Explore topic-wise InterviewSolutions in .

The given equation of the line is

(K – 3) x – (4 – K²) y + K² – 7x + 6 = 0 (i)

Here A = K – 3, B = – (4 – K²), C = K² – 7x + 6

The line will be parallel to y – axis if

B = 0

⇒ – (4 – K²) = 0

⇒ K = ± 2

The given equation is

\(\sqrt{3}\)x + y = 4

⇒ \(\frac {\sqrt{3}x}{\sqrt{{(\sqrt{3})^2+1^2}}}\) + \(\frac {y}{\sqrt{{(\sqrt{3})^2+1^2}}}\) = \(\frac {4}{\sqrt{{(\sqrt{3})^2+1^2}}}\),.

A = \(\sqrt{3}\), B = 1, C = – 4

⇒ \(\frac{\sqrt{3}}{2}\)x + \(\frac{1}{2}\)y = \(\frac{4}{2}\)

⇒ x cos 30° + y sin 30° = 2, is the required equation of line in normal form where P = 2, a = 30°

The given equation is

6x + 2y + 5 = 0

Here, A = 6, B = 2, C = 5

∴ Equation in intercept form is

\(\frac{x}{-\frac{C}{A}} + \frac{y}{-\frac{C}{B}}\) = 1

⇒ \(\frac{x}{-\frac{5}{6}} + \frac{y}{-\frac{2}{6}}\) = 1

⇒ \(\frac{x}{-\frac{5}{6}} + \frac{y}{-\frac{2}{6}}\) = 1,

⇒ \(\frac{x}{-\frac{5}{6}} + \frac{y}{-\frac{1}{3}}\) = 1, where a (intercept on x-axis)

= − \(\frac{5}{6}\) and b (intercept on y – axis) = − \(\frac{1}{3}\)

The given points are A(3, -1) and B = (4, -2)

Slope of AB = (-2 - (-1))/(4 - 3) = - 1/1 = - 1

Let θ be the angle made by the line AB with the positive x-axis then slope of 

AB = tan θ … (2) 

∴ From (1) and (2), we have 

tan θ = -1 = – tan 45° 

= tan (180°-45°) 

= tan (135°) 

= θ =135° 

AB makes an angle of 135° with positive direction of x-axis

Slope of the line passing through the points (-2, 6) and (4, 8) is

m₁ = (8 - 6)/(4 - (-2)) = 2/6 = 1/3

Slope of the line passing through the points (8, 12) and (x, 24) is

m₂ = (24 - 21)/(x - 8) = 12/(x - 8)

Since the lines are perpendicular then m₁ x m₂ = - 1

i.e, 1/3 x 12/(x - 8) = -1

i.e., 4 = - x + 8 ⇒ x = 4

Slope of the line passing through (x₁ > y₁) and (h, k) is = (k - y₁)/(h - x₁) = m  (given)

k - y₁ = m(h - x₁)

Here, A = (2, -1) andB = (4, 3)

Slope of line AB = \(\frac{y_2-y_1}{x_2-x_1} = \frac {3-(-1)}{4-2} = \frac {4}{2} =2\)

Given, A(x₁ , y₁ ), B(h, k) and slope of line AB = m 

Slope of line AB = \(\frac {y_2-y_1}{x_2-x_1}\)

∴ m = \(\frac {k-y_1}{h-x_1}\)

∴ k – y₁ = m (h – x₁)

Given: 

Line ax – by + c = 0 and point (1, 1) 

To prove:

\(\frac{1}{c} + \frac{1}{a} - \frac{1}{b} = \frac{c}{2ab}\) 

Concept Used: 

Distance of a point from a line. 

Explanation: 

The distance of the point (1, 1) from the straight line ax − by + c = 0 is 1

∴ 1 = \(|\frac{a-b+c}{\sqrt{a^2+b^2}}|\)

⇒ a² + b² + c²– 2ab + 2ac – 2bc = a² + b² 

⇒ ab + bc – ac = \(\frac{c^2}{2}\)

Dividing both the sides by abc, we get:

\(\frac{1}{c} + \frac{1}{a} - \frac{1}{b} = \frac{c}{2ab}\) 

Hence proved.

