Question1
Solution:
Join AC. Now we have two triangles, Î” ABC and Î” ACD.
Area(Î” ABC) =
= = = 29
= Area(Î” ACD)
=
= = = 31.5
Now are of quad, ABDC = 29 + 31.5 = 60.5 sq
Question2
Solution:
A.T.Q., two triangles Î” ABC and Î” Aâ€™BC are possible.
Given BC = 2a and midpoint of BC is at 0.
â‡’ OB = OC = a
i.e., coordinate of B and C are (0,a) and (0,a), respectively.
As triangles are equilateral, we have on Î” ABC
AB = BC = CA = 2a
Applying Pythagoras theorem
OA =
=
= =
=
Similarly OAâ€™ =
As A and Aâ€™ lie on Xaxis, coordinates of A and Aâ€™ are (, 0) and (, 0) respectively.
Vertices of
Î” ABC = (0, a), (0, a), (, 0)
Vertices of
Î” Aâ€™BC = (0, a), (0, a), (, 0)
Question3
Solution:
(i) In this case x_{1} = x_{2},
Now PQ =
=
= y_{2} â€“ y_{1}
(ii) Now PQ =
=
=
Question4
Solution:
Let the point on xaxis be P(x, 0)
Let A be (7, 6) and B be (3, 4)
Now PA =
PA^{2} = (x â€“ 7)^{2} + 36
and PB =
PB^{2} = (x â€“ 3)^{2} + 16
Given PA = PB
or PA^{2 }= PB^{2}
or (x â€“ 7)^{2} + 36 = (x â€“ 3)^{2} + 16
or x^{2} â€“ 14x + 49 + 36 = x^{2} + 9 â€“ 6x + 16
or 14x + 6x = 25 â€“ 85
or 8x = 60
or 8x = 60
or 2x = 15 â‡’ x =
Hence the required point is
Question5
Solution:
Let midpoint of A(0, 4) and B(8, 0) be C.
âˆ´ C is or (4, 2)
origin is O(0,0)
Slope of OC =
= = = 
Question6
Solution:
Let the vertices be A(4, 4), B(3, 5) and C(1, 1)
Slope of AB, m_{1} = = = 1
Slope of BC, m_{2} = = =
Slope of AC, m_{3} = = 1
Now m_{1}m_{3} = 1 Ã— (1) = 1
Thus AB âŠ¥ AC or âˆ A = 90^{0}
Hence Î” ABC is a right angled triangle.
Question7
Solution:
As shown in the figure below, the given line makes an angle of 30^{0} with the positive direction of yaxis in anticlockwise direction.
Now the exterior angle
Î¸ = 90^{0 }+ 30^{0} = 120^{0}
Hence slope of AB = tan Î¸ = tan 120^{0}
= tan Î¸ (180^{0} â€“ 60^{0})
= tan 60^{0} = 
Question8
Solution:
Let the points be A(x, 1), B(2, 1) and C(4, 5)
Now as the points are collinear, therefore
m_{AB} = M_{BC }
Now m_{AB} = =
and m_{BC} = = = 2
As m_{AB} = m_{BC }
We have = 2
2 = 2(2 â€“ x) = 4 â€“ 2x
or 2x = 4 â€“ 2
â‡’ 2x = 2 â‡’ x = 1
Question9
Solution:
Let the vertices be A(2, 1), B(4, 0), C(3, 3), D(3, 2)
Now m_{AB} = =
m_{BC} = = 3
m_{AD} = = =
m_{AD} = = 3
Now m_{AB} = m_{CD} â‡’ AB  CD
and m_{BC} = m_{AD} â‡’ BC  AD
Hence ABCD is a parallelogram.
Question10
Solution:
Let the given points be A(3, 1) and B(4, 2)
Here m_{AB} = = = 1
Also slope of xaxis, m = Î¸
Let Î¸ be the angle between these two lines.
Then tanÎ¸ =
or tanÎ¸ = = 1
or Î¸ = 45^{0 }
Now obtuse angle = 180^{0} â€“ 45^{0} = 135^{0}
Hence, the angle between xaxis and the given line is 135^{0}.
