MATHS M.C.Q (1 Marks)

2. $\frac{|\text{c}_1-\text{c}_2|}{\sqrt{1+\text{m}^2}}$

Solution:

Let any point on the line y = mx + c₁ be P(x₁, y₁).

The equation of the other line is: y = mx + c₂

⇒ mx - y + c₂ = 0

Distance of point P from this line, $\text{d}=\frac{|\text{mx}_1-\text{y}_1+\text{c}_2|}{\sqrt{\text{m}^2+1}}$

Since P line on the first line, we get

⇒ y₁ = mx₁ + c₁

⇒ mx₁ - y₁ = -c₁

$\therefore \text{d}=\frac{|\text{c}_1-\text{c}_2|}{\sqrt{\text{m}^2+1}}$​​​​​​​

2. 2

Solution:

Different form of equation of straight line are slope intercept form, y = mx + c, Paramerer = 2

Intercept form, $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1,$ parameter = 2

One-point from, y - y₁ = m(x - x₁), parameter = 2

Normal form, $\text{x}\cos \text{w}+\text{y}\sin\text{w}=\text{P},$ Parameter = 2

Hence, the correct option is (b).

4. None of these.

Solution:

Equation of any line passing through (1, 0) is

⇒ y - 0 = m(x - 1)

⇒ mx - y - m = 0

Distance of the line from origin is $\frac{\sqrt{3}}{2}$

$\therefore \frac{\sqrt{3}}{2} =\Big|\frac{\text{m}\times0-0-\text{m}}{\sqrt{1+\text{m}^2}}\Big|$

$\Rightarrow \frac{\sqrt{3}}{2}=\Big|\frac{-\text{m}}{\sqrt{1+\text{m}^2}}\Big|$

Squaring both sides, we get

$\frac{3}{4}=\frac{\text{m}^2}{1+\text{m}^2}$

$\Rightarrow 4\text{m}^2=3+3\text{m}^2$

$\Rightarrow 4\text{m}^2-3\text{m}^2=3$

$\Rightarrow \text{m}^2=3$

$\therefore \text{m}=\pm\sqrt{3}$

$\therefore$ Required equations are

$\pm\sqrt{3}\text{x}-\text{y}\mp\sqrt{3}=0$

i.e., $\sqrt{3}\text{x}-\text{y}-\sqrt{3}=0=0$ and $-\sqrt{3}\text{x}-\text{y}+\sqrt{3}=0=0$

$\Rightarrow \sqrt{3}\text{x}+\text{y}-\sqrt{3}=0$

Hence, the correct option is (a).

3. (-2, -2).

Solution:

Let ABC be the equilateral triangle with vertex A(h, k).

Also, centroid is G(0, 0).

Now, $\text{AG}\bot\text{BC}$

Slope of line BC or x + y - 2 = 0 is -1.

$\therefore$ Slope of $\text{AG},\frac{\text{k}}{\text{h}}=1$ or h = k.

Now distance of origin from $\text{BC}=\frac{|0+0-2}{\sqrt{1^2+1^2}|}=\sqrt{2}$

$\therefore$ Distance of A form $\text{BC}=3\sqrt{2}=\frac{|\text{h}+\text{k}-2|}{\sqrt{1^2+1^2}}$

$\therefore$ |h + k - 2| = 6

⇒ h + k - 8 = 0 or h + k + 4 = 0

⇒ h + k - 8 = 0 or h + k + 4 = 0

⇒ h = 4 or h = -2

$\therefore$ Vertex is (-2, -2).

1. 2x + 3y = 12

Solution:

Let the given the line meets the axes at A(a, 0) and B(0, b).

Given that C(3, 2) is the mid-point of AB

$\therefore 3=\frac{\text{a}+0}{2}\Rightarrow \text{a}=6$

and $2=\frac{0+\text{b}}{2}\Rightarrow \text{b}=4$

Intercept form of the line AB

$\Rightarrow \frac{\text{x}}{6}+\frac{\text{y}}{4}=1$

$\Rightarrow 2\text{x}+3\text{y}=12$

Hence, the correct option is  (a).

1. $\frac{130}{17\sqrt{29}}$

Solution:

Given lines are:

2x - 3y + 5 = 0 .....(i)

and 3x + 4y = 0 .....(ii)

Solving these lines, we get point of intersection as $\Big(\frac{-20}{17},\frac{15}{17}\Big).$

$\therefore$ Distance of this point from the line '5x - 2y = 0'

$=\frac{\Big|5\times\Big(-\frac{20}{17}\Big)-2\Big(\frac{15}{17}\Big)\Big|}{\sqrt{25+4}}=\frac{\Big|\frac{-100}{17}-\frac{30}{17}\Big|}{\sqrt{25}}$

$=\frac{130}{17\sqrt{29}}$

1. -1

Solution:

Intercept form of a line is

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

$\Rightarrow \frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{a}}=1(\because \text{a}=\text{b})$

⇒ x + y = a

⇒ y =- -x + a

$\therefore$ Slope is -1

Hence, the correct option is (a).

