joining points (x,, y,) & (x, y2) is
e +x, ¥,t+ *2)
2 7 2

Ex9.2,9 teachoo.com
Find the equation of the line passing through (-3, 5) and
perpendicular to the line through the points (2, 5) and (—3, 6).
Let AB be the line passing through (-3, 5)
& perpendicular to the line CD through (2, 5) and (-3, 6)
Let Slope of AB = m, & Slope of CD = m,
Now
Line AB is perpendicular to line CD
If two lines are perpendicular then product of their slopes are
equal to -1

Slope of AB x Slope of CD =-1

So,m,xm,=-1 w(1)

Ex 9.2, 10 teachoo.com
A line perpendicular to the line segment joining the points (1, 0) and
(2, 3) divides it in the ratio 1: n. Find the equation of the line.
Let line CD perpendicular to the line segment
AB joining two points A(1, 0) and B(2, 3) §
i.e. CD 1 AB
A re n B

(1, 0) (2, 3)

We want to equation of line CD
D
We know that if two lines are perpendicular then product of their
slope is equal to -1
Hence,
Slope of AB x Slope of CD= -1

Ex 9.2, 11 teachoo.com
Find the equation of a line that cuts off equal intercepts on the
coordinate axes and passes through the point (2, 3).
Equation of a line by intercept form is

XUV

a + b_ 1
Where a = x-intercept
& b = y-intercept
But given that x & y intercept are equal

Oo a=b

Thus, equation of line is

*4221 (As b =a)

a a

x+y=a

x+y-a=0 (1)

Ex 9.2, 12 teachoo.com
Find equation of the line passing through the point (2, 2) and
cutting off intercepts on the axes whose sum is 9.
Equation of the line in intercept form is
~ 4724
a b
where a =x - intercept
& b= y-intercept
Given that sum of intercepts is 9
a+b=9
b=9-a
Putting value b = 9 — a in equation
x yo
stotazt w(L)

Ex 9,2, 13 teachoo.com
Find equation of the line through the point (0, 2) making an angle a
with the positive x-axis. Also, find the equation of line parallel to it
and crossing the y-axis at a distance of 2 units below the origin.
Let AB be the line passing through P(0, 2) Y
ki le ~= with positive x-axis A
& making an angle z with positive x-axis P(0,2)
an
3
Slope of line AB = tan 8 x (@) x
B
_ 20
=tan (=)
Y
= tan (120°)
= tan (180 - 60° )
=-tan (60°) (tan (180 — 6) =-tan 6)
=-¥3 (tan 60° = V3)

Ex 9.2, 14 teachoo.com
The perpendicular from the origin to a line meets it at the point
(— 2, 9), find the equation of the line. Y
B
C(-2, 9}
X 0 X
A
Y
Let line OC be perpendicular to line AB at point C (-2, 9)
i.e. OC perpendicular AB
We know that
If two lines are perpendicular then product of their slopes is
equal to -1
-. Slope of OC x Slope of AB =-1

Ex 9.2, 15 teachoo.com
The length L (in centimetre} of a copper rod is a linear function of its
Celsius temperature C. In an experiment, if L = 124.942 when C = 20
and L = 125.134 when C = 110, express Lin terms of C.
Assuming C is along x-axis & L is along y-axis

(x, y) =(C, L)
So, we are given two points
(20, 124.942) and (110, 125.134)
So, from two point form

ay 27M (x-x,)
Y7Y¥1 ho — ey 1

Ex 9.2, 16 teachoo.com
The owner of a milk store finds that, he can sell 980 litres of milk
each week at Rs 14/litre and 1220 litres of milk each week at Rs
16/litre. Assuming a linear relationship between selling price and
demand, how many litres could he sell weekly at Rs 17/litre?
Let selling price be P along x-axis
& demand of milk be D along y-axis
We know that the equation of line is

y=mxtc
Here, P is along x-axis and D is along y-axis
So, our equation becomes

D=mP+c _ ...(1)

Ex 9.2, 17 teachoo.com
P (a, b) is the mid-point of a line segment between axes. Show
that equation of the line is - +f =2
Y
Plotting x-axis and y-axis B (0, q)
P (a, b)
A

L ine i i x’ X

et [be a line intersecting oO (p, 0}
x-axis at A and y-axis at B y’ I
Let P(a, b) be midpoint of AB
Here
Let co-ordinates of A be (p, 0)
Let co-ordinates of B be (0, q)

Ex 9.2, 18 teachoo.com
Point R (A, k) divides a line segment between the axes in the ratio
1:2. Find equation of the line.

