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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 x + 2y + 1 = 0, 8x + 12y + k = 0 and 3x - 2y + 5 = 0 are concurrent.

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

⇒ a1 = 1, b1 = 2, c1 = 1, a2 = 8, b2 = 12, c2 = k, a3 = 3, b3 = -2 and c3 = 5.

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:

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

\(\Rightarrow \;\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 1&2&{1}\\ 8&{12}&k\\ 3&-2&{5} \end{array}} \right| \)

= 1(60 + 2k) - 2(40 - 3k) + (-16 - 36) = 0

⇒ 60 + 2k - 80 + 6k - 52 = 0

8k = 72

k = 9

Concept:

The distance formula between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is:

\(D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\)

Calculation:

Given:

x₁ = 2m

y₁ = 3m

z₁= 4m

x₂ = y₂ = z₂ = 0

Now using the distance formula as mentioned in the concept part:

\(D=\sqrt{(2m-0)^2+(3m-0)^2+(4m-0)^2}\)

\(D=\sqrt{4m^2 \ + \ 9m^2 \ + \ 16m^2}\)

\(D=\sqrt{29}m\)

Hence option (1) is the correct answer.

Concept:

The equation of the straight line passing through the point \(\rm (x_1, y_1)\) is given by, 

\(\rm (y - y_1) = m (x -x_1)\), where m is the slope of the line.

Note: The slope of the parallel lines are equal

Calculations:

The equation of straight line passing through the point \(\rm (x_1, y_1)\) is given by, 

\(\rm (y - y_1) = m (x -x_1)\), where m is slope of the line.

The equation of straight line passing through the point (1, -2) is given by, 

\(\rm (y + 2) = m (x -1)\)          ....(1)

The equation of straight line passing through the point (1, -2) and parallel to the line y = 5x - 2

⇒ Slope of the line y = 5x - 2 is m = 5.

The slope of the parallel lines are equal.

So slope of the required line is m =  5

Equation (1) becomes

⇒ \(\rm (y + 2) = 5 (x -1)\)

Concept:

The general equation of a line is y = mx + c 

Where m is the slope and c is any constant

• The slope of parallel lines is equal.
• Slope of the perpendicular line have their product = -1

Equation of a line with slope m and passing through (x1, y1)

(y - y1) = m (x - x1)

Equation of a line passing through (x1, y1) and (x2, y2) is:

\(\rm {y-y_1\over x-x_1}={y_2-y_1\over x_2-x_1}\)

Calculation:

Given line 3x + 4y = 5

⇒ y = \(-3\over4\)x + \(5\over4\)

⇒ Slope(m1) = \(-3\over4\) and c1 = \(5\over4\)

Now for the slope of the perpendicular line (m2)

m1 × m2 = -1

⇒ \(-3\over4\) × m2 = -1 

⇒ m2 = \(4\over3\)

Perpendicular line has the slope \(4\over3\) and passes through (4, 6)

∴ Equation of the perpendicular line is

(y - y1) = m (x - x1)

⇒ y - 6 = \(4\over3\) (x - 4)

⇒ 3y - 4x = 2

The intersection with the line 2x + 3y = 8 is:

(Substracting the 2 equations)

⇒ 6x = 6

⇒ x = 1

Putting value of x in equation of perpendicular line

⇒ y = 2

∴ The point if intersection is (1, 2)

Concept:

The distance of a point (x₁, y₁) from a line ax + by + c = 0 

D = \(\rm \left|ax_1+by_1+c\over\sqrt{a^2+b^2}\right|\)

Calculation:

Given line 3y = 4x + 5

⇒ 4x - 3y + 5 = 0

(x1, y1) = (2, 1)

∴ D = \(\rm \left|ax_1+by_1+c\over\sqrt{a^2+b^2}\right|\)

⇒ D = \(\rm \left|4\times 2 + (-3)\times 1+5\over\sqrt{4^2+(-3)^2}\right|\)

⇒ D = \(\rm \left|10\over5\right|\) = 2

Concept: 

Let the one line has slope m₁ and the second line has slope m₂.

If two straight lines are perpendicular then the multiplication of their slopes will be -1, that is "m₁m₂ = -1".