Question11
Solution:
Let the slopes of the two given lines be m and 2m and Î¸ be the angle between them
The according to questions
tan Î¸ =
Now tan Î¸ =
or = or =
or 1 + 2m^{2} = 3m
or 2m^{2}  2m + 1= 0
or 2m^{2}  2m m + 1= 0
or 2m(m â€“ 1) â€“ 1(m â€“ 1) = 0
or (m â€“ 1) (2m â€“ 1) = 0
or m = 1, m =
Hence, the slopes of the lines are
Question12
Solution:
Let the points be A(x_{1}, y_{1}) and B(h, k)
Now m_{AB} =
and m_{AB} = m
Hence, = m
or k = y_{1} = m(h â€“ x_{1}), which is the required result.
Question13
Solution:
Points A(h,0), B(a, b) and C(0, k) are collinear.
For AB, m_{1} = =
For BC, m_{2} = =
For collinearity of three points,
m_{1} = m_{2}
=
or ab = (a â€“ h)(k â€“ b)
or ab = (ak â€“ hk â€“ ab + hb)
0 = ak â€“ hk + hb
or ak + hb = hk
or = 1
or = 1
Question14
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
â‡’ =
â‡’ 2(y â€“ 97) = 15
â‡’ y â€“ 97 = 7.5
â‡’ y = 104.5
Hence population in 2010 will be 104.5 crores.
Question15
Solution:
On xaxis, y = 0
Thus equation of xaxis is y = 0
On y â€“ axis, x = 0
Thus equation of yaxis is x = 0
Question16
Solution:
Equation of line through (4,3) with slope Â½ is
(y3) = Â½ (x+4)
i.e. x2y+10=0.
Question17
Solution:
The line passes through (0, 0) and its slope = m
Equation of the line is
y â€“ y_{1} = m(x â€“ x_{1})
â‡’ y â€“ 0 = m(x â€“ 0)
â‡’ y = mx
Question18
Solution:
The line passes through (2, 2)
Î¸ = 75^{0}
m = tan 75^{0} = tan(45^{0} + 30^{0}) =
= =
Equation of the line in pointslope form is
y â€“ y_{1} = m(x â€“ x_{1})
y â€“2 = (x â€“ 2)
((y  2) = (+1)(x â€“ 2)
( = x  2+ x â€“ 2)
or (+y  2  2 2 + 6) = 0
(  1)x â€“ (  1)y = 4  4
or ( + 1)x â€“ (  1)y = 4(  1)
Question19
Solution:
Equation of the line intersecting x axis at a distance 3 units to the left of origin with slope â€“2
âˆ´ The line passes through (3,0) the slope â€“2.
Hence the equation is y0 = 2(x+3)
2x â€“ y + 6 = 0.
Question20
Solution:
Equation of line intersecting yaxis at a distance 2 units above the origin and making an angle of 30^{o} with positive xaxis.
The line passes through (0,2) with slope tan30Â° =
Hence the equation of line is
(y2) = (x0)
i.e. xy + 2=0
Question21
Solution:
Equation of line through (1, 1) and (2, 4) is =
i.e =
i.e 5x + 3y + 2 = 0.
Question22
Solution:
The perpendicular distance from the origin is 5 units i.e., p = 5 and Ï‰ = 30^{0}
Now equation of the line in normal form is
x cos Ï‰ + y sin Ï‰ = p
x cos 30^{0} + y sin 30^{0} = 5
x Ã—
= 10 (Required equation).
Question23
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 2point form is
y â€“ y_{1} =
â‡’ y â€“ 2 =
â‡’ y â€“ 2 =
â‡’ 4y â€“ 8 = 3x
3x = 4y â€“ 8
â‡’ 3x â€“ 4y + 8 = 0
Question24
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
(y5) = 5(x+3)
i.e. 5x â€“ y + 20 = 0
Question25
Solution:
A line l with A(1, 0) and B(2, 3) as end points is given
l âŠ¥ 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 m_{1}m_{2} = 1)
equation of l is:
y â€“ y_{1} = n(x â€“ x_{1})
or y =
3 = 
3(n + 1)y â€“ 9n = (n + 1)x + (n+2)
(1 + n)x + 3(1 + n)y = 10n + 2
Question26
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 â‡’ a = 5
Hence the required line is x + y = 5
Question27
Solution:
Let the intercepts on the axes be a and b
Therefore, a + b = 9 â‡’ b = 9 â€“ a
Now, equation of the line in intercept form is
= 1
or = 1
(2, 2) lies on it.
