# Question-1

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

**Solution:**

The point â€“ slope form is y â€“ y

_{1}= m(x â€“ x

_{1})

y + 2 = (4/7)(x + 1)

7y + 14 = 4x + 4

4x â€“ 7y = 10

# Question-2

**Determine the equation of the line with slope 3 and y â€“ intercept 4.**

**Solution:**

The slope â€“ 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 45**

^{o}with x â€“ axis and passes through the point (3, -3). Find its equation.**Solution:**

m = tan45

^{o}= 1

The slope â€“ 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 â€“ 6 = 11(x - 3)

y â€“ 6 = 11x â€“ 33

11x â€“ 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 â€“ 6 = 11(x - 3)

y â€“ 6 = 11x â€“ 33

11x â€“ 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 â€˜aâ€™ and â€˜bâ€™ are x and y intercepts respectively.

a + b = 9 (Given) â€¦â€¦â€¦â€¦â€¦â€¦..(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 + b

^{2}= 9b

b

^{2}- 9b + 18 = 0

b

^{2}- 6b â€“ 3b + 18 = 0

b (b â€“ 6) â€“ 3(b â€“ 6) = 0

(b â€“ 3)(b â€“ 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 â€˜aâ€™ and â€˜bâ€™ are x and y intercepts respectively.

a = 3b (Given) â€¦â€¦â€¦â€¦â€¦â€¦..(i)

= 1

Since (2, 2) lies on the equation, = 1

b = ==

âˆ´ a = 8.

âˆ´ 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 Î” ABC. D, E, F re the mid-points of the sides BC, CA, AB respectively.

âˆ´ D = = = (-1, -2)

E = = = (-2, -3)

F = = = (3, 5)

âˆ´ Equation of AD is

=

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

2(y - 2) = (x - 4)

2y - 6 = x â€“ 4

2y â€“ x = 0

âˆ´ Equation of BE is

=

6(y - 6) = 9(x - 4)

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

2y - 12 = 3x - 12

3y - 3x = 0

âˆ´ Equation of CF is

=

-9(y + 10) = -15(x + 6)

3(y + 10) = 5(x + 6)

3y + 30 = 5x + 30

3y â€“ 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 (x

_{1}, y

_{1}) to the straight line ax + by + c = 0 is given by .âˆ´ 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 â€˜aâ€™ and â€˜bâ€™ 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 â€¦â€¦â€¦â€¦â€¦â€¦â€¦(i)

Mid-point of AB is = (-3, 2)

b = -6 and a = 4.

âˆ´ 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 â€“6 respectively.**

**Solution:**

The intercept form is = 1 where â€˜aâ€™ and â€˜bâ€™ are x and y intercepts respectively.

a + b = 1 â€¦â€¦â€¦â€¦..(i)

ab = -6 â€¦â€¦â€¦â€¦..(ii)

(a - b)

^{2}= 4ab - (a + b)

^{2 }= 4(-6) - (1)

^{2}= -24 â€“ 1 = 25

a â€“ b = 5 â€¦â€¦â€¦â€¦..(iii)

Adding (i) and (iii)

2a = 6

a = 3

âˆ´ b = - 2

âˆ´ The equation of the straight line is = 1.

# Question-12

**Find the intercepts made by the line 7x + 3y â€“ 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 â€“ 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 â€“ axis will have y coordinate as 0.

Let the point on x-axis be P(x

_{1}, 0).

The perpendicular distance from the point P to the given straight line is

= 4

= 4 or = -4

4x

_{1}â€“ 12 = 20 or 4x

_{1}â€“ 12 = -20

4x

_{1}= 32 or 4x

_{1}= -8

x

_{1}= 8 or x

_{1}= -2

Thus the required points are (8, 0) and (-2, 0).