Calculation:

Compare both the given equation with the standard form, y = mx + c.

The slope, m₁ for the line, 2x + 3y - 3 = 0 is, \(- \frac{2}{3}\).

The slope, m2 for the line, x + ky + 7 = 0 is, \( \rm- \frac{1}{k}\).

As both the equations are perpendicular, so m1m2 = -1

\(\rm\left( { - \frac{2}{3}} \right)\left( { - \frac{1}{k}} \right) = - 1\)

⇒ \(\rm \frac{2}{{3k}} = - 1\)

⇒ \(\rm k = - \frac{2}{3}\)

Concept:

The general equation of a line is y = mx + c 

where m is the slope and c is any constant

• Slope of parallel lines is equal.
• Slope of perpendicular line have their product = -1

Equation of a line with slope m and passing through (x1, y1)

(y - y1) = m (x - x1)

Equation of a line passing through (x1, y1) and (x2, y2) is:

\(\rm {y-y_1\over x-x_1}={y_2-y_1\over x_2-x_1}\)

Calculation:

Given line 3x - 2y = 4

⇒ y = \(3\over2\)x + 2 

⇒ Slope(m1) = \(3\over2\) and c1 = 2

Now for the slope of parallel line (m2)

m1 = m2 = \(3\over2\)

Parallel line has the slope \(3\over2\) and passes through (1, 5)

∴ Equation of the parallel line is(y - y1) = m (x - x1)

⇒ y - 5 = \(3\over2\) (x - 1)

⇒ 3x - 2y + 7 = 0

Concept:

The general equation of a line is y = mx + c 

where m is the slope and c is any constant

• Slope of parallel lines are equal.
• Slope of perpendicular line have their product = -1

Equation of a line with slope m and passing through (x₁, y₁)

(y - y₁) = m (x - x₁)

Equation of a line passing through (x1, y1) and (x2, y2) is:

\(\rm {y-y_1\over x-x_1}={y_2-y_1\over x_2-x_1}\)

Calculation:

Given line 4x + 2y - 11 = 0

⇒ y = -2x + 5.5

⇒ Slope(m₁) = -2 and c₁ = 5.5

Now for the slope of perpendicular line (m₂)

m₁ × m₂ = -1

⇒ -2 × m₂ = -1 

⇒ m₂ = \(1\over2\)

Perpendicular line has the slope \(1\over2\) and passes through (3, 4)

∴ Equation of the perpendicular line is

(y - y1) = m (x - x1)

⇒ y - 4 = \(1\over2\) (x - 3)

⇒ 2y - x - 5 = 0

Concept:

The general equation of a line is y = mx + c 

where m is the slope and c is any constant

• Slope of parallel lines are equal.
• Slope of perpendicular line have their product = -1

Equation of a line with slope m and passing through (x1, y1)

(y - y1) = m (x - x1)

Equation of a line passing through (x1, y1) and (x2, y2) is:

\(\rm {y-y_1\over x-x_1}={y_2-y_1\over x_2-x_1}\)

Calculation:

The line has the slope -1 and passes through (2, 4)

∴ Equation of the perpendicular line is

(y - y1) = m (x - x1)

⇒ y - 4 = -1 (x - 2)

⇒ y  = -x + 6   ...(i)

Slope(m1) = -1 and c1 = 6

Now for the slope of perpendicular line (m2)

m1 × m2 = -1

⇒ -1 × m2 = -1 

⇒ m2 = 1

Perpendicular line has the slope 1 and passes through (-1, 3)

∴ Equation of the perpendicular line is

(y - y1) = m (x - x1)

⇒ y - 3 = 1 (x - (-1))

⇒ y = x + 4   ...(ii)

Adding (i) and (ii)

⇒ 2y = 10

⇒ y = 5

Putting y in equation (ii)

5 = x + 4

⇒ x = 1

∴ The point of intersection is (1, 5)

Concept:

The general equation of a line is y = mx + c 

where m is the slope and c is any constant

• Slope of parallel lines is equal.
• Slope of the perpendicular line have their product = -1

Equation of a line with slope m and passing through (x1, y1)

(y - y1) = m (x - x1)

Equation of a line passing through (x1, y1) and (x2, y2) is:

