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Question-1

Draw a quadrilateral in the cartesian plane, whose vertices are(-4,5)(0,7)(5,-5) and (-4,-2). Also find its area.

Solution:
Join AC. Now we have two triangles, Δ ABC and Δ ACD.

Area(Δ ABC) = 

                 = = = 29

                 = Area(Δ ACD)
                
                 =

                 = = = 31.5

Now are of quad, ABDC = 29 + 31.5 = 60.5 sq

 

Question-2

The base of an equilateral triangle with side 2a lies along the y-axis such that the midpoint of the base is at the origin. Find the vertices of triangle.

Solution:

A.T.Q., two triangles Δ ABC and Δ A’BC are possible.

Given BC = 2a and mid-point of BC is at 0.

OB = OC = a

i.e., co-ordinate of B and C are (0,a) and (0,-a), respectively.

As triangles are equilateral, we have on Δ ABC

AB = BC = CA = 2a

Applying Pythagoras theorem

OA =

     =

     = =

     =

Similarly OA’ =

As A and A’ lie on X-axis, coordinates of A and A’ are (, 0) and (-, 0) respectively.

Vertices of

Δ ABC = (0, a), (0, -a), (, 0)

Vertices of

Δ A’BC = (0, a), (0, -a), (-, 0)

Question-3

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

Solution:
(i) In this case x1 = x2,

Now PQ =

            =

            = |y2 – y1|

(ii) Now PQ =

                =
          
                =

 

Question-4

Find a point on the x-axis, which is equidistant from the points(7, 6) and (3,4).

Solution:
Let the point on x-axis be P(x, 0)

Let A be (7, 6) and B be (3, 4)

Now PA =

PA2 = (x – 7)2 + 36

and PB =

PB2 = (x – 3)2 + 16

Given PA = PB

or PA2 = PB2

or (x – 7)2 + 36 = (x – 3)2 + 16

or x2 – 14x + 49 + 36 = x2 + 9 – 6x + 16

or -14x + 6x = 25 – 85

or -8x = -60

or 8x = 60

or 2x = 15 x =

Hence the required point is

Question-5

Find the slope of a line, which passes through the origin, and the midpoint of the line-segment joining and points A(0, -4) and B(8, 0).

Solution:
Let midpoint of A(0, -4) and B(8, 0) be C.

C is or (4, -2)

origin is O(0,0)

Slope of OC =
= = = -

Question-6

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

Solution:
Let the vertices be A(4, 4), B(3, 5) and C(-1, -1)

Slope of AB, m1 = = = -1

Slope of BC, m2 = = =

Slope of AC, m3 = = 1

Now m1m3 = -1 × (1) = -1

Thus AB AC or A = 900

Hence Δ ABC is a right angled triangle.

Question-7

As shown in the figure below, the given line makes an angle of 300 with the positive direction of y-axis in anticlockwise direction.

Solution:
As shown in the figure below, the given line makes an angle of 300 with the positive direction of y-axis in anticlockwise direction.

Now the exterior angle

θ = 900 + 300 = 1200

Hence slope of AB = tan θ = tan 1200
= tan θ (1800 – 600)

= tan 600 = -

Question-8

Find the value of x for which the points(x, -1), (2, 1) and (4, 5) are collinear.

Solution:
Let the points be A(x, -1), B(2, 1) and C(4, 5)

Now as the points are collinear, therefore

mAB = MBC

Now mAB = =

and mBC = = = 2

As mAB = mBC

We have = 2

2 = 2(2 – x) = 4 – 2x

or 2x = 4 – 2

2x = 2 x = 1

Question-9

Without using distance formula, show that points(-2, -1), (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.

Solution:
Let the vertices be A(-2, -1), B(4, 0), C(3, 3), D(-3, 2)

Now mAB = =

mBC = = -3

mAD = = =

mAD = = -3

Now mAB = mCD AB || CD

and mBC = mAD BC || AD

Hence ABCD is a parallelogram.

Question-10

Find the angle between x-axis and the line joining the points(3, -1) and (4, -2).

Solution:
Let the given points be A(3, -1) and B(4, -2)

Here mAB = = = -1

Also slope of x-axis, m = θ

Let θ be the angle between these two lines.

Then tanθ =

or tanθ = = 1

or θ = 450

Now obtuse angle = 1800 – 450 = 1350

Hence, the angle between x-axis and the given line is 1350.

Question-11

The slope of a line is double of the slope of another line. If tangent of the angle between them is find the slopes of the lines.

