Lines

Correct Answer – Option 2 : -2/3

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

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

⇒ \(\rm \frac{2}{{3k}} = – 1\)

⇒ \(\rm k = – \frac{2}{3}\)

Correct Answer – Option 1 : (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 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)

Correct Answer – Option 1 : (1, 2)

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)

Correct Answer – Option 3 : x – y = 0

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.

Correct Answer – Option 1 : √29m

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.