\(\rm {y-y_1\over x-x_1}={y_2-y_1\over x_2-x_1}\)

Calculation:

Given lines are:

2x - y = 1   ...(i)

x + 2y = 3  ...(ii)

Adding 2 × (i) to (ii)

⇒ 5x = 5

⇒ x = 1

Putting value of x in (i)

2(1) - y = 1

⇒ y = 1

The intersection is (1, 1)

So line has the slope -2 and passes through (1, 1)

∴ Equation of the perpendicular line is

(y - y1) = m (x - x1)

⇒ y - 1 = -2 (x - 1)

⇒ y + 2x - 3 = 0

Concept:

If a, b, c are in A.P then 2b = a + c

Calculations:

Given, the straight line ax + by + c = 0 always passes through (1, -2).

⇒The point (1, -2) is satisfying the equation of line ax + by + c = 0.

⇒ a(1) +b (-2)+ c = 0

⇒ 2b = a + c

⇒ we can say that 2b = a + c. then a, b, c are in A.P

Concept:

Angle between two lines is given by, \(\rm \theta =tan^{-1}|\frac{m_2-m_1}{1+m_1m_2}|\), m₁ and m₂ are slopes of lines.

Equatin of straight line: y = mx + c, where m =slope

Calculation:

Here, the pair of straight lines are 2x + y = 1 and 3x - y = 2

2x + y = 1 ⇒y = -2x + 1 so, m₁ = -2

And, 3x - y = 2 ⇒y = 3x - 2 so, m₂ = 3

So, angle between given pair of straight lines = \(\rm \theta =tan^{-1}|\frac{m_2-m_1}{1+m_1m_2}|\)

 \(\rm =tan^{-1}|\frac{3-(-2)}{1+3(-2)}|\\ =tan^{-1} 1\)

Hence, option (1) is correct. 

Concept:

Equation of a Line Parallel to a Line:

Let, ax + by + c = 0 (b ≠ 0) be the equation of the given straight line. So, the equation of a line parallel to a given line is ax + by + k = 0

​Where k is constant.

The slopes of parallel lines are equal.

Calculation:

Hare, equation of the straight line parallel to 2x - 3y = 1

Let, equation of required line be 2x - 3y + k = 0

Now, this line passing through (-3, 2)

∴2(-3) - 3(2) + k = 0

⇒ -6  - 6 + k = 0 

⇒ k = 12

∴ Equation of required line 2x - 3y + 12 = 0

Hence, option (2) is correct. 

Concept:

The general equation of a line is y = mx + c 

where m is the slope and c is any constant

• Slope of parallel lines are equal.
• Slope of perpendicular line have their product = -1

Equation of a line with slope m and passing through (x1, y1)

(y - y1) = m (x - x1)

Equation of a line passing through (x1, y1) and (x2, y2) is:

\(\rm {y-y_1\over x-x_1}={y_2-y_1\over x_2-x_1}\)

Calculation:

Given the line has the slope 3 and passes through (3, 2)

∴ Equation of the line is

(y - y1) = m (x - x1)

⇒ y - 2 = 3 (x - 3)

⇒ y - 3x + 7 = 0

Concept:

The distance between the parallel lines ax + by + c₁ and ax + by + c₂ is:

D = \(\rm \left|c_1-c_2\over\sqrt{a^2+b^2}\right|\)

Calculation;

The 2 given lines are:

3y + 4x - 12 = 0

3y + 4x - 7 = 0

a = 4, b = 3, c₁ = -12 and c₂ = -7

∴ The distance between the lines

D = \(\rm \left|c_1-c_2\over\sqrt{a^2+b^2}\right|\)

⇒ D = \(\rm \left|-12-(-7)\over\sqrt{4^2+3^2}\right|\)

⇒ D = \(\rm \left|-5\over5\right|\) = 1

Concept:

If a point P divides a line joining points A(x1, y1) and B(x2, y2) in a ratio of m:n, then

\(\rm x_P ={nx_1 + mx_2\over n+m}\), \(\rm y_P ={ny_1 + my_2\over n+m}\) and \(\rm z_P ={nz_1 + mz_2\over n+m}\)

Calculation:

