Family of Lines MCQ Quiz - Objective Question with Answer for Family of Lines - Download Free PDF

Explanation:

If one of the line of the pair of lines a2x2 + 4hxy + b2y2 = 0 bisects the angle between positive part of coordinate axes, then a2 + b2​ = -4h.

Option (2) is true.

Explanation:

If one of the line of the pair of lines a2x2 + 4hxy + b2y2 = 0 bisects the angle between positive part of coordinate axes, then a2 + b2​ = -4h.

Option (2) is true.

\(\Rightarrow \left( ax+by+c \right) \left( 5x+y-1 \right) =0\) is the combined equation,

\(\Rightarrow 5a{ x }^{ 2 }+axy-ax+5bxy+b{ y }^{ 2 }-by+5cx+cy-c=0\)

\(\Rightarrow 5a{ x }^{ 2 }+xy\left( a+5b \right) -x\left( a-5c \right) -y\left( b-c \right) +b{ y }^{ 2 }-c=0\)

Given equation is

\(\Rightarrow 5a{ x }^{ 2 }+xy -kx -2y+2=0\)

comparing this with the original equation gives,

\(b=0,a\Rightarrow \ a+5b=1 \Rightarrow a=1,c=-2\),

\(\therefore k=a-5c=11\)

Calculation

4a2 + 9b2 – c2 + 12ab = 0
⇒ (2a + 3b)² − c² = 0

⇒ 2a + 3b - c = 0 or 2a + 3b + c = 0

⇒ c = ±(2a + 3b)

∴ ax + by + c = 0

⇒ ax + by ± (2a + 3b) = 0

⇒ a(x ± 2) + b(y ± 3) = 0 (family of lines)

⇒ (-2, -3) or (2, 3)

Hence option 1 is correct 

Calculation

\(\rm r^2\cos^2\left(\theta-\frac{\pi}{3}\right)=2\)

⇒ r²(cosθ.cos\(\frac{\pi}{3}\) ​+ sinθ.sin\(\frac{\pi}{3}\) )² = 2

⇒ \(r^2 (\frac{cosθ}{2}+\frac{\sqrt{3}}{2}sinθ)^{2} = 2\)

⇒ (rcosθ + √3rsinθ)² = 8

⇒ (x + y√3)² = 8

⇒ (x + y√3 - 2√2)(x + y√3 + 2√2) = 0

Represents a pair of straight lines

Hence option 4 is correct

Concept:

Let L₃ is the line passing through the intersection of lines L₁ and L₂

Then equation of line L₃ is given by L₁ + (λ)L₂ = 0

Calculation:

Equation of line passing through the intersection of lines,

2x - 3y + 7 = 0 and 7x + 4y + 2 = 0 is given by,

2x - 3y + 7 + λ (7x + 4y + 2) = 0                     (1)

Now, this line is also passing through the point (2, 3),

So, it will satisfy the equation of line 2x - 3y + 7 + λ (7x + 4y + 2) = 0

\(\begin{array}{l} \therefore 2\left( 2 \right) - 3\left( 3 \right) + 7 + {\rm{\lambda }}\left( {7\left( 2 \right) + 4\left( 3 \right) + 2} \right) = 0\\ \Rightarrow 4 - 9 + 7 = - {\rm{\lambda }}\left( {14 + 12 + 2} \right) \end{array}\)

\(\Rightarrow - {\rm{\lambda }} = \frac{2}{{28}}\)

\(\Rightarrow {\rm{\lambda }} = - \frac{1}{{14}}\)

Now, putting \({\rm{\lambda }} = - \frac{1}{{14}}\) in (1), we get equation of line,

\(2{\rm{x\;}} - {\rm{\;}}3{\rm{y\;}} + {\rm{\;}}7{\rm{\;}} + \left( { - \frac{1}{{14}}} \right)\left( {7{\rm{x\;}} + {\rm{\;}}4{\rm{y\;}} + {\rm{\;}}2} \right) = {\rm{\;}}0\)

Multiply by 14,

\(\begin{array}{l} \Rightarrow 28{\rm{x}} - 42{\rm{y}} + 98 - \left( {7{\rm{x\;}} + {\rm{\;}}4{\rm{y\;}} + {\rm{\;}}2} \right) = 0\\ \Rightarrow 28{\rm{x}} - 7{\rm{x}} - 42{\rm{y}} - 4{\rm{y}} + 98 - 2 = 0\\ \Rightarrow 21{\rm{x}} - 46{\rm{y}} + 96 = 0 \end{array}\)

Hence, option (2) is correct

Concept:

If a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 and a₃x + b₃y + c₃ = 0 are 3 lines then these lines are said to be concurrent if:

\(\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0\)

Calculation:

Given: the lines 3y + 4x = 1, y = x + 5 and 5y + bx = 3 are concurrent.

