MATHS Assertion (A) & Reason (B) MCQ

1. A is true, R is true; R is acorrect explanation of A.

Solution:

Assertion Area of the triangle, whose vertices are (4, 4), (3, -2) and (-3, 16)

$=\frac{1}{2}4(-2-16)+3(16-4)+(-3)(4+2)$

$=\frac{|72+36-18|}{2}=\frac{|-54|}{2}=\frac{54}{2}=27$

1. A is true, R is true; R is acorrect explanation of A.

Solution:

Assertion Given lines are 4x + 3y = 11 and 4x + 3y = $\frac{15}{2}$

Distance between them

$=\Bigg|\frac{11-\frac{15}{2}}{\sqrt{16}+9}\Bigg|$

$=\Big|\frac{7}{2\times5}\Big|=\frac{7}{10}$

2. A is true, R is true; R is not a correct explanation of A.

Solution:

Assertion Given equation 3x - 4y + 10 = 0 can be written as

$\text{y}=\frac{3}{4}\text{x}+\frac{5}{2}\ ...(\text{i}) $

Comparing Eq. (i) with y = mx + c, we have slope of the given line as $\text{m}=\frac{3}{4}.$

Reason Equation 3x - 4y + 10 = 0 can be written as

$3\text{x}-4\text{y}=-10\text{ or}\frac{\text{x}}{-\frac{10}{3}}+\frac{\text{y}}{\frac{5}{2}}=1\ ...(\text{ii}) $

Comparing Eq. (ii) with $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1,$ we have

x - intercept as $\text{a}=-\frac{10}{3}$ and y-intercept as $\text{b}=\frac{5}{2}.$

4. A is false; R is true.

Solution:

$\because $ ABCD is a parallelogram.

$\therefore$ AB|| CD = Slope of AB = Slope of CD

and BC || AD = Slope of BC = Slope of AD

Reason mid - point of

$\text{Ac}\Big(\frac{-2+3}{2},\frac{-1+3}{2}\Big)$

$=\Big(\frac{1}{2},\frac{2}{2}\Big)=\Big(\frac{1}{2},1\Big)$

and mid-point of $\text{BD}=\Big(\frac{4-3}{2},\frac{0+2}{2}\Big)$

$=\Big(\frac{1}{2},1\Big)$

= Mid - point of AC = Mid-point of BD

2. A is true, R is true; R is not a correct explanation of A.

Solution:

Assertion Centroid of the triaigle is

$\text{G} \equiv\Big(\frac{1+2+\text{c}^2}{3},\frac{\text{a}+\text{b}-3}{2}\Big)$

$\text{i.e.}\Big(\frac{3+\text{c}^2}{3},\frac{\text{a}+\text{b}-3}{3}\Big)$

$\because$ G will lie on Y-axis, then

$\frac{3+\text{c}^2}{3}=0\Rightarrow\text{c}^2=-3\text{ or}\text{ c}\equiv\pm\text{ i}\sqrt{3}$

$\because$ Both values of c are imaginary.

Hence, G cannot lie on Y-axis.

Reason $\because$ G will lies on X-axis, then

$\frac{\text{a}+\text{b}-3}{3}=0$

$\Rightarrow\text{a}+\text{b}-3=0\text{ or}\text{ a}+\text{b}=3$

1. A is true, R is true; R is acorrect explanation of A.

Solution:

Assertion Let m, and m, be the slopes of the straight lines x + 2y - 3 = 0 and 3x + y + 1 = 0.

Then, $\text{m}_1=-\frac{1}{2}$ and m₂ = -3

Let $\theta$ be the angle between the given lines.

Then, $\tan\theta=\pm\Big(\frac{\text{m}_2-\text{m}_1}{1+\text{m}_1\text{m}_2}\Big)$

$=\pm\Bigg(\frac{-3+\frac{1}{2}}{1+\frac{3}{2}}\Bigg)=\pm1$

$\Rightarrow\theta=\tan^1(1)\text{ or }\pi-\tan^{-1}(1)$

3. A is true; R is false

Solution:

Assertion Given equation of line is x - y = 4

On dividing above equation by

$\sqrt{\text{(coefficient of x)}^2+\text{(conffcient of y})^2}$

$\text{i.e.} \sqrt{(1)^2+(-1)^2}=\sqrt{2},$ we get

$\frac{1}{\sqrt{2}}\text{x}-\frac{1}{\sqrt{2}}\text{y}=\frac{4}{\sqrt{2}}$

$\Rightarrow\cos45^\circ\text{x}-\sin45^\circ\text{ y}=2\sqrt{2}$

[ $\because \cos \text{x}$ is positive and $\sin \text{x}$ is negative, it is possible only in fourth quadrant]

$\Rightarrow\text{x}\cos(360^\circ-45^\circ)+\text{y}\sin(360^\circ-45^\circ)=2\sqrt{2}$

$\Big[\because\cos(360^\circ-\theta)=\cos\theta\text{ and }\sin(360^\circ-\theta)=-\sin\theta\Big]$

$\Rightarrow\text{x}=\cos315^\circ+\text{y }315^\circ=2\sqrt{2}$

On comparing with $\text{x}\cos\alpha+\text{y}\sin\alpha=\rho,$ we get

$\alpha=315^\circ$

and $\rho=2\sqrt{2}$

4. A is false; R is true.

Solution:

Assertion Given equation is x + 7y = 0

$\Rightarrow\text{y}=\frac{-\text{x}}{7}+0$

On comparing with y = mx + c, we get

Slope $\text{(m)}=\frac{-1}{7},$ y - intercept = 0

Reason Given equation is 6x + 3y - 5 = 0

$\Rightarrow\text{y}=-2\text{x}+\frac{5}{3}$

On comparing with y = mx + c, we get

Slope (m) = - 2, y - intercept $\frac{5}{3}$

4. A is false; R is true.

Solution:

Assertion Since, the point (A, k) lies on x + y + 1 = 0.

 ⇒ h + k + 1 = 0

and  h² + k²  = 25

⇒ (-1 - k)² + k² = 25

= 2k² + 2k - 24 = 0

k² + k - 12 = 0 ⇒ k = -4 or k =3

[k = 3 rejected as k < 0]

$\therefore$ h = -1 - (-4) = 3