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

Concept:

The slope of the tangent to a curve y = f(x) is m = $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$

The slope of the normal = $-\frac{1}{\mathrm{m}}$ = $-\frac{1}{\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}$

Calculation:

Given curve y = 2x³ - 5x² + x - 2

Differentiating the equation wrt x

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ = 6x² - 10x + 1

Slope at the point (1, -1)

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ at x = 1 = 6(1)² - 10(1) + 1

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ = 6 - 10 + 1

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ = -3

The slope of the normal (m') = $-\frac{1}{\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}$

m' = $\text{boldsymbol}-\frac{1}{-3}$ = $\text{boldsymbol}\frac{1}{3}$

The slope of the normal to the curve at the point (1, -1) is 1/3.

Calculation:

Given,

Line 1: $m^{2}x+ny-1=0$

Line 2: $n^{2}x-my+2=0$

Slope of Line 1, $m_1=-{\displaystyle \frac{m^2}{n}}$

Slope of Line 2, $m_2={\displaystyle \frac{n^2}{m}}$

For perpendicular lines, $m_1\times m_2=-1$.

$\frac{-m^2}{n}\times \frac{-n^2}{m}=-1$

mn -1 = 0

Hence, the correct answer is Option 1.

Calculation:

Only possibility α = 0, β = 1

∴ equation of circle x² + y² - x - y = 0

Image of circle in x + y + 2 = 0 is  

x² + y² + 5x + 5y + 12 = 0

Hence, the correct answer is Option 1. 

Explanation:
If θ be the angle between the lines y = m1x + c1 and y = m2x + c₂ then θ = $\pm {\tan }^{-1}\left(\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right)$

Additional Information

When two lines are perpendicular, the product of their slopes = m₁m₂ = -1

When two lines are parallel then m₁ = m₂

Concept:

When two lines are perpendicular, then the product of their slope is -1.

i.e. m₁ × m₂ = -1

When the slope of a line is m then the slope of a line perpendicular to it = - 1/m

Calculation:

Given:

-2x + 3y + 4 = 0

y = $\frac{2}{3}x-\frac{4}{3}$

The slope of the line = $\frac{2}{3}$

The line perpendicular to the above line will have a slope of = $-\frac{1}{\frac{2}{3}}$ = $-\frac{3}{2}$

option (1) 3x + 2y – 4 = 0 

• The slope of the line is -3/2

option (2) 3x – 2y + 4 = 0

• The slope of the line is 3/2

option (3)  -3x + 2y + 7 = 0

• The slope of the line is 3/2

option (4) 3x – 2y – 7 = 0

• The slope of the line is 3/2

Hence option (1) 3x + 2y – 4 = 0 is correct.

Concept:

The angle between the lines y = m1x + c1 and y = m2x + c2 is given by  tan θ = $\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|$

Calculation:

Given lines are y = x + 4 and y =  2x - 3

Let slope of 1st and 2nd line are m₁ and m₂ respectively,

Therefore, m₁ = 1 and m₂ = 2

As we know, tan θ = $\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|$

⇒ tan θ = $\left|\frac{1-2}{1+1\times 2}\right|=\frac{1}{3}$

∴ θ = ${\tan }^{-1}\left(\frac{1}{3}\right)$

Concept:

Equation of a line in intercept form:

The equation of the line in intercept forms is given by:

${\displaystyle \frac{\mathrm{x}}{\mathrm{a}}}+{\displaystyle \frac{\mathrm{y}}{\mathrm{b}}}=1$

where a and b are the intercepts on the x & y-axis respectively.

Calculation:

The equation of the line in intercept form is given by :

${\displaystyle \frac{\mathrm{x}}{\mathrm{a}}}+{\displaystyle \frac{\mathrm{y}}{\mathrm{b}}}=1$

It is given that both the intercepts are equal so we have a = b.

Therefore, the equation of the line is given as follows:

$$\begin{gathered} {\displaystyle \frac{\mathrm{x}}{\mathrm{a}}}+{\displaystyle \frac{\mathrm{y}}{\mathrm{a}}}=1 \\ {\displaystyle \frac{\mathrm{x}+\mathrm{y}}{\mathrm{a}}}=1 \\ \mathrm{x}+\mathrm{y}=\mathrm{a} \\ \mathrm{y}=-\mathrm{x}+\mathrm{a} \end{gathered}$$

Therefore, comparing with the slope-intercept form y = mx + c we observe that the slope of the line is -1.

We know that slope is given by $\mathrm{m}=\tan \theta$

Where, $\theta$ is the angle made with the positive direction of the x-axis.

Therefore, in the given case we get $\tan \theta =-1⟹\theta ={135}^{\circ }$.

Concept:

• Slope intercept form of line: y = mx + c, Where ‘m’ is the slope of the line and ‘c’ is the y-intercept.
• The "point-slope" form of the equation of a straight line is: y − y₁ = m(x − x₁)
• When two lines are perpendicular, the product of their slope is -1.
• If m is the slope of a line, then the slope of a line perpendicular to it is -1/m.