Given point are (1, 2) and (3, 0)

Midpoint divides the line in ratio 1 : 1

Let the mid-point be (x, y)

x = \(\rm {nx_1 + mx_2\over n+m}\)

⇒ x = \(\rm {1\times1 + 1\times3\over 1+1}\)

⇒ x = \(\rm {1 + 3\over 2}\) = 2

Similarly 

y = \(\rm {ny_1 + my_2\over n+m}\)

⇒ y = \(\rm {1\times2 + 1\times0\over 1+1}\)

⇒ y = \(\rm {2 + 0\over 2}\) = 1

∴ The mid-point is (2, 1)

Concept:

• The co-ordinates of a point P, dividing the line joining the points A(x₁, y₁) and B(x₂, y₂) in the ratio m : n internally, are given by:

  \(\rm P\left(\dfrac{nx_1+mx_2}{m+n},\dfrac{ny_1+my_2}{m+n} \right)\)

• The equation of a line parallel to the line ax + by + c = 0 is k(ax + by) + c = 0, where k is any non-zero number.

Calculation:

The mid-point divides a line in the ratio 1 : 1 internally.

∴ The co-ordinates of the midpoint (M) of points (1, 3) and (1, -7) will be: \(\rm M\left(\dfrac{1\times1+1\times1}{1+1},\dfrac{1\times3+1\times(-7)}{1+1} \right)\) = M (1, -2).

The equation of the line parallel to the line 2x - 3y - 7 = 0 can be assumed to be k(2x - 3y) - 7 = 0.

Since this line passes through M(1, -2), we will get:

k[2(1) - 3(-2)] - 7 = 0

⇒ k(2 + 6) - 7 = 0

⇒ k = \(\dfrac78\).

The equation, therefore, is:

k(2x - 3y) - 7 = 0

⇒ \(\rm\dfrac78(2x-3y)-7=0\)

⇒ 2x - 3y - 8 = 0.

Concept:

Section Formula: Section formula is used to determine the coordinate of a point that divides a line into two parts such that the ratio of their length is m : n

Let A and B be the given two points (x1, y1, z₁) and (x2, y2, z₂) respectively and C(x, y, z) be the point dividing the line- segment AB internally in the ratio m: n

I. Internal Section Formula: When the line segment is divided internally in the ratio m : n, we use this formula.

 \((x,\ y,\ z) = \ (\frac{mx_2\ +\ nx_1}{m\ +\ n},\ \frac{my_2\ +\ ny_1}{m\ +\ n},\ \frac{mz_2\ +\ nz_1}{m\ +\ n})\)

II. External Section Formula: When point C lies on the external part of the line segment.

 \((x,\ y,\ z) = \ (\frac{mx_2\ -\ nx_1}{m\ -\ n},\ \frac{my_2\ -\ ny_1}{m\ -\ n},\ \frac{mz_2\ -\ nz_1}{m\ -\ n})\)

Calculation:

Given that,

y-z plane divides the line joining the points (3, 1, 5) and (-2, -1, 4) in the ratio p/q.

We know that when the line segment is divided internally in the ratio m : n, we use the formula.

 \((x,\ y,\ z) = \ (\frac{mx_2\ +\ nx_1}{m\ +\ n},\ \frac{my_2\ +\ ny_1}{m\ +\ n},\ \frac{mz_2\ +\ nz_1}{m\ +\ n})\)

⇒ \(x \ =\ \frac{3p\ +\ (-2q)}{p\ +\ q}\) 

But, in the y-z plane, the x coordinate is zero. 

\(⇒\ x \ =\ \frac{3p\ -\ 2q}{p\ +\ q}\ =\ 0\)

⇒ 3p - 2q = 0

⇒ p/q = 2/3

Therefore, y-z plane devices given line in ratio of 2 : 3.

⇒ p + q = 2 + 3 = 5

Hence, option 3 is correct.