Now, by comparing the three lines with the standard equation of line ax + by + c = 0 we get:

⇒ a₁ = 4, b₁ = 3, c₁ = -1, a₂ = 1, b₂ = -1, c₂ = 5, a₃ = b, b₃ = 5 and c₃ = -3.

As we know that, if a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 and a₃x + b₃y + c₃ = 0 are 3 lines then these lines are said to be concurrent if:

\(\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0\)

\( \left| {\begin{array}{*{20}{c}} 4&3&{ - \;1}\\ 1&{ - \;1}&5\\ b&5&{ - \;3} \end{array}} \right| = \;14b - 84 = 0\)

Concept:

The intercept form of a line is written as follows:

\(\frac{x}{P}+\frac{y}{Q}=1\)

Here, P and Q are the x-intercept and y-intercept respectively.

Calculation:

Let (h, k) is the arbitrary coordinate of the circumcenter of the triangle OPQ.

As it can be observed that two coordinate axes are acting as the two side of the triangle, so the triangle is right triangle and the circumcenter is the midpoint of sides.

\(\begin{align} & \left( h,k \right)=\left( \frac{P+0}{2},\frac{0+Q}{2} \right) \\ & =\left( \frac{P}{2},\frac{Q}{2} \right) \\ \end{align} \)

h = P/2 ⇒ P = 2h

Similarly, k = Q/2 means Q = 2k

The equation of the line \(\frac{x}{P}+\frac{y}{Q}=1\) can be rewritten as:

\(\frac{x}{2h}+\frac{y}{2k}=1\)

Now, this line passes through (4, 2), so it will satisfy the equation of line.

\(\frac{4}{2h}+\frac{2}{2k}=1\)

\(\frac{2}{h}+\frac{1}{k}=1\)

Now to find the locus replace (h, k) with (x, y).

\(\frac{2}{x}+\frac{1}{y}=1\)

Given equation of line

3x + y + 2 = 0

2x - y + 3 = 0

a²x + 2ay + 6 = 0

If all 3 lines are concurrent then these determinant of coefficient must be zero

\(\begin{vmatrix}3 & 1 &2 \\ 2 &-1&3 \\ a^2 &2a&6\end{vmatrix}=0\)

3(-1 × 6 - 3 × 2a) + 1 (3 × a² - 6 × 2) + 2 (2 × 20 - a² × -1) = 0

3(-6 - 6a) + (3a² - 12) + 2(4a + a²) = 0

-18 - 18a + 3a² - 12 + 8a + 2a² = 0

5a² - 10a - 30 = 0

a² - 2a - 6 = 0

\(a=\frac{-(-2)\pm √{(-2)^2-4\times1\times-6}}{2\times1}\)

\(a = \frac{{2 \pm √ {28} }}{2} = 1 \pm √ 7 \)

a = (1 + √7), (1 - √7)

Concept:

Let L₃ is the line passing through the intersection of lines L₁ and L₂

Then equation of line L₃ is given by L₁ + (λ)L₂ = 0

Calculation:

Equation of line passing through the intersection of lines,

2x - 3y + 7 = 0 and 7x + 4y + 2 = 0 is given by,

2x - 3y + 7 + λ (7x + 4y + 2) = 0                     (1)

Now, this line is also passing through the point (2, 3),

So, it will satisfy the equation of line 2x - 3y + 7 + λ (7x + 4y + 2) = 0

\(\begin{array}{l} \therefore 2\left( 2 \right) - 3\left( 3 \right) + 7 + {\rm{\lambda }}\left( {7\left( 2 \right) + 4\left( 3 \right) + 2} \right) = 0\\ \Rightarrow 4 - 9 + 7 = - {\rm{\lambda }}\left( {14 + 12 + 2} \right) \end{array}\)

\(\Rightarrow - {\rm{\lambda }} = \frac{2}{{28}}\)

\(\Rightarrow {\rm{\lambda }} = - \frac{1}{{14}}\)

Now, putting \({\rm{\lambda }} = - \frac{1}{{14}}\) in (1), we get equation of line,