Solution:
Let the slopes of the two given lines be m and 2m and θ be the angle between them

The according to questions

tan θ =

Now tan θ =

or = or =

or 1 + 2m2 = 3m

or 2m2 - 2m + 1= 0

or 2m2 - 2m -m + 1= 0

or 2m(m – 1) – 1(m – 1) = 0

or (m – 1) (2m – 1) = 0

or m = 1, m =

Hence, the slopes of the lines are

Question-12

A line passes through (x1, y1) and (h, k). If slope of the line is m, show that k – y1 = m(h – x1).

Solution:
Let the points be A(x1, y1) and B(h, k)

Now mAB =

and mAB = m

Hence, = m

or k = y1 = m(h – x1), which is the required result.

Question-13

If three points(h, 0), (a, b) and (0, k) lie on a line, show that = 1.

Solution:
Points A(h,0), B(a, b) and C(0, k) are collinear.

For AB, m1 = =

For BC, m2 = =

For collinearity of three points,

m1 = m2

=

or -ab = (a – h)(k – b)

or -ab = (ak – hk – ab + hb)

0 = ak – hk + hb

or ak + hb = hk

or = 1

or = 1

Question-14

Find the slope of the line AB and using it, find what will be the population in the year 2010.

Solution:

A(1985, 92), B(1995, 97)

Slope of AB = m = = =

Let population for 2010 be y crores.

Thus C is the point having coordinates(2010, y)

Now slope of BC

=

2(y – 97) = 15
y – 97 = 7.5

y = 104.5

Hence population in 2010 will be 104.5 crores.

Question-15

Write the equations for the x-and y-axes.

Solution:
On x-axis, y = 0

Thus equation of x-axis is y = 0

On y – axis, x = 0

Thus equation of y-axis is x = 0

Question-16

Find the equation of the line passing through the point (-4,3) with slope .

Solution:
Equation of line through (-4,3) with slope ½ is

(y-3) = ½ (x+4)

i.e. x-2y+10=0.

Question-17

Find the equation of the line Passing through (0, 0) with slope m.

Solution:
The line passes through (0, 0) and its slope = m

Equation of the line is

y – y1 = m(x – x1)

y – 0 = m(x – 0)

y = mx

Question-18

Find the equation of the line Passing through (2, 2) and inclined with the x-axis at an angle of 75°.

Solution:
The line passes through (2, 2)

θ = 750
m = tan 750 = tan(450 + 300) =
                  = =

Equation of the line in point-slope form is

y – y1 = m(x – x1)

y –2 = (x – 2)

((y - 2) = (+1)(x – 2)

( = x - 2+ x – 2)

or (+y - 2 - 2- 2 + 6) = 0

( - 1)x – ( - 1)y = 4 - 4

or ( + 1)x – ( - 1)y = 4( - 1)

Question-19

Find the equation of the line intersecting x-axis at a distance of 3 units to the left of the origin with slope -2.

Solution:
Equation of the line intersecting x axis at a distance 3 units to the left of origin with slope –2

The line passes through (-3,0) the slope –2.

Hence the equation is y-0 = -2(x+3)

2x – y + 6 = 0.

Question-20

Find the equation of the line intersecting y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of x-axis.

Solution:
Equation of line intersecting y-axis at a distance 2 units above the origin and making an angle of 30o with positive x-axis.

The line passes through (0,2) with slope tan30° =

Hence the equation of line is

(y-2) = (x-0)

i.e. x-y + 2=0

Question-21

Find the equation of the line passing through points (-1,1) and (2, -4).

Solution:
Equation of line through (-1, 1) and (2, -4) is =
i.e =
i.e 5x + 3y + 2 = 0.

Question-22

Find the equation of a line where perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 300.

Solution:
The perpendicular distance from the origin is 5 units i.e., p = 5 and ω = 300

Now equation of the line in normal form is

x cos ω + y sin ω = p

x cos 300 + y sin 300 = 5

x ×

= 10 (Required equation).

Question-23

The vertices of triangle PQR are P(2, 1), Q(-2, 3) and R(4, 5). Find equation of the median through the vertices R.

Solution:
P(2, 1), Q(-2, 3) and R(4, 5)

Midpoint of PQ is

= (0, 2)

Now we have to find the equation of median passing through R i.e. of line AR.

Now equation of line in 2-point form is

y – y1 =
y – 2 =
y – 2 =

4y – 8 = 3x
3x = 4y – 8

3x – 4y + 8 = 0

Question-24

Find the equation of the line passing through (-3,5) and perpendicular to the line through the points (2,5) and (-3,6).