Given, A(4,4), B(3, 5), C (-1, -1).

Slope of AB = \(\frac {y_2-y_1}{x_2-x_1}= \frac {5-4}{3-4} = -1\)

Slope of BC = \(\frac {-1-5}{-1-3}= \frac {-6}{-4}\) = 3/2

Slope of AC = \(\frac {-1-4}{-1-4}=1\)

Slope of AB x slope of AC = – 1 x 1 = – 1

∴ side AB ⊥ side AC 

∴ ∆ABC is a right angled triangle right angled at A.

∴ The given points are the vertices of a right angled triangle.

Let, the given vertices be A(x₁, y₁) = (8, 2), 

B(x₂, y₂) = (5, – 3) and C(x₃, y₃) = (0, 0) 

∴ AB = \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\) 

         = \(\sqrt{(8-5)^2+(2+3)^2}\) =\(\sqrt{34}\) units.

  C =\(\sqrt{(x_2-x_3)^2+(y_2-y_3)^2}\)

      = \(\sqrt{(5-0)^2+(-3-0)^2}\) = \(\sqrt{34}\) units. 

∴ AB = BC (⇒ ∠ACB = ∠BAC). 

     Hence, ∆ABC is isosceles.

Let ax + by + c = 0 be the variable line. It is given that the algebraic sum of the distances of the points (1, 1), (2, 0) and (0, 2) from the line is equal to zero.

∴ \(\frac{a+b+c}{\sqrt{a^2+b^2}}\)+ \(\frac{2a+0+c}{\sqrt{a^2+b^2}}\) + \(\frac{0+2b+c}{\sqrt{a^2+b^2}}\) = 0

⇒ 3a + 3b + 3c = 0 

⇒ a + b + c = 0 

Substituting c = – a – b in ax + by + c = 0, we get: 

ax + by – a – b = 0 

⇒ a(x – 1) + b(y – 1) = 0 

⇒ x - 1 + \(\frac{b}{a}\)(y - 1) = 0

This line is of the form L₁ + λL₂ = 0, which passes through the intersection of L₁ = 0 and L₂ = 0, i.e. x – 1 = 0 and y – 1 = 0. 

⇒ x = 1, y = 1

(a) ↔ (iv) 

(b) ↔ (iii)

(c) ↔ (i), 

(d) ↔ (ii)

(a) ↔ (iii) 

(b) ↔ (i) 

and (c) ↔ (ii)

3x² – 10xy – 8y² = 0

∴ 3x² – 12xy + 2xy – 8y² = 0

∴ 3x(x – 4y) + 2y(x – 4y) = 0

∴ (x – 4y)(3x +2y) = 0

∴ the separate equations of the lines are 

x – 4y = 0 and 3x + 2y = 0.

Given, Inclination of line = θ = 90° the required line is parallel to Y-axis. 

Equation of a line parallel to Y-axis is of the form x = h. 

Since the line passes through (7, 3), h = 7 

∴ The equation of the required line is x = 7.

The required line passes through the points S(2, 1) and T(2, 3). 

Since both the given points have same x coordinates i.e. 2 

the given points lie on a line parallel to Y-axis. 

∴ The equation of the required line is x = 2

(i) Given: AB is parallel to the x - axis.

When AB is parallel to the x - axis, the y co - ordinate of A and B will be the same.

i.e., y₁ = y₂

Distance

= \(\sqrt{(x_2-x_1)^2+(y_1+y_1)^2}\)

⇒ |x₂ – x₁|

Therefore the distance between A and B when AB is parallel to x - axis is |x₂ – x₁|

(ii) Given: AB is parallel to the y - axis.