# Question-14

**Find the distance of the line 4x â€“ y = 0 from the point (4, 1) measured along the straight line making an angle of 135**

^{o}with the positive direction of the x-axis.**Solution:**

m = tan 135

^{o}= -1

Equation of a straight line having slope m = -1 and passing through (4, 1) is

y â€“ y

_{1}= m(x â€“ x

_{1})

y â€“ 1 = (-1)(x â€“ 4)

y â€“ 1 = -x + 4

x + y = 5 â€¦â€¦â€¦â€¦â€¦â€¦.(i)

4x â€“ y = 0 â€¦â€¦â€¦â€¦â€¦â€¦.(ii)

Solving (i) and (ii),

x = 1 and y = 4

âˆ´ 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 m

_{1}= -2.

and slope of the line x + 3y = 5 is m

_{2}= -1/3

Î¸ = tan

^{-1}= tan

^{-1}= tan

^{-1}= tan

^{-1}= tan

^{-1}(-1) = 135

^{o}

# Question-16

**Show that the straight lines 2x + y = 5 and x â€“ 2y = 4 are at right angles.**

**Solution:**

Slope of the line 2x + y = 5 is m

_{1}= -2.

and slope of the line x - 2y = 4 is m

_{2}= Â½

m

_{1}m

_{2}= -2(1/2) = -1

âˆ´ 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 â€“ 7 = 0.**

**Solution:**

The straight line parallel to 3x + 2y â€“ 7 = 0 is of the form 3x + 2y + k = 0 â€¦â€¦â€¦(i)

The point (1, -2) satisfies the equation (i)

Hence 3(1) + 2(-2) + k = 0

â‡’ 3 - 4 + k = 0â‡’ k = 1

âˆ´ 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.

âˆ´ 2 - 1 + k = 0

k = -1

âˆ´ The equation of the required straight line is x - y â€“ 1 = 0.

# Question-19

**Find the point of intersection of the straight lines 5x + 4y â€“ 13 = 0 and 3x + y â€“ 5 = 0.**

**Solution:**

Let (x

_{1}, y

_{1}) be the point of intersection. Then (x

_{1}, y

_{1}) lies on both the straight lines.

âˆ´ 5x

_{1}+ 4y

_{1}â€“ 13 = 0 â€¦â€¦â€¦â€¦â€¦..(i)

3x

_{1}+ y

_{1}â€“ 5 = 0 â€¦â€¦â€¦â€¦â€¦..(ii)

(ii) Ã— 4 12x

_{1}+ 4y

_{1}â€“ 20 = 0 â€¦â€¦â€¦â€¦â€¦.(iii)

(i) â€“ (iii) -7 x

_{1}+ 7 = 0

x

_{1}= 1

Substituting x

_{1}= 1 in (i) 5(1) + 4y

_{1}â€“ 13 = 0

5 + 4y

_{1}â€“ 13 = 0

4y

_{1}â€“ 8 = 0

y

_{1}= 2

âˆ´ The point of intersection is (1, 2).

# Question-20

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

**Solution:**

The two given equations are parallel.

2x â€“ 3y + 9 = 0 â€¦â€¦â€¦â€¦â€¦..(i)

6x + ky + 4 = 0 â€¦â€¦â€¦â€¦â€¦â€¦.(ii)

âˆ´ The coefficients of x and y are proportional

**=**.âˆ´ k = -9.

# Question-21

**Find the distance between the parallel lines 2x + y â€“ 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 â€“ 3p)y + 1 = 0 and px + 8y â€“ 7 = 0 are perpendicular to each other.**

**Solution:**

Slope of the line 8px + (2 â€“ 3p)y + 1 = 0 is m

_{1}= 8p/(2 â€“ 3p).

and slope of the line px + 8y â€“ 7 = 0 is m

_{2}=-p/8

m

_{1}m

_{2}= -1

= -1

-p

^{2}= -(2 â€“ 3p)

p

^{2}+ 3p â€“ 2 = 0

p

^{2}+ 2p + p â€“ 2 = 0

(p + 1)(p â€“ 2) = 0

âˆ´ 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 â€“ 2y + 7 = 0 and is parallel to the straight line 4x + y â€“ 11 = 0.**

**Solution:**

2x + y - 8 = 0 â€¦â€¦â€¦â€¦â€¦â€¦.(i)

3x â€“ 2y + 7 = 0 â€¦â€¦â€¦â€¦.â€¦â€¦(ii)

(i) Ã— 2

4x + 2y â€“ 16 = 0 â€¦â€¦â€¦â€¦â€¦â€¦(iii)

(ii) + (iii)

7x â€“ 9 = 0

x = 9/7

Substituting x = 9/7 in (i)

2(9/7) + y - 8 = 0

18/7 + y - 8 = 0

y = 8 â€“ 18/7 = 38/7

Point of intersection of 2x + y = 8 and 3x â€“ 2y + 7 = 0 is (9/7, 38/7).