Solution:
Slope of line through (2,5) and (-3,6) is =

Slope of the perpendicular is 5

Hence equation of line through (-3,5) with slope 5 is

(y-5) = 5(x+3)

i.e. 5x – y + 20 = 0

Question-25

A line perpendicular to the linesegment joining the points(1, 0) and (2, 3) divides it in the ratio 1 : n. Find the equation of the line.

Solution:
A line l with A(1, 0) and B(2, 3) as end points is given

l AB and AC : CB = 1 : n

Now coordinates of C are

or which is C.

Now slope of AB = = 3

Now slope of l = -

(as m1m2 = -1)

equation of l is:

y – y1 = n(x – x1)

or y- =

3 = -

3(n + 1)y – 9n = -(n + 1)x + (n+2)

(1 + n)x + 3(1 + n)y = 10n + 2

Question-26

Find the equation of a line that cuts off equal intercepts on the coordinates axes and passes through the point(2, 3).

Solution:
Let the equal intercepts on the axes be a.

Equation of line in intercept form is:
= 1

or x + y = a

As it passes through(2, 3), we have

2 + 3 = a a = 5

Hence the required line is x + y = 5

Question-27

Find operation of the line passing through the point(2, 2) and cutting off intercepts on the axes whose sum is 9.

Solution:
Let the intercepts on the axes be a and b

Therefore, a + b = 9 b = 9 – a

Now, equation of the line in intercept form is

= 1

or = 1

(2, 2) lies on it.

= 1

or = 1

or 18 = a(9 – a)

or 18 = 9a – a2

or a2 – 9a + 18 = 0

(a – 6)(a – 3) = 0

Hence a = 3, 6

Case (i), When a = 3 b = 6

Equation of the line is

= 1

2x + y = 6 (Required line)

Case (ii), when a = 6, b = 3

Equation of the line is

= 1

x + 2y = 6 (Required equation)

Question-28

Find the equation of the line through the point, (0, 2) making an angle with the positive x-axis. Also, find the equation of the line parallel to it and crossing the y-axis at a distance of 2 units below the origin.

Solution:
Part I: The line passes through (0, 2) and

m = tan = -

Equation of the line in point-slope form is

y – 2 = (x – 0)

or y – 2 = -x

or x + y - 2 = 0 (Required equation)

Part II: The line passes through(0, -2) and m = -(for parallel lines)

Now equation of the line in point-slope form is

y + 2 = -(x – 0)

x + y + 2 = 0 (Required equation)

Question-29

The perpendicular from the origin to a line meets it at the point(-2, 9). Find the equation of the line.

Solution:
The perpendicular from the origin O(0,0) meets the line AB at A(-3, 9)
 

Hence, mOA = = -

mAB =

Now, equation of the line will be

y – 9 = (x+2)

9y – 81 = 9x + 5

9x – 9y + 85 = 0

Question-30

The length L (in centimeters) 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 = 110, express L is term of C.

Solution:
Given: L - aC + b

Now L = 124.942 and C = 20

Therefore, 20a + b = 124.942 ….(1)

Again when L = 125.134 and C = 110

110a + b = 125.134

From (i) and (ii) on substraction

90a = 0.192

or a =

20 × = 124.942

or b = 124.942 -

Thus, L = +124.942 -

Hence, L =

Question-31

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?

Solution:
Let selling price is denoted by = Rs. Y and demand = x litre.

A.T.Q. y = ax + b …(i)

Now, when x = 980 litre, and y = Rs.14/litre

14 = 980 a + b …(ii)

Again when x = 1220 and y = Rs.16/litre

16 = 1220 a + b …(iii)

From (ii) and (iii), we get

1220a – 980 a = 2

or 240a = 2

or a =

Putting the value of a in (ii)

980 × = 14

b = 14 -

= = =

Now a = and b = and y = 17

From y = ax + b

17 = × x +

× 17 -

-

x = = 1340

Hence, x = 1340 litres.

Question-32

P(a, b) is the midpoint of a line segment between axes. Show that equation of the line is = 2.

Solution:
Let A be (c, 0) and B be (0, d)

Mid-point of AB is

Therefore, =a c = 2a

and = b d = 2b

New equation of AB in intercept form is

= 1

or = 1

= 2

(Required equation)

Question-33

Point R(h, k) divides a line segment between the axes in the ratio 1 : 2. Find the equation of the line.