Y
Let the AB be a line between axis & B(0, b)
point R(h, k) divides AB in the ratio 1: 2 1

R(h, k)
b
. 2
Let AB make x-intercept a (a, 0}
. x’ <*> x
& y-intercept b oO a N

y’
Point A = (a, 0) & B = (0, b)
So, equation of line AB by intercept form is

x y
athe 1_..(1)

Ex 9.2, 19 teachoo.com
By using the concept of equation of a line, prove that the three
points (3, 0), (-2, -2) and (8, 2) are collinear.
We need to prove that the three point
(3, 0), (-2, -2), and (8, 2) are collinear
ie. these three points lie on the same line
We check if (8, 2) lies on the line made by the points (3, 0) & (-2, -2)
We know that equation of line passing through (x,, y,) & (Xp, Yo)

(yi) = x= X)
Equation of line passing though (3, 0) & ( - 2, 2)

~ 2D
(y-0)= (2223) (x- 3)
-2
y=—, (x-3)

Ex 10.2, 8 teackoo.com
Find the equation of the line which is at a perpendicular distance of 5
units from the origin and the angle made by the perpendicular with
the positive x-axis is 30°
We need to calculate equation of line
Perpendicular distance of line from origin is 5 units
& Normal makes an angle of 30° with the positive x-axis
By the normal from
Equation of line is
xcoswt+ysinw=p.
where, p = normal distance from the origin
& w = angle which makes by the normal with positive x-axis

Ex 9.3, 1 teackoo.com
Reduce the following equations into slope-intercept form and find
their slopes and the y-intercepts.
(i) x+7y=0
+ =
x+7y=0 Slope intercept form
Wy=0-x y=mx+c
7y=-x Where mis slope
=x & cis y intercept
ve
=— +40
YEG
--1y49
Y= -OxX+

Reduce the following equations into slope-intercept form and find their slopes and the y-intercepts.
(ii) 6x + 3y – 5 = 0

Reduce the following equations into slope-intercept form and find their slopes and the y-intercepts.
(iii) y = 0

Ex 9,3, 2 teachoo.com
Reduce the following equations into intercept form and find their
intercepts on the axes.
(i) 3x+#2y-12=0.
3x + 2y-12=0 intercept form
x VL
3x + 2y=0+12 styl
Where a is x-intercept
3x + 2y=12 & bis y-intercept
Dividing both sides by 12
3x+2y 12
12° 12
3x 2y _
m2" 1
*y2%24
4°6
The above equation is of the form
~ yr a4
a b

Reduce the following equations into intercept form and find their intercepts on the axes.
(ii) 4x 3y = 6

Reduce the following equations into intercept form and find their intercepts on the axes.
(iii) 3y + 2 = 0.

Ex 9.3, 3 teackoo.com
Find the distance of the point (-1, 1) from the line 12(x + 6) = 5(y— 2).
The distance (d) of a line Ax + By + C = 0 from a point (x,, y,) is
de |Ax, + By, + C|
"fa? +B?
The given line is
12(x + 6) = 5(y— 2)
12x +12 x6=5y-5x2
12x + 72 = 5y-—10
12x -— 5y +82 =0
The above equation is of the form Ax+ By + C=0
where A= 12, B=—-5, and C = 82

Ex 9.3, 4 teackoo.com
Find the points on the x-axis, whose distances from the line
x
*,% = are 4 units.
3.4
We need to find point on the x-axis
Let any point on x-axis be P(x, 0)
Given that perpendicular distance
. . _ x Via,

from point P(x, 0) from given line 3440 1is4
Simplifying equation of line

aren y

304

Ex 9.3,5 teachoo.com
Find the distance between parallel lines
(i) 15x + 8y - 34 =O and 15x + 8y +31=0
We know that,
distance between two parallel lines
Ax + By +C,=0 & Ax + By+ C, =Ois
d= I€,— C2]
VA? +B?
Equation of first line is Equation of second line is
15x + 8y -34=0 15x + 8y+31=0
Above equation is of the form Above equation is of the form
Ax+ By+C,=0 Ax + By+C, =0
where A= 15, B=8&C,= -34 | where A=15,B=8,C,=31

Find the distance between parallel lines
(ii) 𝑙(x + y) + p = 0 and 𝑙(x + y) – r = 0

Ex 9.3, 6 teackoo.com
Find equation of the line parallel to the line 3x — 4y + 2 = 0 and
passing through the point (-2, 3).

