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Slope of ST = 5-2/ 2-1 = 3

Since the required line is perpendicular to ST, slope of required line = -1/3 and lines passes through A(-2,3)

Equation of the line in slope point form is y₁ – y = m(x – x₁) 

The equation of the required line is y- 3 = -1/3 (x+2)

⇒ 3(y – 3) = -(x + 2) 

⇒ 3y – 9 = -x – 2 

⇒ x + 3y = 7

Given, slope = 4, x-intercept = 5 

Since the x-intercept of the line is 5, it passes through (5, 0). 

Equation of the line in slope point form is y – y = m(x – x ) 

Equation of the required line is 

y – 0 = 4(x – 5) 

y = 4x – 20 

4x – y = 20

4x/20 - y/20 = 1

x/5 + y/(-20) = 1

This equation is of the form x/a + y/b = 1

where x-intercept = b, y-intercept = -20

Given, points P(k, – 1), Q (2, 1) and R(4, 5) are collinear. 

∴ Slope of PQ = Slope of QR .

∴ \(\frac {1-(-1)}{2-k} = \frac {5-1}{4-2}\)

∴ \(\frac {2}{2-k} = \frac {4}{2}\)

∴ 4 = 4 (2 – k) 

∴ 1 = 2 – k 

∴ k = 2 – 1 = 1

(i) Formula to be used: y = mx + c where m is the slope of the line and c is the y - intercept.

Here, m = 3 and c = 5.

Hence, y = (3)x + (5)

i.e. y = 3x + 5

(ii) Formula to be used: y = mx + c where m is the slope of the line and c is the y - intercept.

Here, m = - 1 and c = 4.

Hence, y = (- 1)x + (4)

i.e. x + y = 4

(iii) Formula to be used: y = mx + c where m is the slope of the line and c is the y - intercept.

Here, m = -2/5 and c = -3.

Hence, y = (-2/5)x + (-3)

Or, 5y = - 2x - 3

i.e. 2x + 5y + 3 = 0

The equation of the sides AB, BC and CA of ∆ABC are y − x = 2, x + 2y = 1 and 3x + y + 5 = 0, respectively. 

Solving the equations of AB and BC, i.e. y − x = 2 and x + 2y = 1, we get: 

x = − 1, y = 1 

So, the coordinates of B are (− 1, 1). 

The altitude through B is perpendicular to AC. 

∴ Slope of AC = -3 

Thus, slope of the altitude through B is 13. 

Equation of the required altitude is given below: 

y – 1 = 13x + 1 

⇒ x – 3y + 4 = 0

Concept Used:

The equation of the line with intercepts a and b is  \(\frac{x}{a}+\frac{y}{b}=1\) 

Given: 

Here, a – b = 2 

⇒ a = b + 2 ……(i) 

Explanation: 

The line is passing through (3,2). 

\(∴\frac{3}{a}+\frac{2}{b}\)…(ii) 

From equation (i) and (ii)

\(\frac{3}{b+2}+\frac{2}{b}=1\)

⇒ 3b + 2b + 4 = b² + 2b 

⇒ b² – 3b – 4 = 0 

⇒ (b – 4)(b + 1) = 0

⇒ b = 4, -1 

Now, from equation (i) 

For b = 4, a = 4 + 2 = 6 

For b = -1, b = -1 + 2 = 1 

Thus the equation of line is 

   \(\frac{x}{1}+\frac{y}{-1}=1\) or   \(\frac{x}{6}+\frac{y}{4}=1\)

⇒x – y = 1 or 2x + 3y = 12

Given : x-intercept = y-intercept

i.e., a = b, then the equation of the line is

x/a + y/a = 1

⇒ x + y = a  … (1)

Since (1) passes through the point (2, 3), then we have,

2 + 3 = a

⇒ a = 5

∴ From (1). required line is x + y = 5.

Concept Used:

The equation of the line with intercepts a and b is  \(\frac{x}{a}+\frac{y}{b}=1\) 

Given: 

Here a + b = 7, b = 7 – a 

Explanation: 

The line is passing through (-3, 8).