Solution:

Thus h = =

and k = =

Therefore a = and b = 3k.

Now using intercept form, equation of the line AB is

= 1

or = 1

= 3

2kx + hy = 3kh (Required equation)

Question-34

By using the concept of equation of a line, prove that the three points(3, 0),
(-2, -2) and (8, 2) are collinear.

Solution:
The given points are A(3, 0), B(-2, -2), C(8, 2).

First we find equation of AB which is

y – y1 =

or y – 0 =

or 5y – 2x – 6

Now let C(8, 2) lies on it.

Therefore 5 × 2 = 2 × 8 – 6

10 = 10 which is true.

Thus C lies on the equation of AB.

Hence A, B and C are collinear.

Question-35

Reduce the following into slope –intercept form and find their slopes and the y-intercepts.(i) x + 7y = 0 (ii) 6x+3y-5=0 (iii) y=0

Solution:
(i) x + 7y = 0
In the slope intercept form the equation of the line is y = + 0

(ii) 6x + 3y – 5 = 0
In the slope intercept form the equation of the line is y = -2x +

(iii) y = 0

In the slope intercept form the equation of the line y = 0x+0

Question-36

Reduce the following equations into intercepts form and their intercepts on the axes.(i) 3x + 2y – 12 (ii) 4x – 3y – 6 (iii) 3y – 2 = 0

Solution:
(i) 3x + 2y – 12 = 0

3x + 2y – 12

or = 1

or = 1 (Intercepts form)

where a = 4, b = 6

(ii) 4x + 3y = 6

= 1

= 1 (Intercepts form)

where a = 3/2, b = -2

(iii) 3y + 2 = 0

3y = 2

= 1 = 1

Question-37

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 - y+ 8 = 0 (ii) y – 2 = 0 (iii) x – y = 4

Solution:
(i) x - y + 8 = 0

-x + = 8

Now a = 1, b =

Here = = = 2

Thus, =

or = 4

x cos 1200 + y sin 1200 = 4 (normal form)

Where p = 4 and ω = 1200

(ii) y – 2 = 0

y = 2

Now a = 0, b = 1

Here, = = 1

Thus, Ox + 1y = 2

or x cos 900 + y sin 900 = 2 (normal form)

where ω = 900 and p = 2

(iii) x – y = 4

Now a = 1, b = -1

Here, = =

Thus = 4

or x cos 3159 + y sin(3150) = 4 (normal form)

where ω = 3150 and p = 4.

Question-38

Find the distance of the point (-1, 1) from the line 12(x + 6) = 5(y – 2).

Solution:
Given line is

12(x + 6) = 5(y – 2)

12x + 72 = 5y – 10

12x - 5y + 82 = 0

Point is (-1, 1)

Perpendicular distance

d =

  =

  =

  = = 5 minutes

Question-39

Find the points on the x-axis, whose distances from the line = 1 are 4 units.

Solution:
Let the point be (x, 0)

Now = 1

or 4x + 3y = 12

or 4x + 3y – 12 = 0

Therefore d =

4 =

or 10 =

Now 4x1 = 12 or 4x1 = 12 = -20

4x1 = 32 or 4x1 = 8

x1 = 8 or x1 = -2

Hence required points are (8, 0) and (-2, 0)

Question-40

Find the distance between parallel lines

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

(ii) l(x + y) + p = 0 and lx + ly – r = 0

Solution:
(i) 15x + 8y – 34 = 0

and 15x + 8y + 31 = 0

Now A = 15, B = 8, C1 = -34, C2 =31

Distance between parallel lines d =

                                              =

                                              =

(ii) l(x – y) + p = 0

lx + ly - r = 0

Now A = l, B = l, C1 = p, C2 = -r

Distance between parallel lines d =

                                              = = =

Question-41

Fine the equation of the line parellel to the line 3x-4y+2=0 and passing through the point (-2,3)

Solution:
Any line parallel to 3x-4y+2=0 is 3x-4y+k=0

Now, it passes through (-2,3)

So, 3(-2)-4(3)+k=0

-6-12+k=0 k=8

Hence required equation is

3x-4y+18=0

Question-42

Find the equation of the line perpendicular to the line x-7y+5=0 and having x -intercept 3.

Solution:
x-7y+5=0

Any line perpendicular to x-7y+5 = 0 is 7x+y+k=0

Now (3,0) lies on it.