Let the equation of line AB be 3x-—4y +2 =0

& let point P be P(-2, 3)

Let line CD be parallel to line AB & passing through the point P(—2,3)
We have to find the equation of line CD

We know that if two lines are parallel then their slopes are equal ,
Since line CD is parallel to line AB

Thus, Slope of line CD = Slope of line AB (1)

Ex9.3,7 teachoo.com
Find equation of the line perpendicular to the line x - 7y+5=0
and having x intercept 3.
Y B

Let equation of line AB be x - 7y +5 =0 c 67°

x
Let line CD be perpendicular to line AB cM
and having x-intercept 3 x A (3, 0) x

(@) |<—_—_—+"
3

D
Since Line CD has x-intercept 3 Y
So, line CD passes through the point (3, 0)
We have to find equation of line CD,

Ex 9.3, 8 teackoo.com
Find angles between the lines V3x + y=landx+ v3 yal
Given equation of lines,
V3x+y=1 (AL)
x+¥3 y=l (2)
We know that
angle between 2 lines (8) can be found by using formula
tan @ =|
L+mgmy,
Let slope of line (1) be m,
& slope of line (2) be m,

Ex 9.3, 9 teachoo.com
The line through the points (h, 3) and (4, 1) intersects the line 7x
—9y—- 19 =0. at right angle. Find the value of h.
Let the equation of line AB be 7x-9y-19=0
Let point (h, 3) be C(h, 3)
& point (4, 1) be D(4, 1)
Let CD be the line passing through points C(h,3) & D(4,1)
Given line CD intersects line AB at right angle
So, line AB and CD are perpendicular

Ex 9.3, 10 teachoo.com
Prove that the line through the point (x,, y,) and parallel to the line Ax
+ By+C=0 is A(x—x,) +Bl(y-y,) =0.
The line passing through (x,, y,) and parallel to the line Ax + By + C=0
has the same slope as the line Ax + By + C=0
Finding slope of

Ax+ By+C=0

By =-Ax-C

-Ax-€
rr
-A -C

ve(=)«+(F)
The above equation is of the form y = mx +c
Where m = slope of line
Slope of line (Ax + By + C= 0) = =

Ex 9.3, 11 teachoo.com
Two lines passing through the point (2, 3) intersects each other at an
angle of 60°. If slope of one line is 2, find equation of the other line.
We know that Angle between 2 lines be

tan6 ||

1+m gm,

Here

m, = Slope of one line = 2

8 = 60° (given)
We need to find m,
Putting the values

mz-—2
tan 60° =|-2=2 |
1+2Xx m2

Ex 9.3, 12 teachoo.com

Find the equation of the right bisector of the line segment joining

the points (3, 4) and (-1, 2).
c

Let AB be the line joining points A(-1, 2) & B(3, 4)

Let CD be the right bisector of line AB A B

P(1, 3) 12

We have to find equation of line CD (3, 4) (-1,2)
D

Since CD is the right bisector of line AB,

Point P is the mid-point of line AB

We know that co-ordinates of mid-point is given by (AR a)

: . -14+3 244 2 6
So, co-ordinates of point P = FS) = G , 5) = (1, 3)

Ex 9.3, 13 teachoo.com
Find the coordinates of the foot of perpendicular from the point
(—1, 3) to the line 3x —-4y-16=0.
C{-1, 3)
Let the equation of line AB be 3x — 4y —- 16 =0
& Point C be (-1, 3)
3x-4y-16=0
A
CD is perpendicular to the line AB D{a, b) B
& we need to find coordinates of point D
Let coordinates of point D be (a, b)
Also point D(a, b) lies on the line AB
i.e. point (a, b) satisfy the equation of line AB 3x — 4y - 16 =0
Putting x = a & y = bin equation
3a-4b-16=0
3a-—4b=16 (1)

Ex 9.3, 14 teachoo.com
The perpendicular from the origin to the line y = mx +c meets it at
the point (—1, 2). Find the values of m and c.
Let the equation of line AB be y = mx +c y
Be
x
Let OM be perpendicular to the line AB Ss
M(-1,2)_ 19"
i.e.OM L AB
x Xx
Also, M =(-1, 2) A 9
Y
Point M passes through the line AB
So, M(-1,2) must satisfy the equation of line AB
Putting values in equation
y=mx+c
2=m(-1)+c
2=-mtc

teachoo.com

Ex 9.3, 15
If p and q are the lengths of perpendiculars from the origin to the
lines x cos @-y sin 8 = k cos 26 and x sec@ +ycosec@ =k,
respectively, prove that p? + 4q? = k?
The given lines are

xcos @—y sin @=kcos 20 (1)

xsec@ + ycosec0=kcos 26 a (2)
The distance (d) of a line Ax + By +C =0 froma point (x,, y,) is
given by

d= |Ax, + By, +C|

"fa? +B?

teachoo.com
Ex 9.3, 16
In the triangle ABC with vertices A (2, 3), B (4, -1) and C (1, 2), find
the equation and length of altitude from the vertex A.
A(2, 3)

Let ABC be the triangle with
vertex A(2, 3), B(4, -1) & C(1, 2)
& AM be the altitude of triangle ABC

B(4,-1) M Cc (1, 2)
We need to calculate length & equation of altitude AM
Now,
Altitude AM is perpendicular to BC