Calculation:

Given: 3x + y = 7

⇒ y = -3x + 7

The slope of line = m = -3

Then the slope of a line perpendicular to it is -1/m = 1/3

The equation of line passing through (1, 1) with slope "1/3" is

y – 1 = (1/3) (x – 1)

⇒ 3y – 3 = x – 1

⇒ 3y = x + 2

⇒ 3y - x = 2

For x-intercept, y = 0

∴ x = -2

So, the x-intercept of line is -2

Concept:

The angle between the lines y = m1x + c1 and y = m2x + c2 is given by  tan θ = $\left|\frac{{\mathrm{m}}_1-{\mathrm{m}}_2}{1+{\mathrm{m}}_1{\mathrm{m}}_2}\right|$

Calculation:

Let slope of 1st and 2nd line are m1 and m2 respectively,

Therefore, m1 = $\sqrt{3}$ and m2 = $\frac{1}{\sqrt{3}}$

As we know, tan θ = $\left|\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3}\times \frac{1}{\sqrt{3}}}\right|$

⇒ tan θ = $\left|\frac{\frac{3-1}{\sqrt{3}}}{1+1}\right|=\frac{1}{\sqrt{3}}$

∴ θ = ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$ = $\frac{\pi }{6}$

Concept:

• The slope of a line passing through the distinct points (x₁, y₁) and (x₂, y₂) is $\frac{y_2-y_1}{x_2-x_1}$
• When two lines are perpendicular, the product of their slope is -1. If m is the slope of a line, then the slope of a line perpendicular to it is -1/m.

Calculation:

Let the slope of the line 2x – 5y + 4 = 0 be m₁ and the slope of the line joining the points (1, 5) and (α, 3) be m₂ 

${\mathrm{m}}_2=\frac{3-5}{\alpha -1}=\frac{-2}{\alpha -1}$

Now, the slope of the line = m₁ = 2/5

Given lines are perpendicular to each other,

∴ m1 m2 = -1

$\Rightarrow \frac{-2}{\alpha -1}\times \frac{2}{5}=-1$

⇒ -4 = -5 × (α -1)

⇒ (α -1) = 4/5

⇒ α = (4/5) + 1 = 9/5

Concept:

If three points (x1, y1), (x2, y2) and (x3, y3) are collinear then the area of the triangle determined by the three points is zero.

$\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\right|=0$

or

If three or more points are collinear then the slope of any two pairs of points is same.

For example, let three points A, B and C are collinear then

slope of AB = slope of BC = slope of AC

Slope of the line if two points $({\mathrm{x}}_1,{\mathrm{y}}_1)\mathrm{a}\mathrm{n}\mathrm{d}\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$ are given by: 

$\Rightarrow \left(\mathbf{m}\right)=\frac{{\mathbf{y}}_2-{\mathbf{y}}_1}{{\mathbf{x}}_2-{\mathbf{x}}_1}$

Calculation:

Given: the points (-2, -5), (2, -2) and (8, a) are collinear

$$\begin{gathered} \left|\begin{array}{ccc}-2& -5& 1\\ 2& -2& 1\\ 8& \mathrm{a}& 1\end{array}\right|=0 \\ \Rightarrow -2(-2-\mathrm{a})-(-5)(2-8)+1(2a+16)=0 \\ \Rightarrow 4+2\mathrm{a}-30+2\mathrm{a}+16=0 \\ \Rightarrow 4\mathrm{a}-10=0 \\ \Rightarrow 4\mathrm{a}=10 \\ \therefore \mathrm{a}=\frac{5}{2} \end{gathered}$$

Concept:

• Slope intercept form of line: y = mx + c, Where ‘m’ is the slope of the line and ‘c’ is the y-intercept.
• The "point-slope" form of the equation of a straight line is: y - y₁ = m(x - x₁)
• When two lines are perpendicular, the product of their slope is -1.
• If m is the slope of a line, then the slope of a line perpendicular to it is -1/m.

Calculation:

Given: 3x + y = 3

⇒ y = -3x + 3

The slope of line = m = -3

Then the slope of a line perpendicular to it is -1/m = 1/3

The equation of line passing through (2, 2) with slope "1/3" is

y - 2 = (1/3) (x - 2)

⇒ 3y - 6 = x - 2

⇒ 3y = x + 4

$\Rightarrow \mathrm{y}=\frac{1}{3}\mathrm{x}+\frac{4}{3}$

So, the y-intercept of line is 4/3

Concept:

The slope of the line joining the points (x1, y1) and (x2, y2) is given by: $\mathrm{m}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}$

Calculation:

Given: The slope of a line joining the points A (1, x) and B (3, 2) is 8

As we know, The slope of the line joining the points (x1, y1) and (x2, y2) is given by: $\mathrm{m}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}$

⇒ 8 = $\frac{2-\mathrm{x}}{3-1}$

⇒ 8 × 2 = 2 - x

⇒ 16 = 2 - x

∴ x = -14

Concept:

If θ is the inclination of the line l, then

The slope of a line is denoted by m = tan θ, θ ≠ 90°

Calculation:

Given: Line makes 30° with respect to the x-axis in a positive direction

∵ the inclination of the line is 30° i.e θ = 30° 

As we know that slope of a line is given by: m = tan θ 

Concept:

Intercept form of line: $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$, Where, a = x-intercept and b = y-intercept

Calculation:

We have, intercept form of line $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$

Here, line making equal intercepts on the coordinate axes

So, a = b

⇒ x  + y = a

Straight line passing through (4, 3)

∴ (4) + (3) = a

⇒ a = 7

So, equation of required line: x + y = 7

Hence, option (1) is correct.