Concept:

Let second-degree equation be ax2 + by2 + 2hxy + 2gx + 2fy + c = 0

It will represent a pair of straight lines, If discriminant (Δ) of this equation equal to zero (Δ = 0) 

Discriminant = Δ = \(\left| {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{h}}&{\rm{g}}\\ {\rm{h}}&{\rm{b}}&{\rm{f}}\\ {\rm{g}}&{\rm{f}}&{\rm{c}} \end{array}} \right| = {\rm{abc}} + 2{\rm{fgh}} - {\rm{a}}{{\rm{f}}^2} - {\rm{b}}{{\rm{g}}^2} - {\rm{c}}{{\rm{h}}^2}\)

Calculation:

Given: Second degree equation, 2x2 + 7xy + 3y2 + 8x + 14y + λ = 0

Compare with second-degree equation ax2 + by2 + 2hxy + 2gx + 2fy + c = 0

So, a = 2, b = 3, h = \(\frac 72\), g = 4, f = 7 and c = λ

Given equation represents a pair of straight lines

So, Δ = abc + 2fgh - af2 - bg2 - ch² = 0

⇒ 2 × 3 × λ + 2 × 7 × 4 × \(\frac 72\) - 2 × (7)² - 3 × (4)2 - λ × (\(\frac 72\))2 = 0

⇒ 6λ + 196 - 98 - 48 - \(\frac {49λ}{4}\) = 0

⇒ 50 - \(\frac {25λ}{4}\)= 0

⇒ 200 - 25λ = 0

∴ λ = 8

Concept:

Let m1 and m2 be the slope of the line ax2 + 2hxy + by2 = 0 

⇒ m1 + m2 = \(\rm \dfrac {-2h}{b}\)

⇒ m1 . m2 = \(\rm \dfrac {a}{b}\)

Calculations:

Given equation is x2 - 2cxy - 7y2 = 0  

Comparing with the equation ax2 + 2hxy + by2 = 0

⇒ a = 1, h = - c, and b = - 7

Let m₁ and m2 be the slope of the line x2 - 2cxy - 7y2 = 0  

⇒ m1 + m2 = \(\rm \dfrac {-2h}{b}= \dfrac{2c}{-7}\)

⇒ m1 . m2 = \(\rm \dfrac {a}{b}= \dfrac{1}{-7}\)

Given, the sum of the slopes of the lines given by x2 - 2cxy - 7y2 = 0 is four time their products.

⇒\(\rm \dfrac{2c}{-7} = \dfrac {4}{-7}\)

 ⇒ c = 2

Concept:

Equation of family of lines passing through the intersection of two lines S₁ = 0 and S₂ = 0 is given by S₁ + λS₂ = 0

Calculations:

Given, the equation of lines are \(\dfrac{x}{a}+\dfrac{y}{b}=1\) and \(\dfrac{x}{b} + \dfrac{y}{a}=1\)

The equation to the straight line joining the origin to the point of intersection of the lines \(\dfrac{x}{a}+\dfrac{y}{b}=1\) and \(\dfrac{x}{b} + \dfrac{y}{a}=1 \) is 

\((\dfrac{x}{a}+\dfrac{y}{b}-1) + λ (\dfrac{x}{b} + \dfrac{y}{a}-1) = 0\)    ....(1)

which is passing through the origin.

⇒ \((\dfrac{0}{a}+\dfrac{0}{b}-1) - λ (\dfrac{0}{b} + \dfrac{0}{a}-1) = 0\)

⇒ \(λ = -1\)

Equation of line becomes, 

⇒ \((\dfrac{x}{a}+\dfrac{y}{b}-1) +(-1) (\dfrac{x}{b} + \dfrac{y}{a}-1) = 0\)

⇒ \((\dfrac{x}{a}+\dfrac{y}{b}-1)-(\dfrac{x}{b} + \dfrac{y}{a}-1) = 0\)

⇒ \(\dfrac{x}{a}+\dfrac{y}{b}-1 - \dfrac{x}{b} - \dfrac{y}{a}+1 = 0\)

⇒ \(\rm x(\dfrac{1}{a}-\dfrac{1}{b})- y(\dfrac{1}{a} - \dfrac{1}{b}) = 0\)

⇒ x - y = 0

Hence, the equation to the straight line joining the origin to the point of intersection of the lines \(\dfrac{x}{a}+\dfrac{y}{b}=1\) and \(\dfrac{x}{b} + \dfrac{y}{a}=1 \) is x - y = 0.