Ex 9.3, 17 teachoo.com
If p is the length of perpendicular from the origin to the line
whose intercepts on the axes are a and b, then show that
1 1 1
pe = @ + ye :
Equation of line whose intercept on the axes are a & b is

eed

a b
The perpendicular distance (d) of a line Ax + By + C=0 froma
point (x,, y,) is given by

de |Ax, + By, +C|

Val + Be

feachoo.com
Ex10.3, 3
Reduce the following equations into normal form. Find their
perpendicular distances from the origin and angle between
perpendicular and the positive x-axis.
(i) x-VBy + 8=0 Normal form is
xcos@ + ysinWw=p
Where p is the perpendicular
x- v3 y+8=0 distance from origin
8=-x+4 v3y & w is the angle between
perpendicular & the positive x-axis
-x + V3By=8
2
Divide equation by |(—1}* + (v3) =V1 + 3=V4=2
—x+ VBy _8
2 ~2
=x v3
at aye4
-1 v3\_
(>) +v)-4 (1)

Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.
(ii) y 2 = 0

Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.
(iii) x y = 4

teachoo.com
Example 1
Find the slope of the lines:
(a) Passing through the points (3, -2) and (-1, 4)
We know that slope between two points (x,, y,) & (X>, Yo) is
y27—¥1
m=——
X27 %4
Here x, = 3, y, =-2,
&x%=-Ly, =4
Putting values
240%)
Slope = 23

Example 1
Find the slope of the lines:
(b) Passing through the points (3, –2) and (7, –2),

Example 1
Find the slope of the lines:
(c) Passing through the points (3, – 2) and (3, 4),

Example 1
Find the slope of the lines:
(d) Making inclination of 60° with the positive direction of x-axis.

Example 2 teachoo.com
If the angle between two lines is . and slope of one of the lines is 3
find the slope of the other line.
We know that angle between two lines are
m, —™m.
tan0= [P|
1+ mm,
| m 180
Putting 8 = — =—— = 45°
4 4
Let m, and m, be the slope of 2 lines
5 _1
0, M1 = 5

Example 3 teachoo.com
Line through the points (—2, 6) and (4, 8) is perpendicular to the
line through the points (8, 12) and (x, 24). Find the value of x.
Let points be A(-2, 6), B(4, 8) , C(8, 12) and D(x, 24)
If two lines are perpendicular , then product of their slope is -—1
So, Slope of AB x Slope of CD = -1 (1)
We know that slope of a line through the points (x,, y,) , (Xo, ya)is

¥2 7591

m=——
Xz 4

Example 4 teachoo.com

Find the equations of the lines parallel to axes and passing through

(-2, 3). Y

Plotting point (-2, 3) y=3
(-2, 3)

Drawing line which is parallel to x-axis x" le) x

and passing through (-2, 3) y’

Hence,

y = 3 for all points on line

Hence equation is y = 3

Example 5 teachoo.com
Find the equation of the line through (-2, 3) with slope — 4.
We know that
equation of line passing through point (x, y>) with slope m is
Y¥— Vo = M(X— Xp)
Here Slope = m = -4
Point (Xo, Yo) = (-2, 3)
Hence x, =-2 ,Y) =3
Putting the values
(y- 3) = -4(x- (-2))
(y- 3) =-4(x + 2)

Example 6 teachoo.com
Write the equation of the line through the points (1, -1) and (3, 5).
We know that equation of line through two points (x,, y,) & (x, y2) is

Y¥~ Vy = (xX)
Since equation of the line through the points (1, -1) and (3, 5).
Here,

xX=Ly,=-1

&x, =3 andy, =5
Putting values
ate Dy
(y-(-1)) =-S=* (x-1)
yt1="=*(x-1)

Example 7 teachoo.com
Write the equation of the lines for which tan @ = ; , where 6 is the
inclination of the line and
(i) y-intercept is - 5
We know that equation of line is
y=mxt+c
where m is slope of a line & c is y-intercept
Here
1
m=tan@=-
2
| 3
&c=y-intercept =— 3

Example 9
Write the equation of the lines for which tan θ = 1/2 , where θ is the inclination of the line and
(ii) x-intercept is 4

Example 8 teackoo.com
Find the equation of the line, which makes intercepts—3 and 2 on
the x- and y-axes respectively.
We know that

~ yr a4

a b
where a = x-intercept

b =y-intercept
Here, a=-3
&b=2

Putting values

~ 47a

-3°2

2x+(-3)y =1
=3x2

Example 9 teachoo.com
Find the distance of the point (3, -5) from the line 3x - 4y-26 =0.
We know that distance (d) of a point (x,, y,) from a
line Ax + By + C = Ois

de |Ax, + By2 + Cl

Vaz + B2

Now, our equation is

3x—4y-26=0
The above equation is of the form

Ax + By+C=0
where A= 3, B=-4, C=-26

Example 10 teackoo.com
Find the distance between the parallel lines 3x — 4y + 7 =O and 3x
—4y+5=0
We know that,
distance between two parallel lines
Ax + By +C,=0 & Ax + By + C, =Ois
d= ley — Gal
VA? + B?
Equation of first line is Equation of second line is
3x-4y+7=0 3x-4y+5=0
Above equation is of the form Above equation is of the form
Ax+ By+C=0 Ax+ By+C=0
Where A =3, B=-4,C,=7 Where A=3,B = -4,C, =5