\(2{\rm{x\;}} - {\rm{\;}}3{\rm{y\;}} + {\rm{\;}}7{\rm{\;}} + \left( { - \frac{1}{{14}}} \right)\left( {7{\rm{x\;}} + {\rm{\;}}4{\rm{y\;}} + {\rm{\;}}2} \right) = {\rm{\;}}0\)

Multiply by 14,

\(\begin{array}{l} \Rightarrow 28{\rm{x}} - 42{\rm{y}} + 98 - \left( {7{\rm{x\;}} + {\rm{\;}}4{\rm{y\;}} + {\rm{\;}}2} \right) = 0\\ \Rightarrow 28{\rm{x}} - 7{\rm{x}} - 42{\rm{y}} - 4{\rm{y}} + 98 - 2 = 0\\ \Rightarrow 21{\rm{x}} - 46{\rm{y}} + 96 = 0 \end{array}\)

Hence, option (2) is correct

Given :

(a + 2b)x + (a - 3b)y + a - 8b = 0

Point P(2, 2)

Formula:

If (x1, y1) and (x2, y2) are two points on a line then

\(Slope(m)= \frac{y_2 - y_1}{x_2 - x1}\)      -----(1)

Two-line are perpendicular if the product of their slope is -1 and vice-versa.

The equation of line having slope m and passes through the point (x1, y1) is

(y - y1​) = m(x - x1)

Calculations :

Let B be (2, 2)

(a + 2b)x + (a - 3b)y + a - 8b = 0

⇒ ax + 2bx + ay - 3by + a - 8b = 0

⇒ a(x + y + 1) + b(2x - 3y - 8) = 0

⇒ \(L_1 + \lambda L_2 = 0\)

These lines are concurrent at point of intersection of the lines L1 &  L2

⇒ L1 = x + y + 1          ----(2)

⇒ L2 = 2x - 3y - 8        ----(3)

Multiply equation (2) by 3 and subtract from equation (3), we get

⇒ 5x - 5 = 0

⇒ x = 1 and y = -2

A (1, -2) is the point of intersection of L1 &  L2

Using equation (1), we get

⇒ \(slope \ of \ AB(m_1) = \frac {2 + 2} {2 - 1}\)

⇒ slope(m1) = 4

Now, The line through A (1, -2) which is farthest from the point 

B (2, 2) is perpendicular to AB 

So, m1 m2 = -1

⇒ m2 = -1/4

Now, We have the slope and point of the line so,

Using equation (1)

⇒ (y - y1) = m(x - x1)

⇒ \(y + 2 = - \frac {1} {4} (x- 1)\)

⇒ The equation of the line L is x + 4y + 7 = 0. 

⇒ Q(h, k) is the image point.

⇒ \(\frac{h-x_1}{a}=\frac{k-y_1}{b}=-2\frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}} \) 

⇒ \(\frac{h-2}{1}=\frac{k-2}{4}=-2(\frac{2 + 8 +7}{ 17}) \)

⇒ h = 0 and k = -6

∴ The image of the point(2,2) from the line L is (0, -6).

Given:

Equation of line L is

(a + 2b)x + (a - 3b)y + a - 8b = 0 

x - 2y + 1 = 0       -----(1)

3x - 4y + λ = 0       -----(2)

Formula:

If (x1, y1) and (x2, y2) are two points on a line then

\(Slope(m)= \frac{y_2 - y_1}{x_2 - x1}\)      -----(3)

Two-line are perpendicular if the product of their slope is -1 and vice-versa.

The equation of line having slope m and passes through the point (x1, y1) is

(y - y1​) = m(x - x1)

Calculations :

Let B be (2, 2)

(a + 2b)x + (a - 3b)y + a - 8b = 0

⇒ ax + 2bx + ay - 3by + a - 8b = 0

⇒ a(x + y + 1) + b(2x - 3y - 8) = 0

⇒ \(L_1 + \lambda L_2 = 0\)

These lines are concurrent at point of intersection of the lines L1 &  L2

⇒ L1 = x + y + 1          ----(4)

⇒ L2 = 2x - 3y - 8        ----(5)

Multiply equation (4) by 3 and subtract from equation (5), we get

⇒ 5x - 5 = 0

⇒ x = 1 and y = -2

A (1, -2) is the point of intersection of L1 &  L2

Using equation (1), we get

⇒ \(slope \ of \ AB(m_1) = \frac {2 + 2} {2 - 1}\)

⇒ slope(m1) = 4

Now, The line through A (1, -2) which is farthest from the point 

B (2, 2) is perpendicular to AB 

So, m1 m2 = -1

⇒ m2 = -1/4

Now, We have the slope and point of the line so,

Using equation (1)

⇒ (y - y1) = m(x - x1)

⇒ \(y + 2 = - \frac {1} {4} (x- 1)\)

The equation of L = x + 4y + 7 = 0       ----(6)

According to the question

Line L is concurrent with the given lines

If means they intersect each other at same point.