So, 7 x 3+0+k=0 or k = -21

Thus required equation is 7x+y-21=0 or 7x+y = 21(Required equation)

Question-43

Find the angles between the linesx+y=1 and x+y=1.

Solution:
x + y = 1 .. (1)

x +y = 1 .. (2)

From (1) m1= - and m2=

Now, c

tanθ =
tanθ =

θ = 30°

Now, acute angle θ = 30° and thus obtuse angle θ = 180-30 = 150°.

Question-44

The line through the points (h,3) and (4,1) intersects line 7x-9y-19=0 at right angle. Find the value of k.

Solution:
Let A(h,3) and B(4,1) be the given point.

We have, m1=

Now, for 7x-9y-19=0

M2=

Using tanθ =

For perpendicular lines 1+m1m2= 0

i.e. 1+()()=0

9(4-h)-14=0

36-9h-14=0

9h=22

h=

Question-45

Prove that the line through the point (x1,y1) and parallel to the line Ax+By+C=0 is A(x-x1)+b(y-y1) = 0

Solution:
Any line parallel to Ax+By+C = 0 is Ax+By+K=0 – (1)

Now, (x1,y1) lies on it.

Therefore Ax1+By1+k=0 – (2)

From (1) and (2)

Ax+By = Ax1+By1

A(x-x1)+B(y-y1)=0 (Required equation)

Question-46

Two lines passing through the point (2,3) intersect each other at an angle of 60° . If slope of one line is 2, find the equation of the other line.

Solution:
Now, θ = 60° , m1= 2. Now we find m2(the slope of the other line)
We have tanθ =

Part I

tan60° =

=

+2m2=2-m2

m2(2+1)=2-

Therefore equation of the other line is

y - 3 =

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

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

( - 2)x + (2 + 1)y = 6 + 3 - 4 + 2

( - 2)x + (2 + 1)y = 8 - 1.

Part II

tan 60 =

=

+ 2m2 = m2 - 2

m2(2 - 1) = (-2 - )

m2 =

Equation of the other line is

y - 3 = (x - 2)

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

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

(2 +)X + (2 - 1)Y = 4 + 2 + 6

(2 +)x + (2 - 1)y = 8 + 1 is the equation required.

Question-47

Find the equation of the right bisector of the line segment joining the points (3,4) and (-1,2).

Solution:
Slope of the line joining the points (3, 4) and (-1, 2)

=

Therefore the slope of the perpendicular bisector is = -2

Midpoint of the line joining the points (3, 4) and (-1, 2)

()

= (1, 3)

The equation of the right bisector is

y - 3 = -2(x - 1)

y - 3 = -2x + 2

y + 2x = 5

Question-48

(-1,3) to the line 3x-4y-16=0

 


Solution:
The equation of the straight line perpendicular to the given straight line

3x-4y-16=0 (1)

Will be of the form 4x+3y+k=0.

Since, (-1,3) is a point on it, 4(-1)+3(3)+k=0

Therefore k = -5

Substituting in the equation above, the equation of the straight line passing through(-1,3) and perpendicular to the equation 3x-4y=16 is

4x+3y-5=0 …(2)

Solving equation (1) and equation (2)

4x+3y=5

3x-4y=16

Therefore the foot of the perpendicular is = ()

Question-49

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.

Solution:
Now(-1,2) lies on y=mx+c

2 = -m+c

m-c = -2

Slope of the perpendicular =

Therefore the slope of the required line perpendicular the line from the origin m =

Substituting in the equation m-c = -2

-c=-2

c=

m=, c=

Question-50

If p and q are the lengths of perpendiculars from the origin to the lines xcosθ -ysinθ =kcos2θ and xsecθ +ycosθ =k, respectively, prove that p2+4q2=k2

Solution:
Given lines are xcosθ -ysinθ = -kcos2θ

And xsecθ +ycosecθ =k

Now, we find perpendicular of these lines from the origin

P = and q=

P=kcos2θ and q = kcosθ sinθ

P=kcos2θ and q =

Cos2θ = , q= i.e. sin2θ =

Since cos22θ +sin22θ =1

We have,

Or p2+4q2=k2

Question-51

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

Solution:
Equation of the line joining the points B(4,-1) and C(-1,2)

5(y+1) = -3(x-4)

5y+5 = -3x+12

5y+3x=7

Equation of the altitude from A would be of the form

3x-5y+k = 0

Since point A(2,3) passes through this equation

3(2)-5(3) +k = 0

k = 9

Therefore the equation of the altitude is = 3x-5y+9 = 0

Length of the altitude =





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