Example 11 teachoo.com
If the lines 2x + y-3 = 0, 5x +ky-—3 =O and 3x -y-2=Oare
concurrent, find the value of k.
Three lines are concurrentif they pass through a common point
i.e. point of intersection of any two lines lies on the third line
It is given that lines

x+y-3=0 (1)

5x+ky-3=0 (2)

3x-y-2=0 (3)
are concurrent
So, finding point of intersection of lines (1) & (3)

Example 12 teachoo.com
Find the distance of the line 4x — y = 0 from the point P (4, 1)
measured along the line making an angle of 135° with the positive
x-axis. OLB
4
Cc >
iy
There are two lines
. P (4,1)
1. Line AB 4x-y=0 °
Se
2. Line CD making an angle 135° x 7 x
with positive x-axis A D
y
Both lines meet at Q
Point P(4, 1) is on line CD
We need to find distance PQ.

Example 13 teachoo.com
Assuming that straight lines work as the plane mirror for a point,
find the image of the point (1, 2) in the line x - 3y +4 =0.
Y
Let line AB be x-3y+4=0 B
y Q(h, k)
& point P be (1, 2)
B%
Let Q (h, k) be the image of wD P (1, 2)
. a x; x
point P (1, 2) in line AB A e)
y’
Since line AB is mirror
1. Point P & Q are at equal distance from line AB,
ie. PR = QR, i.e. Ris the mid point of PQ
2. Image is formed perpendicular to mirror
i.e. line PQ is perpendicular to line AB

Example 14 teachoo.com
Show that the area of the triangle formed by the lines y = m,x + c,
= =Qis $2 cow
»Y=M)x +c, and x = O is Zhucml’
Q(0, c>)
Poe
20%
yz
R
There are three lines given in the graph
a
x=0 he
P(O, cy) S
This line will lie be y-axis x’ x
oO
y'
ysm, xtc, y=m,x+c,
Putting x=0 Putting x =0
y=Otq=c y=O0+c,=¢,
Hence, point is P(0, c,) Hence point is Q(0, c,)

Example 15 teachoo.com
A line is such that its segment between the lines 5x-—y+4=0and
3x + 4y — 4 = Ois bisected at the point (1, 5). Obtain its equation.
Y
Given lines are (a,,B,)
5x-y+4=0. «-(1) t
3x+4y-4=0_ ... (2) ey (1,5)
x
xs TSR Be)
an
y’ SO
Let AB be the segment between the lines (1) & (2)
& point P(1, 5) be the mid-point of AB
We need to find equation of line AB
Let the points be A(a,, B,) & B(a,, B)

Example 16 teachoo.com
Show that the path of a moving point such that its distances from
two lines 3x — 2y = 5 and 3x + 2y =5 are equal is a straight line.
Given lines are

3x-2y=5ie.3x-2y-5=0 ...(1)
& 3xt+2y=Sie.3x+2y-5=0 --(2)
Let point (h, k) be any point
whose distance from line (1) & (2) equal
We know that distance of a point (x,, y,) from line Ax + By + C = 0 is

de |Ax, + By, +C|
VA? + Be

Example 4 teachoo.com
Three points P (h, k), Q (x, y,) and R (x,, y>} lie on a line. Show that
(h=,) (Yo — va) = (k= ya} (X) — X)-
Since 3 points P, Q and R are collinear
So, Slope of QP = Slope of RQ
We know that slope of a line passing through (x,, y,){X2, Y2)is
m= ¥2— Vi
X2 — xy
Slope of a line OP passing Slope of line RQ passing
through Q(x,, y;), P(h, k) through Q(x, y,) & R(X, Yo)
Here x, =h&y,=k Here, x,=%, &y,=¥y
2 =X & Y= Vi X= X RY =Yo
Putting values Putting values
m=27* m = 22294
yh X2 — xy

feachoo.com
Example 5
In Fig 10.9, time and distance graph of a linear motion is given. Two
positions of time and distance are recorded as, when T =0, D=2
and when T = 3, D = 8. Using the concept of slope, find law of
motion, i.e., how distance depends upon time.
Y

_ Cc

é
Let A=(0,2),B=(3,8),C=(1,D) § m™?)