â‡’ = 1
or = 1
or 18 = a(9 â€“ a)
or 18 = 9a â€“ a^{2}
or a^{2} â€“ 9a + 18 = 0
â‡’(a â€“ 6)(a â€“ 3) = 0
Hence a = 3, 6
Case (i), When a = 3 âˆ´ b = 6
Equation of the line is
= 1
2x + y = 6 (Required line)
Case (ii), when a = 6, âˆ´ b = 3
Equation of the line is
= 1
x + 2y = 6 (Required equation)
Question28
Solution:
Part I: The line passes through (0, 2) and
m = tan = 
Equation of the line in pointslope form is
y â€“ 2 = (x â€“ 0)
or y â€“ 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 pointslope form is
y + 2 = (x â€“ 0)
x + y + 2 = 0 (Required equation)
Question29
Solution:
The perpendicular from the origin O(0,0) meets the line AB at A(3, 9)
Hence, m_{OA} = = 
â‡’ m_{AB} =
Now, equation of the line will be
y â€“ 9 = (x+2)
9y â€“ 81 = 9x + 5
9x â€“ 9y + 85 = 0
Question30
Solution:
Given: L  aC + b
Now L = 124.942 and C = 20
Therefore, 20a + b = 124.942 â€¦.(1)
Again when L = 125.134 and C = 110
110a + b = 125.134
From (i) and (ii) on substraction
90a = 0.192
or a =
â‡’ 20 Ã— = 124.942
or b = 124.942 
Thus, L = +124.942 
Hence, L =
Question31
Solution:
Let selling price is denoted by = Rs. Y and demand = x litre.
A.T.Q. y = ax + b â€¦(i)
Now, when x = 980 litre, and y = Rs.14/litre
14 = 980 a + b â€¦(ii)
Again when x = 1220 and y = Rs.16/litre
16 = 1220 a + b â€¦(iii)
From (ii) and (iii), we get
1220a â€“ 980 a = 2
or 240a = 2
or a =
Putting the value of a in (ii)
980 Ã— = 14
â‡’ b = 14 
= = =
Now a = and b = and y = 17
From y = ax + b
17 = Ã— x +
â‡’ Ã— 17 

x = = 1340
Hence, x = 1340 litres.
Question32
Solution:
Let A be (c, 0) and B be (0, d)
Midpoint of AB is
Therefore, =a â‡’ c = 2a
and = b â‡’ d = 2b
New equation of AB in intercept form is
= 1
or = 1
= 2
(Required equation)
Question33
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)
Question34
(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 â€“ y_{1} =
or y â€“ 0 =
or 5y â€“ 2x â€“ 6
Now let C(8, 2) lies on it.
Therefore 5 Ã— 2 = 2 Ã— 8 â€“ 6
10 = 10 which is true.
Thus C lies on the equation of AB.
Hence A, B and C are collinear.
Question35
Solution:
(i) x + 7y = 0
In the slope intercept form the equation of the line is y = + 0
(ii) 6x + 3y â€“ 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
Question36
Solution:
(i) 3x + 2y â€“ 12 = 0
3x + 2y â€“ 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 â‡’ = 1
Question37
Solution:
(i) x  y + 8 = 0
x + = 8
Now a = 1, b =
Here = = = 2
Thus, =
or = 4
x cos 120^{0} + y sin 120^{0} = 4 (normal form)
Where p = 4 and Ï‰ = 120^{0}
(ii) y â€“ 2 = 0
y = 2
Now a = 0, b = 1
Here, = = 1
Thus, Ox + 1y = 2
or x cos 90^{0} + y sin 90^{0} = 2 (normal form)
where Ï‰ = 90^{0} and p = 2
(iii) x â€“ y = 4
Now a = 1, b = 1
Here, = =
Thus = 4
or x cos 315^{9} + y sin(315^{0}) = 4 (normal form)
where Ï‰ = 315^{0 }and p = 4.