When AB is parallel to the y - axis, the x co - ordinate of A and B will be the same.

i.e., x₂ = x₁

Distance

= \(\sqrt{(x_1-x_1)^2+(y_2+y_1)^2}\)

⇒ |y₂ – y₁|

Therefore the distance between A and B when AB is parallel to y - axis is |y₂ – y₁|

The given is 

Y = 3x – 1 ⇒     3x – y – 1 = 0    …(1) 

∴ Any line parallel to (1) is written as 

3x – y + γ = 0     …(2) 

Since the line passes through (1, 2), 

3(1) – 1 + γ = 0 

⇒ γ = – 1 

∴ The line is 3x – y – 1 = 0, which is itself.

The given equation is

 3x – 4y – 16 = 0 

⇒ y = \(\frac{3}{4}\) x – 4 

∴ Slope of the line, m₁ = \(\frac{3}{4}\) 

∴ Slope of the perpendicular line, m₂ = – \(\frac{4}{3}\) . 

Since the line passes through (– 1, 3), so the equation of the line is 

y – y₀ = m(x – x₀) 

⇒ y – 3 = m₂ (x+ 1) 

⇒ y – 3 = –\(\frac{4}{3}\)(x + 1) 

⇒ 3y – 9 = – 4x – 4 

⇒ 4x + 3y – 5 = 0

The given parallel lines are 

9x + 6y – 7 = 0    …(1) 

3x + 2y + 6 = 0. 

⇒ 9x + 6y + 18 = 0   …(2)  

Let the line which is equidistant from (1) and (2) be 9x + 6y + \(\lambda\)= 0   …(3) 

Distance between (1) and (2) = \(^\frac{|7 - \lambda|}{\sqrt{9^2+6^2}}\) = \(^\frac{|7+ \lambda|}{\sqrt{117}}\)

and distance between (2) and (3) = \(^\frac{|18 - \lambda|}{\sqrt{9^2+6^2}}\)=\(^\frac{|18 - \lambda|}{\sqrt{117}}\)

Given,

\(^\frac{|7+ \lambda|}{\sqrt{117}}\)=\(^\frac{|18 - \lambda|}{\sqrt{117}}\)

⇒ 7 + \(\lambda\) = 18 – \(\lambda\) ⇒ \(\lambda\) = \(\frac{11}{2}\)

The equation of the required line is 

9x + 6y +\(\frac{11}{2}\) = 0

⇒ 18x + 12y + 11 = 0

Correct option is : (b) 2

Given two lines are 

2x + 3y = 4 ……(i) 

3x + 4y = 5 …….(ii) 

Multiplying (i) by 3 and (ii) by 2 and then subtracting, we get y = 2 

Substituting y = 2 in (i), we get x = -1 

∴ Point of intersection of lines (i) and (ii) is (-1, 2). 

Given that the line kx + 4y = 6 passes through (-1, 2). 

k(-1) + 4(2) = 6 

∴ k = 2

The number of lines intersecting at a point are called concurrent lines and their point of intersection is called the point of concurrence. Equations of the given lines are 

3x – 4y + 5 = 0 …(i) 

7x-8y + 5 = 0 …(ii)

4x + 5y – 45 = 0 …(iii) 

By (i) x 2 – (ii), we get – x + 5 = 0 

∴ x = 5 

Substituting x = 5 in (i), we get 3(5) – 4y + 5 = 0 

∴ -4y = – 20 

∴ y = 5 

∴ The point of intersection of lines (i) and (ii) is given by (5, 5). 

Substituting x = 5 and y = 5 in L.H.S. of (iii), 

we get L.H.S. = 4(5) + 5(5) – 45 

= 20 + 25 – 45 

= 0 

= R.H.S. 

∴ Line (iii) also passes through (5, 5). Hence, the given three lines are concurrent and the point of concurrence is (5, 5).

Given: 

The parallel lines are 

4x + 3y – 11 = 0 and 8x + 6y = 15 

To find: 

Distance between the givens parallel lines 

Explanation: 

The given parallel lines can be written as 

4x + 3y − 11 = 0 … (1) 

4x + 3y − \(\frac{15}{2}\) = 0 … (2) 

Let d be the distance between the given lines.