EI B

z G, 8)

Z
Points A,B, C lie on the line A

0,2
So, A, B & C are collinear oO x
oO Time (T)
-. Slope of AB = Slope of BC
We know that
slope of a line through the points (x,, y,}(X», ya)is
m= 20%
x —%,

teachoo.com
Example 11
Find the equation of the line whose perpendicular distance from the
origin is 4 units and the angle which the normal makes with positive
direction of x-axis is 15°.
We need to find equation of line
Perpendicular distance of AB from origin is 4 units
& angle which the normal makes with (+)ve direction of x-axis is 15°
By Normal from
Equation of line is
xcosW + ysinw=p
where, p = normal distance from the origin

& w = angle which makes by the normal with positive x-axis

Here p=4&w=15°

Example 12 teackoo.com
The Fahrenheit temperature F and absolute temperature K satisfy a
linear equation. Given that K = 273 when F = 32 and that K = 373 when
F = 212. Express K in terms of F and find the value of F, when K = 0.
Assuming F along x-axis
& K along y-axis

Given that K = 273 when F = 32

& K = 373 when F = 212
We have two points (32, 273) & (212, 373) in XY plane
By two point form

Example 13 teackoo.com
Equation of a line is 3x — 4y + 10 = 0. Find its
(i) slope
3x—4y+10=0 We convert it to
—4y = -3x - 10 yomete
—4y = — (3x +10)
4y = 3x +10
_ 3x +10
ye"
3. 10
y= q xXt+ 7
3.05
y= x + 3
The above equation is of the form y = mx +c

Example 14 teachoo.com
Reduce the equation ¥3x + y — 8 = 0 into normal form. Find the
values of p and w.
Vix+y-8=0 Normal form is
xcosw +ysinw=p
V3xty=8 Where p is the perpendicular
distance from origin
Dividing by || (v3) + & w is the angle between
=/3 + 1=V4=2 perpendicular & the positive x-
axis
vax y _8
2 2 = 2
we + =4

Example 15 teachoo.com
Find the angle between the lines y - V3x - 5 =O and V3y-x+6=0.
Let the lines be
y-v3x-5=0 w(1)
V3 y-x+6=0 (2)
We know that
angle between 2 lines (8) can be found by using formula
tan6 -|a— |
L+m zm,
Let the slope of line (1) be m,
& slope of line (2) be m,

teachoo.com
Example 16
Show that two lines a,x + b,y +c, =O anda,x+b,y+c,=0, where
b,, b, # O are:
(i) Parallel if 4 ==
by b,
The given lines are
ax+by+c,=0 (1)
&a,)x+b,y+c,=0 (2)
Let slope of line (1) be m,
& slope of line (2) be m,
If two lines are parallel, then their slopes are equal

teachoo.com
Example 17
Find the equation of a line perpendicular to the line x - 2y + 3 =O and
passing through the point (1, — 2).
Let equation of line ABbe x-2y +3=0
And let point P be P(1,-2)
Let line CD be perpendicular to line AB & passing through point P(1,-2)
Lets first calculate slope of line AB
x-2y + 3=0
-2y=-x -3
-2y = —(x+3)
2y=x+3
_x +3
yao
=*,3
yaa" 3

. teachoo.com
Misc 1
Find the values of k for which the line
(k-3)x-(4-k’)y+k?-7k+6=0 is
(a) Parallel to the x-axis, Y
Any line parallel to x-axis is of the form 1
y=p x ) x
where p is constant Mi
So, there is no x term
Since Line (k-— 3) x—(4—k?) y + k?- 7k + 6 = Ois parallel to x-axis
Hence,
(k- 3)x=0

Misc 3 teachoo.com
Find the equations of the lines, which cut-off intercepts on the
axes whose sum and product are 1 and —6, respectively.
Equation of a line by intercept form is

xyrvat

a ob

where a is x — intercept

& b is y— intercept
Given that sum of intercept is 1

he.atb=t -({1)
Product of intercept is — 6

iLe.axb= -6 (2)

teachoo.
Misc 4 enieneie
What are the points on the y-axis whose distance from the line ; +
“= 1is 4 units.
4
Let any point on y-axis
be P(O, k) (at y-axis, x is always zero)
Given that distance of point on y-axis from the
line= + ~=1is 4units
3 4
Given line is
X 4 Me
3 4
Ax + BY _ 1
12
4x + 3y=12
4x + 3y - 12=0

teachoo.com

Misc 5
Find the perpendicular distance from the origin to the line joining the
points (cos 9, sin 6) and (cos ¢, sin :) .
Frist we find equation of line
We know that equation of line joining two point (x,, y,) & {X, y2) is

(y= vi) = X=)
Equation of line passing through (cos 9, sin 8) & (cos &, sin q) is

. _ sing —sin®d _

(y—sin 6) = cos 2 cos (x— cos 9)
(cos =) — cos 8) (y—sin 8) = (sind — sin 6)(x - cos 8)
cos ¢ (y—sin 8) — cos 8 (y—sin 9) = sin c (x — cos 8) — sin O(x - cos 8)
cos dy — cos d sin 8 — cos 8y + cos 8 sin 8