 \(\frac{-3}{a}+\frac{8}{b}=1\) 

Substituting b =7 – a, we get

 \(\frac{x}{a}+\frac{y}{7-b}=1\)  

⇒ -3(7 - a) + = 7a - a²

⇒ a² + 4a – 21 = 0 

⇒ (a – 3)(a + 7)= 0

⇒ a = 3 ( since, a can only be positive) 

Substituting a = 3 in equation (i) we get, b = 7 – 3 = 4 

Hence, the equation of the line is  \(\frac{x}{3}+\frac{y}{4}=1\) 

Given: A line passing through (1, -2) 

Assuming: 

The equation of the line cutting equal intercepts at coordinates of length ‘ a ‘ is

Explanation: 

Formula used:

 \(\frac{x}{a}+\frac{y}{b}=1\) 

  \(\frac{x}{a}+\frac{y}{a}=1\) 

⇒ x + y = a 

The line x + y = a passes through (1, -2 ) So the point satisfy the equation 1 -2 = a 

⇒ a = -1 

Hence the equation of the line is x + y = -1

Let the points (− 3, − 4) and (1, − 2) be divided by y-axis at (0, t) in the ratio m:n.

 ∴ \(\Big(\frac{m-3n}{m+n},\frac{-2m-4n}{m+n}\Big) \) = (0,t)

⇒ 0 = \(\frac{m-3n}{m+n}\) 

⇒ m:n = 3:1

Let the points be A(x₁, y₁) = (– 1, 1) and B(x₂, y₂) = (5, 7) and P(x₃, y₃) be the point which divides AB in the ratio m: n.

∴ Co-ordinates of P are \(\big(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}\big)\)

= \(\big(\frac{5m - n}{m + n}, \frac{7m + n}{m + n}\big)\)

Since the point P lies on line x + y = 4.

∴ \(\frac{5m - n}{m + n}, \frac{7m + n}{m + n}\) = 4

⇒ \(\frac{12m}{m+n}\) = 4

⇒ 8m = 4n

⇒ m: n = 1: 2.

Let the line segment joining the points (1, 2) and (− 2, 1) be divided by the line 3x + 4y = 7 in the ratio m:n.

Then, the coordinates of this point will be \(\Big(\frac{-2m+n}{m+n},\frac{m+2n}{m+n}\Big)\) that lie on the line 3x + 4y = 7 

\(3 \times \frac{-2m+n}{m+n}+4\times\frac{m+2n}{m+n}\) = 7

⇒ - 2 + 11 n = 7 m + 7 n 

⇒ - 9 m = - 4 n 

⇒ m: n = 4 : 9

The combined equation of the lines 2x + y = 0 and 3x – y = 0 is

(2x + y)( 3x – y) = 0

∴ 6x² – 2xy + 3xy – y² = 0

∴ 6x² – xy – y² = 0.

Let L₁ and L₂ be the lines passing through the point (2, 3) and perpendicular to the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0 respectively.

Slopes of the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0 are -3/2 and -1/-3 = 1/3 respectively.

∴ slopes of the lines L₁ and L₂ are 2/3 and -3 respectively.

Since the lines L₁ and L₂ pass through the point (2, 3), their equations are

y – 3 = \(\frac{2}{3}\)(x – 2) and y – 3 = -3 (x – 2)

∴ 3y – 9 = 2x – 4 and y – 3= -3x + 6

∴ 2x – 3y + 5 = 0 and 3x – y – 9 = 0

∴ their combined equation is

(2x – 3y + 5)(3x + y – 9) = 0

∴ 6x² + 2xy – 18x – 9xy – 3y² + 27y + 15x + 5y – 45 = 0

∴ 6x² – 7xy – 3y² – 3x + 32y – 45 = 0.

The combined equation of the lines x + 2y – 1 = 0 and x – 3y + 2 = 0 is

(x + 2y – 1)(x – 3y + 2) = 0

∴ x² – 3xy + 2x + 2xy – 6y² + 4y – x + 3y – 2 = 0

∴ x² – xy – 6y² + x + 7y – 2 = 0.

Let m₁ be the slope of the line x – 2y – 7 = 0

∴ m₁ = \(\frac {-coefficient\, of\, x}{coefficient\, of\, y} = \frac {-1}{2}= \frac {1}{2}\)

Let m₂ be the slope of the line 2x + y + 1 = 0.

