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Question-1

Determine the equation of the straight line passing through the point (-1, -2) and having slope 4/7.

Solution:
The point – slope form is y – y1 = m(x – x1)
                                        y + 2 = (4/7)(x + 1)
                                     7y + 14 = 4x + 4
                                      4x – 7y = 10

Question-2

Determine the equation of the line with slope 3 and y – intercept 4.

Solution:
The slope – intercept form is y = mx + c.
Therefore the equation of the straight line is y = 3m + 4.

Question-3

A straight line makes an angle of 45o with x – axis and passes through the point (3, -3). Find its equation.

Solution:
m = tan45o = 1
The slope – intercept form is y = mx + c.
(-3) = 1(3) + c
    c = -6
Therefore the equation of the straight line is y = x - 6.

Question-4

Find the equation of the straight line joining the points (3, 6) and (2, -5).

Solution:
The equation of a straight line passing through two points is =.
Substituting the points (3, 6) and (2, -5),
=
=

y – 6 = 11(x - 3)
y – 6 = 11x – 33
11x – y = 27 is the required equation of the straight line.
The equation of a straight line passing through two points is =.
Substituting the points (3, 6) and (2, -5),
=
=

y – 6 = 11(x - 3)
y – 6 = 11x – 33
11x – y = 27 is the required equation of the straight line.

Question-5

Find the equation of the straight line passing through the point (2, 2) and having intercepts whose sum is 9.

Solution:
The intercept form is = 1 where ‘a’ and ‘b’ are x and y intercepts respectively.
a + b = 9 (Given) ………………..(i)
  = 1

Since (2, 2) lies on the equation, = 1
2(a + b) = ab
        ab = 18 (from i)
          a = 18/b

Substituting in (i)
18 + b2 = 9b
b2 - 9b + 18 = 0
b2 - 6b – 3b + 18 = 0
b (b – 6) – 3(b – 6) = 0
(b – 3)(b – 6) = 0
b = 3 or 6
Therefore a = 6 or 3.

Therefore the required equation of straight line is = 1 or = 1.

Question-6

Find the equation of the straight line whose intercept on the x-axis is 3 times its intercept on the y-axis and which passes through the point (-1, 3).

Solution:
The intercept form is = 1 where ‘a’ and ‘b’ are x and y intercepts respectively.
     a = 3b (Given) ………………..(i)
= 1
Since (2, 2) lies on the equation, = 1
     b = ==
  a = 8.
The required equation of straight line is = 1 i.e., x + 3y = 8.

Question-7

Find the equations of the medians of the triangle formed by the point (2, 4), (4, 6) and (-6, -10).

Solution:
Let A(2, 4), B(4, 6) and C(-6, -10) be the given vertices of a Δ ABC. D, E, F re the mid-points of the sides BC, CA, AB respectively.  

D = = =  (-1, -2)
   E = = = (-2, -3)
   F = = = (3, 5)                
Equation of AD is
     =
6(y - 2) = 3(x - 4)
2(y - 2) = (x - 4)
  2y - 6 = x – 4
  2y – x = 0

Equation of BE is
=
6(y - 6) = 9(x - 4)
2(y - 6) = 3(x - 4)
 2y - 12 = 3x - 12

3y - 3x = 0

Equation of CF is
=
-9(y + 10) = -15(x + 6)
 3(y + 10) = 5(x + 6)
   3y + 30 = 5x + 30
    3y – 5x = 0

Question-8

Find the length of the perpendicular from (3, 2) to the straight line 3x + 2y + 1 = 0.

Solution:
The perpendicular distance from (x1, y1) to the straight line ax + by + c = 0 is given by . The length of the perpendicular from (3, 2) to the straight line 3x + 2y + 1 = 0 is =.--

Question-9

The portion of straight line between the axes is bisected at the point (-3, 2). Find its equation.