⇒ d =\(\Big|\frac{-11+\frac{15}{2}}{\sqrt{4^2+3^2}}\Big|\) = \(\frac{7}{2\times5}\) = \(\frac{7}{10}\)  units

Hence, distance between givens parallel line is \(\frac{7}{10}\)  units

We know that the equation of the line in symmetric form passing through the point (x1, y1) and making an angle 0 with the positive direction of x-axis is:

    [x - x₁] / cos 0 = [y - y₁] / sin 0

Here, x₁ = -2 , y_{​​​​​1} = 1 and 0 = 45°, therefore, required equation of the line is:

[x + 2] / cos 45° = [y - 1] / sin 45°

x + 2 = y - 1      {•.• cos 45° = sin 45°}

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

Given: 

The line \(\frac{x}{a}+\frac{y}{b}=1\) intersects the axes (a,0) and (0,b). 

Explanation: 

So, (-4,3) divides the line segment AB and the ratio 5:3

-4 =   \(\frac{5+3a}{5+3},3=\frac{5b}{5+3}=1\) 

⇒ a = \(-\frac{32}{3},b=\frac{24}{5}\)

So, the equation of the line is   \(\frac{x}{\frac{32}{5}}+\frac{y}{\frac{24}{5}}=1\) 

⇒ 9x – 20y = -96 

Hence, the equation of line is 9x – 20y = -96

Given, A(2, 1) and B(3,2) Equation of the line in two point form is

\(\frac {y-y_1}{y_2-y_1} = \frac {x-s_1}{x_2-x_1}\)

∴ The equation of the required line is \(\frac {y-1}{2-1}= \frac{x-2}{3-2}\)

∴\(\frac {y-1}{1}= \frac {x-2}{1}\)

∴ y – 1 = x – 2

∴ y = x – 1 Comparing this equation with y = mx + c, 

we get m = 1 and c = – 1

Alternate Method: 

Points A(2, 1) and B(3, 2) lie on the line y = mx + c. 

∴ They must satisfy the equation. 

∴ 2m + c = 1 …(i) 

and 3m + c = 2 …(ii) 

equation (ii) – equation (i) gives m = 1 

Substituting m = 1 in (i), we get 2(1) + c = 1 

∴ c = 1 – 2 = – 1

Concept Used:

The equation of a line in intercept form is  \(\frac{x}{a}+\frac{y}{b}=1\) 

Given: 

The line passes through (2, 1) 

 \(\frac{2}{a}+\frac{1}{b}=1\)  ……(i) 

Assuming: 

The line 3x – 5y = 15 intercept the x-axis and the y-axis at A and B, respectively. 

Explanation: 

At x = 0 we have, 

0– 5y = 15 

⇒ 5y = -15 

⇒ y = -3

At y = 0 we have, 

3x – 0 =15 

⇒ x = 5 

A= (0, -3) and B = (5, 0) 

The midpoint of AB is (\(\frac{5}{2},-\frac{3}{2}\)) lies on the line \(\frac{x}{a}+\frac{y}{b}=1\) 

 \(\frac{5}{2a}-\frac{3}{2b}=1\)  ……(ii) 

Using\(\frac{3}{2}\times\) eq(i) + eq(ii) we get,

 \(\frac{3}{a}+\frac{5}{2a}=\frac{3}{2}+1\)  

⇒ a =  \(\frac{11}{5}\)

For a =  \(\frac{11}{5}\) we have, 

   \(\frac{10}{11}+\frac{1}{b}=1\)

⇒ b = 11 

Therefore, the equation of the required line is: 

  \(\frac{x}{\frac{11}{5}}+\frac{y}{11}=1\)

  \(\frac{5x}{11}+\frac{y}{11}=1\)

⇒ 5x + y = 11 

Hence, the equation of line is 5x + y = 11