 ∴ m₂ = \(\frac {-coefficient\, of\, x}{coefficient\, of\, y} = \frac {-2}{1}= -2\)

Since m x m = 1/2 x (-2) = -1,

the given lines are perpendicular to each other. Consider

x – 2y – 7 = 0 …(i) 

2x + y + 1 =0 …(ii) 

Multiplying equation (ii) by 2, we get 4x + 2y + 2 = 0 …(iii) 

Adding equations (i) and (iii), we get 5x – 5 = 0 ∴ x = 1 

Substituting x = 1 in equation (ii), we get 2 + y + 1 = 0 

∴ y = – 3 

∴ The point of intersection of the given lines is (1,-3).

Given, A line which intersects at y–axis at a distance of 2 units and makes an angle of 30° with the positive direction of x–axis. 

To Find: The equation of that line. 

Formula used: The equation of line is [y – y₁ = m(x – x₁)] 

Explanation: Here, Angle = 30° (Given) 

So, The slope of the line, m = tan θ 

m = tan 30°

m = \(\frac{1}{\sqrt{3}}\)

Now, The coordinates are (x₁, y₁) = (0, 2)

The equation of line = y – y₁ = m(x – x₁)

y – 2 =  \(\frac{1}{\sqrt{3}}\) (x – 0)

\(\sqrt{3y}\) + \(2\sqrt{3}\) = x

 \(\sqrt{3y}\) + \(2\sqrt{3}\) - x = 0

Hence, The equation of line is  \(\sqrt{3y}\) + \(2\sqrt{3}\) - x = 0

Equations of the coordinate axes are x = 0 and y = 0.

∴ the equations of the lines passing through (2, 3) and parallel to the coordinate axes are x = 2 and

i.e. x – 2 = 0 and y – 3 = 0.

∴ their combined equation is

(x – 2)(y – 3) = 0. 

∴ xy – 3x – 2y + 6 = 0.

Let a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 be the two lines. 

(a) The lines intersect if \(\frac{a_1}{a_2}≠\frac{b_1}{b_2}\) is true. 

(b) The lines are parallel if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}\) is true. 

(c) The lines are coincident if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) is true.  

(i) Given: 2x + y − 1 = 0 and 3x + 2y + 5 = 0 

Explanation:  

Here, 

\(\frac{2}{3}≠\frac{1}{2}\)

Therefore, the lines 2x + y − 1 = 0 and 3x + 2y + 5 = 0 intersect. 

(ii) Given: x − y = 0 and 3x − 3y + 5 = 0 

Explanation: 

Here, \(\frac{1}{3}=-\frac{1}{-3}≠\frac{0}{5}\)

Therefore, the lines x − y = 0 and 3x − 3y + 5 = 0 are parallel. 

(iii) Given: 3x + 2y − 4 = 0 and 6x + 4y − 8 = 0 

Explanation:

Here, \(\frac{3}{6}=\frac{2}{4}=-\frac{4}{-8}\)

Therefore, the lines 3x + 2y − 4 = 0 and 6x + 4y − 8 = 0 are coincident.

Correct option is: (C) xy = 0

Comparing the equation x² + 6xy + 9y² = 0

with ax² + 2hxy + by² = 0, we get,

a = 1, 2h = 6, i.e. h = 3 and b = 9

Since h² – ab = (3)² – 1(9)

= 9 – 9 = 0, .

the lines represented by x² + 6xy + 9y² = 0 are coincident.

Correct option is: (B) Sq. 8/√3 units 

[Hint : Area = p²/√3 , where p is the length of perpendicular from the origin to x – y – 4 = 0]

Correct option is: (A) 2

Comparing the equation 3x² – 4 xy – 3y² = 0 with ax² + 2hxy + by² = 0, we get, 

a = 3, 2h = -4, b = -3 

Since a + b = 3 + (-3) = 0, the lines represented by 3x² – 4xy – 3y² = 0 are perpendicular to each other.

Correct option is: (B) k = -6

Correct option is: (B) 9m² – 3m – 2 = 0 

Correct option is: (A) 5x² + 4xy – 3y² = 0

The auxiliary equation of the lines given by 

3x² – 4xy + ky² = 0 is km² – 4m + 3 = 0.

Given, slope of one of the lines is 1.

∴ m = 1 is the root of the auxiliary equation km² – 4m + 3 = 0.