Solution:
The intercept form is = 1 where ‘a’ and ‘b’ are x and y intercepts respectively.
The straight line make the x intercept OA = a, and y intercept OB = b.
Then, A is (0, a) and B is (b, 0).
= 1 …………………(i)
Mid-point of AB is = (-3, 2)
b = -6 and a = 4.
The equation of straight line is = 1.

Question-10

Find the equation of the diagonals of quadrilateral whose vertices are (1, 2), (-2, -1), (3, 6) and (6, 8).

Solution:
Let A(1, 2), B(-2, -1), C(3, 6) and D(6, 8) be the vertices of quadrilateral ABCD.
Equation of the diagonal AC is =
    -2(y - 2) = -4(x - 1)
     -2y + 4 = -4x + 1
4x - 2y + 3 = 0
Equation of the diagonal BD is =
   -16(y + 1) = -9(x + 2)
     -16y - 16 = -9x - 18
9x - 16y + 2 = 0

Question-11

Find the equation of the straight line, which cut of intercepts on the axes whose sum and product are 1 and –6 respectively.

Solution:
The intercept form is = 1 where ‘a’ and ‘b’ are x and y intercepts respectively.
a + b = 1 …………..(i)
    ab = -6 …………..(ii)
(a - b)2 = 4ab - (a + b)2 = 4(-6) - (1)2 = -24 – 1 = 25
   a – b = 5 …………..(iii)
Adding (i) and (iii)
 2a = 6
   a = 3
b = - 2
The equation of the straight line is = 1.

Question-12

Find the intercepts made by the line 7x + 3y – 6 = 0 on the coordinate axis.

Solution:
If y = 0 then x = 6/7
   If x = 0 then y = 6/3 = 2
x - intercept is 6/7 and y – intercept is 2.

Question-13

What are the points on x-axis whose perpendicular distance from the straight line = 1 is 4?

Solution:
= 1
4x + 3y = 12
Any point on x – axis will have y coordinate as 0.
Let the point on x-axis be P(x1, 0).
The perpendicular distance from the point P to the given straight line is
= 4
= 4         or    = -4
4x1 – 12 = 20     or     4x1 – 12 = -20
4x1 = 32            or     4x1 = -8
x1 = 8               or     x1 = -2
Thus the required points are (8, 0) and (-2, 0).

Question-14

Find the distance of the line 4x – y = 0 from the point (4, 1) measured along the straight line making an angle of 135o with the positive direction of the x-axis.

Solution:
m = tan 135o = -1
Equation of a straight line having slope m = -1 and passing through (4, 1) is
y – y1 = m(x – x1)
y – 1 = (-1)(x – 4)
y – 1 = -x + 4
x + y = 5 ……………….(i)
4x – y = 0 ……………….(ii)
Solving (i) and (ii),
x = 1 and y = 4
The distance between the points (1, 4) and (4, 1) is
= === 3units

Question-15

Find the angle between the straight lines 2x + y = 4 and x + 3y = 5

Solution:
Slope of the line 2x + y = 4 is m1 = -2.
and slope of the line x + 3y = 5 is m2 = -1/3
θ = tan-1= tan-1= tan-1= tan-1= tan-1(-1) = 135o

Question-16

Show that the straight lines 2x + y = 5 and x – 2y = 4 are at right angles.

Solution:
Slope of the line 2x + y = 5 is m1 = -2.
and slope of the line x - 2y = 4 is m2 = ½
m1 m2 = -2(1/2) = -1
The two straight lines are at right angles.

Question-17

Find the equation of the straight line passing through the point (1, -2) and parallel to the straight line 3x + 2y – 7 = 0.

Solution:
The straight line parallel to 3x + 2y – 7 = 0 is of the form 3x + 2y + k = 0 ………(i)
The point (1, -2) satisfies the equation (i)
Hence 3(1) + 2(-2) + k = 0
3 - 4 + k = 0 k = 1
3x + 2y + 1 = 0 is the equation of the required straight line.