The equation of lines parallel the straight line 4x + y â€“ 11 = 0 is 4x + y + k = 0.

4() + + k = 0

k = -âˆ´ The required equation of straight line is 4x + y - = 0 i.e., 28x + 7y â€“ 74 = 0.

# Question-24

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

**Solution:**

Equation of line through the intersection of straight lines 5x â€“ 6y = 1 and

3x + 2y + 5 = 0 is

(5x â€“ 6y â€“ 1) + k(3x + 2y + 5) = 0

(5 + 3k)x + (â€“ 6 + 2k)y + (â€“ 1 + 5k) = 0

Slope of the above equation is -(5 + 3k)/(â€“ 6 + 2k).

Above equation is perpendicular to 3x â€“ 5y + 11 = 0.

Slope of the above equation is 3/5.

= -1

= 1

15 + 9k = -30 + 10k

k = 45

âˆ´ The required equation of straight line is (5 + 3Ã— 45)x + (â€“ 6 + 2Ã— 45)y + (â€“ 1 + 5Ã— 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 â€“ y + 7 = 0 and x + y â€“ 1 = 0.**

**Solution:**

2x - y + 7 = 0 â€¦â€¦â€¦â€¦â€¦â€¦.(i)

x + y â€“ 1 = 0 â€¦â€¦â€¦â€¦.â€¦â€¦(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 â€“ x = 1 and 2x + y + 2 = 0.**

**Solution:**

3x + 2y + 1 = 0 â€¦â€¦â€¦â€¦â€¦â€¦.(i)

x + y = 3 â€¦â€¦â€¦â€¦.â€¦â€¦(ii)

(ii) Ã— -2

-2x - 2y + 6 = 0 â€¦â€¦â€¦â€¦.â€¦â€¦(iii)

(i) + (iii)

x + 7 = 0

x = -7

Substituting x = -7 in (ii)

x + y â€“ 3 = 0

-7 + y â€“ 3 = 0

y = 10

âˆ´ The point of intersection of lines 3x + 2y + 1 = 0 and x + y = 3 is (-7, 10).

y â€“ x â€“ 1 = 0 â€¦â€¦â€¦â€¦â€¦â€¦.(i)

2x + y + 2 = 0 â€¦â€¦â€¦â€¦.â€¦â€¦(ii)

(i) â€“ (ii)

-3x - 3 = 0

x = -1

Substituting x = -1 in (ii)

y + 1 â€“ 1 = 0

y = 0

âˆ´ The point of intersection of lines y â€“ 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.

âˆ´ The required equation of straight line is 5x + 3y + 5 = 0.

# Question-27

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

**Solution:**

Slope of the line 3x + 2y = 0 is m

_{1}= -3/2.

Slope of the line 4x â€“ y = 0 is m

_{2}= 4.

Angle between the straight line 3x + 2y = 0 and 4x â€“ y = 0 is tan Î¸

_{1}

= = = = 11/10

Slope of the line 2x + y = 0 is m

_{1}= -2.

Slope of the line 9x + 32y = 41 is m

_{2}= -9/32.

Angle between the straight line 2x + y = 0 and 9x + 32y = 41 is tan Î¸

_{2}=

= = = 11/10âˆ´ tan Î¸

_{1}= tan Î¸

_{2}âˆ´ Î¸

_{1}= Î¸

_{2}

# Question-28

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

**Solution:**

Slope of the line y = 2x + 7 is m

_{1}= 2.

Slope of the line x â€“ 3y â€“ 6 = 0 is m

_{2}= 1/3.

Slope of the line x + 2y = 8 is m

_{3}= -1/2.

Angle between the straight line y = 2x + 7 and x â€“ 3y â€“ 6 = 0 is

tan Î¸

_{1}= = = = 1 \ Î¸

_{1}= 45

^{o}

Angle between the straight line x â€“ 3y â€“ 6 = 0 and x + 2y = 8 is

tan Î¸

_{2}= == = 1

âˆ´ Î¸

_{2}= 45

^{o}

Angle between the straight line x + 2y = 8 and y = 2x + 7 is

tan Î¸

_{3}= = = Î¸

_{3}= 90

^{o}

âˆ´ The triangle is right angled isosceles.