∴ k(1)² – 4(1) + 3 = 0

∴ k – 4 + 3 = 0 

∴ k = 1.

Comparing the given equation with

ax² + 2 hxy + by² + 2gx + 2fy + c = 0,

we get, a = 0, h = k/2 , b = 0, g = 5, f = 3, c = 4

Now, given equation represents a pair of lines.

∴ abc + 2fgh – af² – bg² – ch² = 0

∴ (0)(0)(4) + 2(3)(5) (k/2) – 0(3)² – 0(5)² – 4

(k/2)² = 0

∴ 0 + 15k – 0 – 0 – k² = 0

∴ 15k – k² = 0

∴ -k(k – 15) = 0

∴ k = 0 or k = 15.

If k = 0, then the given equation becomes 10x + 6y + 4 = 0 which does not represent a pair of lines.

∴ k ≠ o

Hence, k = 15.

Given, inclination, θ = 45°

∴ Slope of the line, m = tan θ

= tan 45° = 1.

And inclination, θ = 135°

∴ Slope, m = tan 135°

= tan (90° + 45°)

= – cot 45°

= – 1

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Let A(3, 0), B(- 2, 2) and C(8, 2) be the given points Using two-point form, we can find the equation of AB as

y - 0 = (- 2 - 0)/(- 2 - 3)(x - 3)

⇒ y = 2/5(x - 3)

⇒ 5y = 2x - 6 ....(1)

If C(8, 2) lies on the line (1) then we can conclude A, B and C are collinear. 

Now putting x = 8 and y =2 in (1), we get 

5(2) = 2(8) – 6 

10 = 16 – 6= 10 

10 – 10 = 0 

= Clieson(1) 

= A, B and C are collinear

(i) Cutting off intercepts 3 and 2 from the axes.

Given:

a = 3, b = 2

Let us find the equation of line cutoff intercepts from the axes.

By using the formula,

The equation of the line is x/a + y/b = 1

x/3 + y/2 = 1

By taking LCM,
2x + 3y = 6

∴ The equation of line cut off intercepts 3 and 2 from the axes is 2x + 3y = 6

(ii) Cutting off intercepts -5 and 6 from the axes.

Given:

a = -5, b = 6

Let us find the equation of line cutoff intercepts from the axes.

By using the formula,

The equation of the line is x/a + y/b = 1

x/-5 + y/6 = 1

By taking LCM,
6x – 5y = -30

∴ The equation of line cut off intercepts 3 and 2 from the axes is 6x – 5y = -30

Equation of the line in two point form is \(\frac {y-y_1}{y_2-y_1} = \frac {x-x_1}{x_2-x_1}\)

∴ The equation of the side BC is \(\frac {y-0}{6-0} = \frac{x-2}{-1-2}\)

\(\frac {y}{6} = \frac {x-2}{3}\)

∴ – 3y = 6x – 12 

∴ 6x + 3y – 12 = 0 

∴ 2x + y – 4 = 0

Given: 

Lines 3x – y = 5 and x + 3y = 1 

To find: 

The equation of the straight line which passes through the point of intersection of the lines 3x – y = 5 and x + 3y = 1 and makes equal and positive intercepts on the axes.

Explanation: 

The equation of the straight line passing through the point of intersection of 

3x − y = 5 and x + 3y = 1 is 3x − y − 5 + λ(x + 3y − 1) = 0 

⇒ (3 + λ)x + (− 1 + 3λ)y − 5 − λ = 0 … (1) 

⇒ y = \(-\Big(\frac{3+λ}{-1+λ}\Big)x\) + \(\Big(\frac{5+λ}{-1+λ}\Big)\)

The slope of the line that makes equal and positive intercepts on the axis is − 1. From equation (1), we have:

\(-\Big(\frac{3+λ}{-1+3λ}\Big)\) = -1

⇒ λ = 2 

Substituting the value of λ in (1), we get the equation of the required line. 

⇒ 3 + 2x + -1 + 6y – 5 – 2 = 0 

⇒ 5x + 5y – 7 = 0 

Hence, equation of required line is 5x + 5y – 7 = 0

Let u ≡ x + y – 3 = 0 and v ≡ 2x – y + 1 = 0 

Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0. 