Question-18

Find the equation of the straight line passing through the point (2, 1) and perpendicular to the straight line x + y = 9.

Solution:
The equation of the straight line perpendicular to the straight line x + y = 9 is of the form x - y + k = 0.
The point (2, 1) lies on the straight line x - y + k = 0.
2 - 1 + k = 0
              k = -1
The equation of the required straight line is x - y – 1 = 0.

Question-19

Find the point of intersection of the straight lines 5x + 4y – 13 = 0 and 3x + y – 5 = 0.

Solution:
Let (x1, y1) be the point of intersection. Then (x1, y1) lies on both the straight lines.
5x1 + 4y1 – 13 = 0 ……………..(i)
       3x1 + y1 – 5 = 0 ……………..(ii)
(ii) × 4 12x1 + 4y1 – 20 = 0 …………….(iii)
        (i) – (iii) -7 x1+ 7 = 0
                              x1 = 1
Substituting x1 = 1 in (i) 5(1) + 4y1 – 13 = 0
5 + 4y1 – 13 = 0
        4y1 – 8 = 0
               y1 = 2
The point of intersection is (1, 2).

Question-20

If the two straight lines 2x – 3y + 9 = 0, 6x + ky + 4 = 0 are parallel, find k.

Solution:
The two given equations are parallel.
2x – 3y + 9 = 0 ……………..(i)
6x + ky + 4 = 0 ……………….(ii)
The coefficients of x and y are proportional = . k = -9.

Question-21

Find the distance between the parallel lines 2x + y – 9 = 0 and 4x + 2y + 7 = 0.

Solution:
The distance between the parallel lines is ===units

Question-22

Find the values of p for which the straight lines 8px + (2 – 3p)y + 1 = 0 and px + 8y – 7 = 0 are perpendicular to each other.

Solution:
Slope of the line 8px + (2 – 3p)y + 1 = 0 is m1 = 8p/(2 – 3p).
and slope of the line px + 8y – 7 = 0 is m2 =-p/8
 m1 m2 = -1
= -1
-p2 = -(2 – 3p)
p2 + 3p – 2 = 0
p2 + 2p + p – 2 = 0
(p + 1)(p – 2) = 0
The two straight lines are at right angles.

Question-23

Find the equation of the straight line which passes through the intersection of the straight lines 2x + y = 8 and 3x – 2y + 7 = 0 and is parallel to the straight line 4x + y – 11 = 0.

Solution:
2x + y - 8 = 0 ……………….(i)
 3x – 2y + 7 = 0 ………….……(ii)
(i) × 2
4x + 2y – 16 = 0 ………………(iii)

(ii) + (iii)
7x – 9 = 0
       x = 9/7

Substituting x = 9/7 in (i)
2(9/7) + y - 8 = 0
  18/7 + y - 8 = 0
 y = 8 – 18/7 = 38/7

Point of intersection of 2x + y = 8 and 3x – 2y + 7 = 0 is (9/7, 38/7).
The equation of lines parallel the straight line 4x + y – 11 = 0 is 4x + y + k = 0.
4() + + k = 0
                 k = - The required equation of straight line is 4x + y - = 0 i.e., 28x + 7y – 74 = 0.

Question-24

Find the equation of the straight line passing through intersection of the straight lines 5x – 6y = 1 and 3x + 2y + 5 = 0 and perpendicular to the straight line 3x – 5y + 11 = 0.

Solution:
Equation of line through the intersection of straight lines 5x – 6y = 1 and
3x + 2y + 5 = 0 is

         (5x – 6y – 1) + k(3x + 2y + 5) = 0
(5 + 3k)x + (– 6 + 2k)y + (– 1 + 5k) = 0
Slope of the above equation is -(5 + 3k)/(– 6 + 2k).
Above equation is perpendicular to 3x – 5y + 11 = 0.
Slope of the above equation is 3/5.
= -1
    = 1
   15 + 9k = -30 + 10k
            k = 45
The required equation of straight line is (5 + 3× 45)x + (– 6 + 2× 45)y + (– 1 + 5× 45) = 0
140x + 84y + 224 = 0
   20x + 12y + 32 = 0
        5x + 3y + 8 = 0

Question-25

Find the equation of the straight line joining (4, -3) and the intersection of the straight lines 2x – y + 7 = 0 and x + y – 1 = 0.