# Question-29

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

**Solution:**

Slope of the line 3x + y + 4 = 0 is m

_{1}= -3.

Slope of the line 3x + 4y - 15 = 0 is m

_{2}= -3/4.

Slope of the line 24x â€“ 7y â€“ 3 = 0 is m

_{3}= 24/7.

Angle between the straight line 3x + y + 4 = 0 and 3x + 4y - 15 = 0 is

tan Î¸

_{1}= = = = 9/13

Angle between the straight line 3x + 4y - 15 = 0 and 24x â€“ 7y â€“ 3 = 0 is

tan Î¸

_{2}= = = = 117/44

Angle between the straight line 24x â€“ 7y â€“ 3 = 0 and 3x + y + 4 = 0 is

tan Î¸

_{3}= = = = 9/13

tan Î¸

_{1 }= tan Î¸

_{3}

âˆ´ The triangle is isosceles.

# Question-30

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

**Solution:**

3x + 4y = 13 â€¦â€¦â€¦â€¦â€¦..(i)

2x â€“ 7y + 1 = 0 â€¦â€¦â€¦â€¦â€¦..(ii)

5x â€“ y = 14 â€¦â€¦â€¦â€¦â€¦..(iii)

(iii) Ã— 4

20x â€“ 4y = 56 â€¦â€¦â€¦â€¦â€¦..(iv)

(i) + (iv)

23x = 69

x = 3

Substitute x = 3 in (iii)

15 â€“ y = 14

y = 1

âˆ´ 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 â€˜aâ€™ so that the straight lines x â€“ 6y + a = 0, 2x + 3y + 4 = 0 and x + 4y + 1 = 0 may be concurrent.**

**Solution:**

x â€“ 6y + a = 0 â€¦â€¦â€¦â€¦â€¦.(i)

2x + 3y + 4 = 0 â€¦â€¦â€¦â€¦â€¦.(ii)

x + 4y + 1 = 0 â€¦â€¦â€¦â€¦â€¦.(iii)

2 Ã— (iii)

2x + 8y + 2 = 0 â€¦â€¦â€¦â€¦â€¦.(iv)

(ii) â€“ (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),

â€“ 6Ã— + a = 0

- 25 + 5a = 0

a = 5

# Question-32

**Find the values of â€˜aâ€™ for which the straight lines x + y â€“ 4 = 0, 3x + 2 = 0 and x â€“ y + 3a = 0 are concurrent.**

**Solution:**

x + y â€“ 4 = 0 â€¦â€¦â€¦â€¦â€¦.(i)

3x + 2 = 0 ..â€¦â€¦â€¦â€¦.(ii)

x â€“ y + 3a = 0 â€¦â€¦â€¦â€¦â€¦.(iii)

x = -2/3 â€¦â€¦â€¦â€¦â€¦.(iv)

(-2/3) + y â€“ 4 = 0

y = 4 + 2/3 = 14/3

Substituting (-2/3, 14/3) in (iii),

(-2/3) â€“ (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 â€¦â€¦â€¦â€¦â€¦â€¦..(i)

Line perpendicular to x + 2 = 0 is x + k = 0

If it passes through (2, 5), then 2 + k = 0, k = -2

âˆ´ x â€“ 2 = 0 is one altitude. â€¦â€¦â€¦â€¦â€¦â€¦..(ii)

Line BC is =

=

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

2y + 3x = 16 â€¦â€¦â€¦â€¦â€¦â€¦..(iii)

Line perpendicular to 2y + 3x = 16 is 2x - 3y + k = 0.