(x + y – 3) + k(2x – y + 1) = 0 …..(i) 

x + y – 3 + 2kx – ky + k = 0

x + 2kx + y – ky – 3 + k = 0 

(1 + 2k)x + (1 – k)y – 3 + k = 0 

But, this line is parallel to X-axis Its slope = 0

⇒ \(\frac {-(1+2k)}{1-k} =0\)

⇒ 1 + 2k = 0

⇒ k = -1/2

Substituting the value of k in (i), we get

(x + y – 3) + -1/2 (2x – y + 1) = 0

⇒ 2(x + y – 3) – (2x – y + 1 ) = 0 

⇒ 2x + 2y – 6 – 2x + y – 1 = 0 

⇒ 3y – 7 = 0, which is the equation of the required line.

Given: 

The equation is parallel to x + 7y + 2 = 0 and at unit distance from the point (1, -1)

The equation of given line is

x + 7y + 2 = 0 … (1)

The equation of a line parallel to line x + 7y + 2 = 0 is given below:

x + 7y + λ = 0 … (2)

The line x + 7y + λ = 0 is at a unit distance from the point (1, − 1).

So,

1 = 1 – 7 + λ1 + 49

λ – 6 = ± 5√2

λ = 6 + 5√2, 6 – 5√2
Now, substitute the value of λ back in equation x + 7y + λ = 0, we get

x + 7y + 6 + 5√2 = 0 and x + 7y + 6 – 5√2

∴ The required lines:

x + 7y + 6 + 5√2 = 0 and x + 7y + 6 – 5√2

(i) √3x – y + 2 = 0

⇒ y = √3x + 2

Comparing with y = mx + c

m = √3

⇒ m = tan θ

tan θ = √3 = tan 60°

Thus tangent of required angle is 60°.

(ii) x + √3y – 2√3 = 0

⇒ √3y = - x + 2√3

⇒ y = (- 1/√3)x + 2

Comparing with y = mx + c

m = tanθ = - 1/√3

⇒ tanθ = tan150°

Thus, tangent of required angle is 150°.

If these three points lie on a line, the slope will be equal. 

So, slope of A(h, 0) and P(a, b) = Slope of A(h, 0) and B(0, k) 

Slope of AP = \(\Big(\frac{b-a}{a-h}\Big)\)

Slope of AB = \(\Big(\frac{k-0}{0-h}\Big)\)

Now,

\(\Big(\frac{b-a}{a-h}\Big)\) = \(\Big(\frac{k-0}{0-h}\Big)\) 

 \(\frac{b}{a-h}\) = \(-\frac{k}{h}\) 

bh = - ka + kh 

ak + bh = kh 

Dividing both sides by kh, we get,

 \(\frac{a}{h}\)+\(\frac{b}{k}\) = 1

Given equation is 3x + 2y – 6 = 0. 

Substituting x = 2 and y = 3 in L.H.S. of given equation, we get 

L.H.S. = 3x + 2y – 6 

= 3(2) + 2(3) – 6 

= 6 

≠ R.H.S. 

∴ Point A does not lie on the given line.

Let the given points be A(x₁, y₁) = (3, 4) and B(x₂, y₂) = (– 2, 1). Let m : n be the ratio in which the line segment is divided by y-axis (the coordinates of the point are (0,y))

∴ (0, y) = \(\big(\frac{mx_2 + mx_1} {m + n} , \frac{my_2 + my_1}{m + n}\big)\)

⇒ (0, y) = \(\big(\frac{3m + 3m} {m + n} , \frac{m + 4n}{m + n}\big)\)

⇒ 0 = \(\frac{−2m + 3n}{m + n}\) y = \(\frac{m + 4n}{m + n}\)

⇒ – 2m + 3n = 0

⇒ 2m = 3n

⇒ m : n = 3 : 2.

Given: The equation which makes an angle of tan^–1(3) with the x–axis and cuts off an intercept of 4 units on the negative direction of y–axis. 

To Find: The equation of the line? 

The formula used: The equation of the line is y = mx + c 

Explanation: Here, angle θ = tan^–1(3)

So, tan θ = 3 

The slope of the line is, m = 3

And, Intercept in the negative direction of y–axis is (0, -4) 

Now, The required equation of the line is y = mx + c

y = 3x – 4 

Hence, The equation of the line is y = 3x – 4.