Solution:
2x - y + 7 = 0 ……………….(i)
    x + y – 1 = 0 ………….……(ii)

(i) + (ii)
3x + 6 = 0
       x = -2

Substituting x = -2 in (i)
  2(-2) - y + 7 = 0
      -4 - y + 7 = 0
                  y = 3

The equation of the line joining (4, -3) and (-2, 3) is
=
6(y + 3) = -6(x - 4)
y + 3 + x - 4 = 0
x + y = 1

Question-26

Find the equation of the straight line joining the point of the intersection of the straight lines 3x + 2y + 1 = 0 and x + y = 3 to the point of intersection of the straight lines y – x = 1 and 2x + y + 2 = 0.

Solution:
3x + 2y + 1 = 0 ……………….(i)
             x + y = 3 ………….……(ii)

(ii) × -2
-2x - 2y + 6 = 0 ………….……(iii)

(i) + (iii)
x + 7 = 0
      x = -7

Substituting x = -7 in (ii)
  x + y – 3 = 0
-7 + y – 3 = 0
            y = 10
The point of intersection of lines 3x + 2y + 1 = 0 and x + y = 3 is (-7, 10).
   y – x – 1 = 0 ……………….(i)
2x + y + 2 = 0 ………….……(ii)

(i) – (ii)
-3x - 3 = 0
        x = -1

Substituting x = -1 in (ii)
y + 1 – 1 = 0
           y = 0
The point of intersection of lines y – x = 1 and 2x + y + 2 = 0 is (-1, 0).
The equation of the line joining (-7, 10) and (-1, 0) is
=
-6(y - 10) = 10(x + 7)
-3(y - 10) = 5(x + 7)
 -3y + 30 = 5x + 35
5x + 3y + 5 = 0.
The required equation of straight line is 5x + 3y + 5 = 0.

Question-27

Show that the angle between 3x + 2y = 0 and 4x – y = 0 is equal to the angle between 2x + y = 0 and 9x + 32y = 41.

Solution:
Slope of the line 3x + 2y = 0 is m1 = -3/2.
     Slope of the line 4x – y = 0 is m2 = 4.
Angle between the straight line 3x + 2y = 0 and 4x – y = 0 is tan θ1
= = = = 11/10
   Slope of the line 2x + y = 0 is m1 = -2.
Slope of the line 9x + 32y = 41 is m2 = -9/32.
Angle between the straight line 2x + y = 0 and 9x + 32y = 41 is tan θ 2 =
 = = = 11/10 tan θ 1 = tan θ 2 θ 1 = θ 2

Question-28

Show that the triangle whose sides are y = 2x + 7, x – 3y – 6 = 0 and x + 2y = 8 is right angled. Find its other angles.

Solution:
Slope of the line y = 2x + 7 is m1 = 2.
Slope of the line x – 3y – 6 = 0 is m2 = 1/3.
     Slope of the line x + 2y = 8 is m3 = -1/2.
     Angle between the straight line y = 2x + 7 and x – 3y – 6 = 0 is
tan θ 1 = = = = 1  \ θ 1 = 45o
Angle between the straight line x – 3y – 6 = 0 and x + 2y = 8 is
tan θ 2 = == = 1
  θ 2 = 45o
Angle between the straight line x + 2y = 8 and y = 2x + 7 is
tan θ 3 = = =       θ 3 = 90o
The triangle is right angled isosceles.