⇒ Subtract equation (6) from (1), we get

⇒ x = -3 and y = -1         ----- (7)

Using equation (7) in equation (2)

⇒ 3x - 4y = - λ

⇒ - 9 + 4 = - λ 

∴ The value of λ is 5.

Concept:

If a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 and a₃x + b₃y + c₃ = 0 are 3 lines then these lines are said to be concurrent if:

\(\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0\)

Calculation:

Given: the lines 3y + 4x = 1, y = x + 5 and 5y + bx = 3 are concurrent.

Now, by comparing the three lines with the standard equation of line ax + by + c = 0 we get:

⇒ a₁ = 4, b₁ = 3, c₁ = -1, a₂ = 1, b₂ = -1, c₂ = 5, a₃ = b, b₃ = 5 and c₃ = -3.

As we know that, if a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 and a₃x + b₃y + c₃ = 0 are 3 lines then these lines are said to be concurrent if:

\(\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0\)

\( \left| {\begin{array}{*{20}{c}} 4&3&{ - \;1}\\ 1&{ - \;1}&5\\ b&5&{ - \;3} \end{array}} \right| = \;14b - 84 = 0\)

Given:

Equation of line L is

(a + 2b)x + (a - 3b)y + a - 8b = 0 

Formula:

If (x₁, y₁) and (x₂, y₂) are two points on a line then

\(Slope(m)= \frac{y_2 - y_1}{x_2 - x1}\)      -----(1)

Two-line are perpendicular if the product of their slope is -1 and vice-versa.

The equation of line having slope m and passes through the point (x1, y1) is

(y - y1​) = m(x - x1)

Calculations :

Let B be (2, 2)

(a + 2b)x + (a - 3b)y + a - 8b = 0

⇒ ax + 2bx + ay - 3by + a - 8b = 0

⇒ a(x + y + 1) + b(2x - 3y - 8) = 0

⇒ \(L_1 + \lambda L_2 = 0\)

These lines are concurrent at point of intersection of the lines L1 &  L2

⇒ L₁ = x + y + 1          ----(2)

⇒ L₂ = 2x - 3y - 8        ----(3)

Multiply equation (2) by 3 and subtract from equation (3), we get

⇒ 5x - 5 = 0

⇒ x = 1 and y = -2

A (1, -2) is the point of intersection of L1 &  L2

Using equation (1), we get

⇒ \(slope \ of \ AB(m_1) = \frac {2 + 2} {2 - 1}\)

⇒ slope(m1) = 4

Now, The line through A (1, -2) which is farthest from the point 

B (2, 2) is perpendicular to AB 

So, m1 m2 = -1

⇒ m2 = -1/4

Now, We have the slope and point of the line so,

Using equation (1)

⇒ (y - y1) = m(x - x1)

⇒ \(y + 2 = - \frac {1} {4} (x- 1)\)

∴ The equation of the line L is x + 4y + 7 = 0.

Concept:

The intercept form of a line is written as follows:

\(\frac{x}{P}+\frac{y}{Q}=1\)

Here, P and Q are the x-intercept and y-intercept respectively.

Calculation:

Let (h, k) is the arbitrary coordinate of the circumcenter of the triangle OPQ.

As it can be observed that two coordinate axes are acting as the two side of the triangle, so the triangle is right triangle and the circumcenter is the midpoint of sides.

\(\begin{align} & \left( h,k \right)=\left( \frac{P+0}{2},\frac{0+Q}{2} \right) \\ & =\left( \frac{P}{2},\frac{Q}{2} \right) \\ \end{align} \)

h = P/2 ⇒ P = 2h

Similarly, k = Q/2 means Q = 2k

The equation of the line \(\frac{x}{P}+\frac{y}{Q}=1\) can be rewritten as:

\(\frac{x}{2h}+\frac{y}{2k}=1\)

Now, this line passes through (4, 2), so it will satisfy the equation of line.

\(\frac{4}{2h}+\frac{2}{2k}=1\)

\(\frac{2}{h}+\frac{1}{k}=1\)

Now to find the locus replace (h, k) with (x, y).

\(\frac{2}{x}+\frac{1}{y}=1\)