=sin dx -— sindcos®@ —- sin@x+sin@cos8

Misc 6 teachoo.com
Find the equation of the line parallel to y-axis and drawn through
the point of intersection of the lines x— 7y + 5 =O and 3x + y=0.
First we calculate point of intersection of lines

x-7y+5=0 (1)

&3x+y=0 (2)
Solving (1)

x-7y+5=0

x=7y-5
Putting value of x in (2)

3x+y=0

Misc 7 teachoo.com
Find the equation of a line drawn perpendicular to the line : + z
= 1 through the point, where it meets the y-axis.
D
. . x y Y
Let equation of line AB be i+ 3e 1 A
, 6)
Cc ee
Above equation is of the form 6 x
Y
ey %uy °
a b 7
. . ars) x
Where a = x-intercept of line = 4 4 8
b =y-intercept of line = 6 Y’
Since line AB makes y-intercept 6
-. Line AB meets y-axis at point (0, 6}
Let line CD be drawn perpendicular to the line AB through the
point where AB meet at the y-axes

Misc 8 teachoo.com
Find the area of the triangle formed by the lines y— x =0,x+y=0
andx-k=0.
There are three lines given in the graph
Y x
+
7
y-x=0_ ..(1) a
(0, 0) & (1, 1) satisfy this line (1,1
‘ (k, 1)
x’ Xx
xt+y=0 (2) (k, 0)
1,-1
(0,0) & (1, -1) satisfy this line ( ) B
a4
y’ “Fs
x-k=0. -.(3) x=k 9
(k, 0) & (k, 1} satisfy this line

Misc 9 teachoo.com
Find the value of p so that the three lines 3x + y— 2 = 0, px+2y—3=0
and 2x — y— 3 =O may intersect at one point.
Let lines be

3x+y—-2=0 (1)

px + 2y-3=0 (2)

2x-y—3=0 (3)
Three line may intersect at one point
Finding point of intersection of line (1) & (3)

Misc 10 teachoo.com
If three lines whose equations are y=m,x + C,,y=mM,x + C, and
y=m,xX +c; are concurrent, then show that m, (c,—c3) + m,(c3—
c,) + m,(c, -c,) = 0.
Three lines are concurrent if they pass through a common point
i.e. point of intersection of any two lines lies on the third line
It is given that lines

y=m,x + cy (1)

y=m,x + (2)

y=m,3X + C3 .-{3)
are concurrent
So, finding point of intersection of lines (1) & (3)
Subtracting (1) from (2)

y-y=(m,x + ¢,) - (mx + ¢)

O=m,x + Cc, — MX - Cc,

teachoo.com
Misc 11
Find the equation of the lines through the point (3, 2) which make an
angle of 45° with the line x — 2y = 3.
Let the equation line AB be
x-2y=3

Let line CD pass through point (3, 2)
& make an angle of 45° with line AB
i.e. angle between CD & AB is 45°
Let m, be the slope of line AB

& m, be the slope of line CD

Misc 12 teachoo.com
Find the equation of the line passing through the point of
intersection of the lines 4x + 7y— 3 = 0 and 2x — 3y + 1 = 0 that has
equal intercepts on the axes.
Given lines are

4x +7y-3=0 (1)

2x-3y+1=0 (2)
We need to calculate Equation of line
that passes through point of intersection of lines (1) & {2)
& make equal intercepts on the axes
Calculating point of intersection of lines (1) & (2)

teachoo.com
Misc 13
Show that the equation of the line passing through the origin and
+ tan@
making an angle 6 with the line y = mx + cis y-maem.
x 1+mtand
Let OP be the line passing through origin Y
ig
20rd P
Let PQ be the line y = mx +c y
Whose slope is m x’ 5 x
and makes an angle @ with line OP Y
. . _y mttand
We need to show equation line OP is — = —-————
x 1+mtanéeé

teachoo.com

Misc 14
In what ratio, the line joining (—1, 1) and (5, 7) is divided by the line
x+y=4?

C xty=4

k 1
Let line AB is the line joining A P B
the points A(-1, 1) & B(5, 7) (-1,1) (5,7)
D

&let line CD bex+y=4
Let line AB be divided by the line CD at point P
Let k :1 be the ratio line AB is divided by the line CD
We need to find value of k

Misc 15 teackoo.com
Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2)
along the line 2x — y = 0. Y D
fe)
“4
There are two lines ay
A
1. Line AB 4x+7y+5=0 Men
Ly P (1,2)
2. Line CD 2x-y=0 ON
x’ xX
oO
Both lines meet at Q
c B
Point P(1, 2) is on line CD Y’
We need to find distance PQ.
In PQ, P is (1, 2)
We need to find point Q
To find PQ, we must find coordinates of point Q

Misc 16 teachoo.com
Find the direction in which a straight line must be drawn through the
point (—1, 2) so that its point of intersection with the line x + y =4
may be at a distance of 3 units from this point.
Y
Let line AB bex+y=4 A
& point P be (—1, 2) tx
Sy
Q (h, k

P(-1, 2) Os
Qis the point of intersection of line X’*—@ x
through P(-1, 2) to AB Y’ 8
& PQ = 3 units
We need to find the equation of line PQ
First we will find the co-ordinates of point 0

teachoo.com

Misc 17
The hypotenuse of a right angled triangle has its ends at the points
(1, 3) and (— 4, 1). Find an equation of the legs (perpendicular sides)
of the triangle.