Question-29

Show that the straight lines 3x + y + 4 = 0, 3x + 4y - 15 = 0 and 24x – 7y – 3 = 0 form an isosceles triangle.

Solution:
Slope of the line 3x + y + 4 = 0 is m1 = -3.
Slope of the line 3x + 4y - 15 = 0 is m2 = -3/4.
Slope of the line 24x – 7y – 3 = 0 is m3 = 24/7.
Angle between the straight line 3x + y + 4 = 0 and 3x + 4y - 15 = 0 is
tan θ 1 = = = = 9/13
Angle between the straight line 3x + 4y - 15 = 0 and 24x – 7y – 3 = 0 is
tan θ 2 = = = = 117/44
Angle between the straight line 24x – 7y – 3 = 0 and 3x + y + 4 = 0 is
tan θ 3 = = = = 9/13
tan θ 1 = tan θ 3
The triangle is isosceles.

Question-30

Show that the straight lines 3x + 4y = 13, 2x – 7y + 1 = 0 and 5x – y = 14 are concurrent.

Solution:
3x + 4y = 13 ……………..(i)
2x – 7y + 1 = 0 ……………..(ii)
        5x – y = 14 ……………..(iii)
(iii) × 4
20x – 4y = 56 ……………..(iv)
(i) + (iv)
23x = 69
   x = 3
Substitute x = 3 in (iii)
15 – y = 14
       y = 1
The point of intersection is (3, 1).
3(3) + 4(1) = 13
        9 + 4 = 13
The point (3, 1) satisfies equation (i). Hence they are concurrent.

Question-31

Find ‘a’ so that the straight lines x – 6y + a = 0, 2x + 3y + 4 = 0 and x + 4y + 1 = 0 may be concurrent.

Solution:
x – 6y + a = 0 …………….(i)
 2x + 3y + 4 = 0 …………….(ii)
   x + 4y + 1 = 0 …………….(iii)
2 × (iii)
2x + 8y + 2 = 0 …………….(iv)
(ii) – (iv)
-5y + 2 = 0
         y = 2/5
Substituting y = 2/5 in (iii)
x + 4(2/5) + 1 = 0
                   x = -13/5
Substituting (-13/5, 2/5) in (i),
– 6× + a = 0
       - 25 + 5a = 0
                  a = 5

Question-32

Find the values of ‘a’ for which the straight lines x + y – 4 = 0, 3x + 2 = 0 and x – y + 3a = 0 are concurrent.

Solution:
x + y – 4 = 0 …………….(i)
       3x + 2 = 0 ..………….(ii)
  x – y + 3a = 0 …………….(iii)
               x = -2/3 …………….(iv)
(-2/3) + y – 4 = 0
   y = 4 + 2/3 = 14/3
Substituting (-2/3, 14/3) in (iii),
(-2/3) – (14/3) + 3a = 0
          (-16/3) + 3a = 0
                          a = 16/9

Question-33

Find the coordinates of the orthocentre of the triangle whose vertices are the points (-2, -1), (6, -1) and (2, 5).

Solution:
Let A(-2, -1), B(6, -1) and C(2, 5) be the vertices of the triangle ABC.
Line AB is =
              x + 2 = 0 ………………..(i)

Line perpendicular to x + 2 = 0 is x + k = 0
  If it passes through (2, 5), then 2 + k = 0, k = -2
x – 2 = 0 is one altitude. ………………..(ii)

Line BC is =
                =
          2(y + 1) = -3(x - 6)
           2y + 3x = 16 ………………..(iii)

Line perpendicular to 2y + 3x = 16 is 2x - 3y + k = 0.
If it passes through (-2, -1), then 2(-2) – 3(-1) + k = 0 i.e., -4 + 3 + k = 0 i.e., k = 1
2x - 3y + 1 = 0 is another altitude. ………………..(iv)

Solving (ii) and (iv)
                 x = 2
2(2) – 3y + 1 = 0
    4 – 3y + 1 = 0
              -3y = -5
                 y = 5/3
Orthocentre is (2, 5/3).