Question38
Solution:
Given line is
12(x + 6) = 5(y â€“ 2)
12x + 72 = 5y â€“ 10
12x  5y + 82 = 0
Point is (1, 1)
Perpendicular distance
d =
=
=
= = 5 minutes
Question39
Solution:
Let the point be (x, 0)
Now = 1
or 4x + 3y = 12
or 4x + 3y â€“ 12 = 0
Therefore d =
4 =
or 10 =
Now 4x_{1} = 12 or 4x_{1} = 12 = 20
4x_{1} = 32 or 4x_{1} = 8
x_{1} = 8 or x_{1} = 2
Hence required points are (8, 0) and (2, 0)
Question40
(i) 15x + 8y â€“ 34 = 0 and 15x + 8y + 31 = 0
(ii) l(x + y) + p = 0 and lx + ly â€“ r = 0
Solution:
(i) 15x + 8y â€“ 34 = 0
and 15x + 8y + 31 = 0
Now A = 15, B = 8, C_{1} = 34, C_{2} =31
Distance between parallel lines d =
=
=
(ii) l(x â€“ y) + p = 0
lx + ly  r = 0
Now A = l, B = l, C_{1} = p, C_{2} = r
Distance between parallel lines d =
= = =
Question41
Solution:
Any line parallel to 3x4y+2=0 is 3x4y+k=0
Now, it passes through (2,3)
So, 3(2)4(3)+k=0
612+k=0 k=8
Hence required equation is
3x4y+18=0
Question42
Solution:
x7y+5=0
Any line perpendicular to x7y+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+y21=0 or 7x+y = 21(Required equation)
Question43
Solution:
x + y = 1 .. (1)
x +y = 1 .. (2)
From (1) m_{1}=  and m_{2}=
Now, c
tanÎ¸ =
tanÎ¸ =
Î¸ = 30Â°
Now, acute angle Î¸ = 30Â° and thus obtuse angle Î¸ ^{â€™}= 18030 = 150Â°.
Question44
Solution:
Let A(h,3) and B(4,1) be the given point.
We have, m_{1}=
Now, for 7x9y19=0
M_{2}=
Using tanÎ¸ =
For perpendicular lines 1+m_{1}m_{2}= 0
i.e. 1+()()=0
9(4h)14=0
369h14=0
9h=22
h=
Question45
Solution:
Any line parallel to Ax+By+C = 0 is Ax+By+K=0 â€“ (1)
Now, (x_{1},y_{1}) lies on it.
Therefore Ax_{1}+By_{1}+k=0 â€“ (2)
From (1) and (2)
Ax+By = Ax_{1}+By_{1 }
A(xx_{1})+B(yy_{1})=0 (Required equation)
Question46
Solution:
Now, Î¸ = 60Â° , m_{1}= 2. Now we find m_{2}(the slope of the other line)
We have tanÎ¸ =
Part I
tan60Â° =
=
+2m_{2}=2m_{2 }
m_{2}(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 =
=
+ 2m_{2 }= m_{2 } 2
m_{2}(2  1) = (2  )
m_{2 }=
Equation of the other line is
y  3 = (x  2)
(21)(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.
Question47
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
Question48
Solution:
The equation of the straight line perpendicular to the given straight line
3x4y16=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 3x4y=16 is
4x+3y5=0 â€¦(2)
Solving equation (1) and equation (2)
4x+3y=5
3x4y=16
Therefore the foot of the perpendicular is = ()
Question49
Solution:
Now(1,2) lies on y=mx+c
2 = m+c
mc = 2
Slope of the perpendicular =
Therefore the slope of the required line perpendicular the line from the origin m =
Substituting in the equation mc = 2
c=2
c=
m=, c=
Question50
Solution:
Given lines are xcosÎ¸ ysinÎ¸ = kcos2Î¸
And xsecÎ¸ +ycosecÎ¸ =k
Now, we find perpendicular of these lines from the origin
P = and q=
P=kcos2Î¸ and q = kcosÎ¸ sinÎ¸
P=kcos2Î¸ and q =
Cos2Î¸ = , q= i.e. sin2Î¸ =
Since cos^{2}2Î¸ +sin^{2}2Î¸ =1
We have,
Or p^{2}+4q^{2}=k^{2}
Question51
Solution:
Equation of the line joining the points B(4,1) and C(1,2)
5(y+1) = 3(x4)
5y+5 = 3x+12
5y+3x=7
Equation of the altitude from A would be of the form
3x5y+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 = 3x5y+9 = 0
Length of the altitude =