Given, A line segment joining (2, – 5) and (1, 2) if it cuts off an intercept – 4 from y–axis. 

To Find: The equation of that line. 

Formula used: The equation of line is y = mx + C 

Explanation: Here, The required equation of line is y = mx + c 

Now, c = – 4 (Given) 

Slope of line joining (x₁ – x₂) and (y₁ – y₂) ,m = \(\frac{y_2-y_1}{x_2-x_1}\)

So, Slope of line joining (2, – 5) and (1, 2), m = \(\frac{2-(-5)}{1-2}\) = \(\frac{7}{-1}\)

Therefore, m = – 7 

Now, The equation of line is y = mx + c 

y = –7x – 4 

y + 7x + 4 = 0 

Hence, The equation of line is y + 7x + 4 = 0.

Given: 

A (1, 3) and B (3, 1) be the points joining the perpendicular bisector

Let C be the midpoint of AB.

So, coordinates of C = [(1+3)/2, (3+1)/2] 

= (2, 2)
Slope of AB = [(1-3) / (3-1)] 

= -1
Slope of the perpendicular bisector of AB = 1

Thus, the equation of the perpendicular bisector of AB is given as,

y – 2 = 1(x – 2)

y = x

x – y = 0

∴ The required equation is y = x.

Given line is 3x - 4y + 2 = 0

∴ Slope = - a/b = - (3/-4) = 3/4 = m

∴ Slope of the required line 3/4

(∴ required line is parallel to given line)

Thus equation of the required line passing through (-2, 3) with slope 3/4 is, of the form

y - y₁ = m(x - x₁)

i.e, y - 3 = 3/4(x + 2)

i.e., 4y - 12 = 3x + 6

i.e., 3x - 4y + 18 = 0

For the given graph,

Slope of line AB = \(\Big(\frac{97-92}{1995-1985}\Big)\)

Slope of line AB = \(\Big(\frac{1}{5}\Big)\)

Now, Slope of AB = Slope of AC 

Therefore, 

Slope of AC = \(\Big(\frac{(P-92)}{2010-1985}\Big)\) = \(\frac{1}{5}\)

5p – 460 = 25 

5p = 485 

P = 97 Crores.

Let (h, k) be the third vertex of the triangle. 

It is given that the area of the triangle with vertices (h, k), (− 2, − 1) and (3, 2) is 4 square units. 

\(\frac{1}{2}\) |h(-1-2)-3(-1-k)-2(2-k)| = 4

⇒ 3h – 5k + 1 = ± 8 

Taking positive sign, we get, 

3h – 5k + 1 = 8 

3h – 5k – 7 = 0 … (1) 

Taking negative sign, we get, 

3h – 5k + 9 = 0 … (2) 

The vertex (h, k) lies on the line x + y = 5. 

H + k – 5 = 0 … (3) 

On solving (1) and (3), we find (4, 1) to be the coordinates of the third vertex. 

Similarly, on solving (2) and (3), we find (2, 3) to be the coordinates of the third vertex.

Answer is (B) 4x + 3y + 2 = 0

Equation of line perpendicular to

3x – 4y + 7 = 0

4x + 3y + c = 0

It passes through point (1,2)

4(1) + 3(-2) + c = 0

⇒ 4 – 6 + c = 0

⇒ c – 2=0

⇒ c = 2

Thus, equation is 4x + 3y + 2 =0

Concept Used:

The equation of the line passing through the origin is y = mx 

To find: 

Equation of the straight line passing through the origin and bisecting the portion of a line intercepted between the coordinate axes. 

Assuming: 

The line ax + by + c = 0 meets the coordinate axes at A and B. 

Explanation: 

So, the coordinate of A and B are A(-\(\frac{c}{a}\),0) and B (0, -\(\frac{c}{a}\)) 

Now,

The midpoint of AB is \(\Big(-\frac{c}{2a},-\frac{c}{2b}\Big)\)

∴ \(-\frac{c}{2b}\) = \( m\times-\frac{c}{2a}\) 

⇒ m =  \(\frac{a}{b}\) 

Hence, the equation of the required line is 

y =   \(\frac{a}{b}x\) 

⇒ ax – by = 0