Question-34

If ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 are concurrent, show that a3 + b3 + c3 = 3abc.

Solution:
The condition for three lines concurrency is = 0
a(cb – a2) – b(b2 - ca) + c(ab – c2) = 0
      abc – a3 – b3 + abc + abc – c3 = 0
      abc – a3 – b3 + abc + abc – c3 = 0
                            \ a3 + b3 + c3 = 3abc.

Question-35

Find the coordinates of the orthocentre of the triangle formed by the straight lines x + y – 1 = 0, x + 2y – 4 = 0 and x + 3y – 9 = 0.

Solution:
Let the equation of sides AB, BC and CA of a Δ BC be represented by
x + y – 1 = 0 …………(i)
x + 2y – 4 = 0 …………(ii)
x + 3y – 9 = 0 …………(iii)

(i) × 2
2x + 2y – 2 = 0 …………(iv)

(ii) - (iv)
- x – 2 = 0
        x = -2
Substituting x = -2 in (ii)
- 2 + 2y – 4 = 0
             2y = 6
               y = 3
The vertex A is (-2, 3).
The equation of the straight line CA is x + 3y – 9 = 0. The straight line perpendicular to its is of the form 3x - y + k = 0 ………………..(v)

A(-2, 3) satisfies the equation (v)
3x - y + k = 0
3(-2) - 3 + k = 0
                k = 9
The equation of AD is 3x - y + 9 = 0. ………………(vi)

(ii) – (iii)
- y + 5 = 0
y = 5

Substituting y = 5 in (ii)
x + 2(5) – 4 = 0
x + 10 – 4 = 0
x = -6
The vertex C is (-6, 5).
The equation of the straight line AB is x + y – 1 = 0. The straight line perpendicular to its is of the form x - y + k = 0 ………………..(vii)

C(-6, 5) satisfies the equation (vii)
x - y + k = 0
  -6 - 5 + k = 0
              k = 11
The equation of CE is x – y + 11 = 0. ………………(viii)

(vi) – (viii)
2x – 2 = 0
       x = 1
Substituting x = 1 in (viii)
1 – y + 11 = 0
             y = 12
The orthocentre O is (1, 12).

Question-36

The equation of the sides of a triangle are x + 2y = 0, 4x + 3y = 5 and 3x + y = 0. Find the coordinates of the orthocentre of the triangle.

Solution:
Let the equation of sides AB, BC and CA of a Δ BC be represented by
  x + 2y = 0 …………(i)
4x + 3y = 5 …………(ii)
  3x + y = 0 …………(iii)

Substituting x = -2y (ii)
4(-2y) + 3y = 5
    -8y + 3y = 5
              y = -1
           x = 2
The vertex B is (2, -1).

The equation of the straight line AC is 3x + y = 0. The straight line perpendicular to its is of the form x - 3y + k = 0 ………………..(iv)

B(2, -1) satisfies the equation (iv)
2 – 3(-1) + k = 0
        2 + 3 + k = 0
                    k = -5
The equation of BD is x – 3y - 5 = 0. ………………(v)

Substituting x = -2y (iii)
3(-2y) + y = 0
    -6y + y = 0
         -5y = 0
            y = 0
         x = 0
The vertex A is (0, 0).The equation of the straight line BC is 4x + 3y = 5. The straight line perpendicular to its is of the form 3x - 4y + k = 0 ………………..(vi)
A(0, 0) satisfies the equation (vi)
3(0) – 4(0) + k = 0
                       k = 0
The equation of AE is 3x – 4y = 0 ………………(vii)

3 × (v) – (vii)
3x – 9y - 15 = 0

  -5y – 15 = 0
    y = -15/5 = -3
 Substitute y = –3 in (vii)
    3x – 4(-3) = 0
       3x + 12 = 0
                x = -4 The orthocentre O is (-4, -3).




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