If it passes through (-2, -1), then 2(-2) â€“ 3(-1) + k = 0 i.e., -4 + 3 + k = 0 i.e., k = 1

âˆ´ 2x - 3y + 1 = 0 is another altitude. â€¦â€¦â€¦â€¦â€¦â€¦..(iv)

Solving (ii) and (iv)

x = 2

2(2) â€“ 3y + 1 = 0

4 â€“ 3y + 1 = 0

-3y = -5

y = 5/3

âˆ´ 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 a**

^{3}+ b^{3}+ c^{3}= 3abc.**Solution:**

The condition for three lines concurrency is = 0

a(cb â€“ a

^{2}) â€“ b(b

^{2}- ca) + c(ab â€“ c

^{2}) = 0

abc â€“ a

^{3}â€“ b

^{3}+ abc + abc â€“ c

^{3}= 0

abc â€“ a

^{3}â€“ b

^{3}+ abc + abc â€“ c

^{3}= 0

\ a

^{3}+ b

^{3}+ c

^{3}= 3abc.

# Question-35

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

**Solution:**

Let the equation of sides AB, BC and CA of a Î” BC be represented by

x + y â€“ 1 = 0 â€¦â€¦â€¦â€¦(i)

x + 2y â€“ 4 = 0 â€¦â€¦â€¦â€¦(ii)

x + 3y â€“ 9 = 0 â€¦â€¦â€¦â€¦(iii)

(i) Ã— 2

2x + 2y â€“ 2 = 0 â€¦â€¦â€¦â€¦(iv)

(ii) - (iv)

- x â€“ 2 = 0

x = -2

Substituting x = -2 in (ii)

- 2 + 2y â€“ 4 = 0

2y = 6

y = 3

The vertex A is (-2, 3).

The equation of the straight line CA is x + 3y â€“ 9 = 0. The straight line perpendicular to its is of the form 3x - y + k = 0 â€¦â€¦â€¦â€¦â€¦â€¦..(v)

A(-2, 3) satisfies the equation (v)

âˆ´ 3x - y + k = 0

3(-2) - 3 + k = 0

k = 9

The equation of AD is 3x - y + 9 = 0. â€¦â€¦â€¦â€¦â€¦â€¦(vi)

(ii) â€“ (iii)

- y + 5 = 0

y = 5

Substituting y = 5 in (ii)

x + 2(5) â€“ 4 = 0

x + 10 â€“ 4 = 0

x = -6

The vertex C is (-6, 5).

The equation of the straight line AB is x + y â€“ 1 = 0. The straight line perpendicular to its is of the form x - y + k = 0 â€¦â€¦â€¦â€¦â€¦â€¦..(vii)

C(-6, 5) satisfies the equation (vii)

âˆ´ x - y + k = 0

-6 - 5 + k = 0

k = 11

The equation of CE is x â€“ y + 11 = 0. â€¦â€¦â€¦â€¦â€¦â€¦(viii)

(vi) â€“ (viii)

2x â€“ 2 = 0

x = 1

Substituting x = 1 in (viii)

1 â€“ y + 11 = 0

y = 12

âˆ´ 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 Î” BC be represented by

x + 2y = 0 â€¦â€¦â€¦â€¦(i)

4x + 3y = 5 â€¦â€¦â€¦â€¦(ii)

3x + y = 0 â€¦â€¦â€¦â€¦(iii)

Substituting x = -2y (ii)

4(-2y) + 3y = 5

-8y + 3y = 5

y = -1

âˆ´ 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 â€¦â€¦â€¦â€¦â€¦â€¦..(iv)

B(2, -1) satisfies the equation (iv)

âˆ´ 2 â€“ 3(-1) + k = 0

2 + 3 + k = 0

k = -5

The equation of BD is x â€“ 3y - 5 = 0. â€¦â€¦â€¦â€¦â€¦â€¦(v)

Substituting x = -2y (iii)

3(-2y) + y = 0

-6y + y = 0

-5y = 0

y = 0

âˆ´ 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 â€¦â€¦â€¦â€¦â€¦â€¦..(vi)

A(0, 0) satisfies the equation (vi)

âˆ´ 3(0) â€“ 4(0) + k = 0

k = 0

The equation of AE is 3x â€“ 4y = 0 â€¦â€¦â€¦â€¦â€¦â€¦(vii)

3 Ã— (v) â€“ (vii)

3x â€“ 9y - 15 = 0

âˆ´ -5y â€“ 15 = 0

y = -15/5 = -3

Substitute y = â€“3 in (vii)

3x â€“ 4(-3) = 0

3x + 12 = 0

x = -4âˆ´ The orthocentre O is (-4, -3).