A(1, 3)
Let A ABC be a right angle triangle
where AC is the hypotenuse
& ZB = 90°

B C(-4, 1)
Given that the hypotenuse has its ends at points (1, 3) & {—4, 1)
A=(1,3) &C=(-4, 1)
We need to calculate equations of the legs of triangle
i.e. we need to find equation of line AB & BC
Let Slope of line AB =m

Misc 18 teackoo.com
Find the image of the point (3, 8) with respect to the line x + 3y =7
assuming the line to be a plane mirror. Y
Let line AB be x + 3y = 7 A p (3, 8)
& point P be (3, 8) R X43
, y= 7
xX X
° B
. Q (h, k)
Let Q (h, k) be the image of Y’
point P (3, 8) in the line AB x + 3y =7
Since line AB is mirror
1. Point P & Qare at equal distance from line AB, i.e. PR = QR, i.e.
R is the mid point of PQ.
2. Image is formed perpendicular to mirror
i.e. line PQ is perpendicular to line AB

Misc 19 teachoo.com
If the lines y = 3x + 1 and 2y =x + 3 are equally inclined to the line y =
mx + 4, find the value of m.
Let line AB be y = 3x+1

,line CD be 2y=x+3

& line PQ be y= mx +4
Lines AB & CD are equally inclined to the line PQ
First we find slopes of lines
Slope of line AB
y=3xt+1

Misc 20 teachoo.com
If sum of the perpendicular distances of a variable point P (x, y}
from the lines x + y—5 =O and 3x— 2y + 7 = Ois always 10. Show
that P must move on a line.
Given lines are
x+y-5=0 (1)
& 3x-2y+7=0 (2)
We know that ,
Perpendicular distance of a point (x,, y,) from line Ax + By + C = Ois
d= |Ax, + By, +C|
~Vae + Be

Misc 21 teachoo.com
Find equation of the line which is equidistant from parallel lines
9x + By —7 =O and 3x + 2y+6=0.
Let the given parallel lines be
Line AB: 9x + 6y-7=0
Line CD : 3x + 2y+6=0
Multiplying equation of line CD by 3
3(3x + 2y + 6) = 3(0)
9x + 6y +18 =0
So equation of line CD is 9x + 6y +18 =0

Misc 22 teachoo.com
A ray of light passing through the point (1, 2) reflects on the x-axis
at point A and the reflected ray passes through the point (5, 3).
Find the coordinates of A.

Y
There is a point A on x-axis
on which ray reflects Q{5, 3)

P (1,
x’ xX
oO A(k, 0)

y’
A ray passing through P(1, 2) reflects on point A
On reflection, the ray passes through point Q(5, 3)
We need to find coordinate of A

Misc 23 teachoo.com
Prove that the product of the lengths of the perpendiculars
drawn from the points (Va? — b2,0) and(- va? — b2,0) to
the line = cos @ +F sin 6 =1isb?.
Let point A be (va? — b2, 0)
& point B be (-Va? — b2, 0}
The given line is - cos 0+ ; sin@ = 1
We need to show that
Product of the length of perpendiculars from point A
& point B to the line ~ cos 6 +2 sin 8 =1is b?

Misc 24 teachoo.com
A person standing at the junction (crossing) of two straight paths
represented by the equations 2x — 3y + 4=0 and 3x+4y-—5=0
wants to reach the path whose equation is 6x — 7y + 8 = O in the least
time. Find equation of the path that he should follow.
mn 6x—7y+8=0 B
Let equation of lines
% S
OA: 2x- 3y+4=0 (1) Y ow
/
OB: 3x+4y-5=0 (2) *e a
X *
° +
AB: 6x—7y+8=0 (3) ”
[e)
The two paths cross at point O
-. The person is standing at point O

Misc 2 teachoo.com
Find the values of 8 and p, if the equation is the normal form of the
line V3x+y+2=0.
Normal form is
V3xty+2=0 xcosw +ysinw=p
Where p is the perpendicular
2=-V3x-y
distance from origin
—~V3x — y=2 & w is the angle between
perpendicular & the positive x-
Dividing by |(-v3)? + (-1) axis
= V3 + 1=~¥4 =2 both sides
=v3x _y _2
2 2 2
ae -F=1 (1)