1. 5y - 3x + 15 = 0

Solution:

Since the lines cut off intercepts -3 on y-axis then the line is passing through the point (0, -3).

Given that: $\tan\theta=\frac{3}{5}$

⇒ Slope of the line $\text{m}=\frac{3}{5}$

So, the equation of the line is

y - y₁ = m(x - x₁)

$\Rightarrow \text{y}+ 3 = \frac{3}{5}(\text{x} -0)$

⇒ 5y + 15 = 3x

⇒ 3x - 5y - 15 = 0

⇒ 5y - 3x + 15 = 0

Hence, the correct option is (a).

2. 3 : 7

Solution:

Given lines are:

3x + 4y + 5 = 0 .....(i)

3x + 4y - 5 = 0 .....(ii)

The third line is: 3x + 4y + 2 = 0

Distance between the line (i) and (iii) $=\frac{|5-2|}{\sqrt{9+16}}=\frac{3}{5}$

Distance between the lines (ii) and (iii) $=\frac{|-5-2|}{\sqrt{9+16}}=\frac{7}{5}$

Hence, the required ratio is $\frac{3}{5}:\frac{7}{5}$ or 3 : 7.

1. $\sqrt{3}\text{x}+\text{y}-\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}-\sqrt{3}=0$

Solution:

Slope of the given line $\sqrt{3}\text{x}+\text{y}=1$ is, $\text{m}=-\sqrt{3}.$

Let the slope of the required line which makes an angle of 60° with the above line is m.

$\therefore \tan 60^\circ=\bigg|\frac{-\sqrt{3}-\text{m}}{1-\sqrt{3}\text{m}}\bigg|$

$\Rightarrow \bigg|\frac{-\sqrt{3}-\text{m}}{1-\sqrt{3}\text{m}}\bigg|=\sqrt{3}$

$\Rightarrow -\sqrt{3}-\text{m}=\sqrt{3}-3\text{m}$ or $-\sqrt{3}-\text{m}=-\sqrt{3}+3\text{m}$

$\Rightarrow \text{m}=\sqrt{3}$ or m = 0

Line is passing throught the point (3, -2).

Thus, the equation of the required line is: $\text{y}+2=\sqrt{3}(\text{x}-3)$ or y + 2 = 0

$\Rightarrow \sqrt{3}\text{x}-\text{y}-2-3\sqrt{3}=0$ and y + 2 = 0

2. $\frac{4}{3}.$

Solution:

Any line perpendicular to 3x + y = 3

$\text{x}-3\text{y}=\lambda \ (\lambda=\text{constant})$

If is passes through the point (2, 2) then

$2-3(2)=\lambda\Rightarrow \lambda=-4$

$\therefore$ Required equation is x - 3y = -4

$\Rightarrow 3\text{y}=-\text{x}-4$

$\Rightarrow \text{y}=\frac{1}{3}\text{x}+\frac{4}{3}\big[\because \text{y}=\text{mx}+\text{c}\big]$

So, the y-intercept is $\frac{4}{3}.$

Hence, the correct option is (d).

2. $\frac{-1}{10},\frac{37}{10}$

Solution:

Let the foot of perpendicular from the point P(2, 3) on the line 3x - y + 4 = 0 be M(h, k).

M(h, k) lies on the given line,

$\therefore$ 3h - k + 4 = 0 .....(i)

Also, slopw of the given line is 3.

$\therefore$ Slope of $\text{PM}=-\frac{1}{3}=\frac{\text{k}-3}{\text{h}-2}$ or h + 3k - 11 = 0 .....(ii)

Solving (1) and (ii), we get $(\text{h},\text{k})\equiv\Big(-\frac{1}{10},\frac{37}{10}\Big)$

2. (0, 0)

Solution:

Given equation are

4x + 3y + 10 = 0 .....(i)

5x - 12y + 26 = 0 .....(ii)

and 7x + 27y - 50 = 0 .....(iii)

Let (x₁, y₁) be any point equidistant from eq. (i), eq. (ii) and eq. (iii).

Distance of (x₁, y₁) from eq. (i)

$=\Big|\frac{4\text{x}_1+3\text{y}_1+10}{\sqrt{16+9}}\Big|=\Big|\frac{4\text{x}_1+3\text{y}_1+10}{5}\Big|$

Distance of (x₁, y₁) from eq. (iii)

$=\Big|\frac{5\text{x}_1-12\text{y}_1+26}{\sqrt{25+144}}\Big|=\Big|\frac{5\text{x}_1-12\text{y}_1+26}{13}\Big|$

Distance of (x₁, y₁) from eq. (iii)

$=\Big|\frac{7\text{x}_1+24\text{y}_1-50}{\sqrt{49+576}}\Big|=\Big|\frac{7\text{x}_1+24\text{y}_1-50}{25}\Big|$

If the point (x₁, y₁) is equidistant from the given lines, then

$\Big|\frac{4\text{x}_1+3\text{y}_1+10}{5}\Big|=\Big|\frac{5\text{x}_1-12\text{y}_!+26}{13}\Big|=\Big|\frac{7\text{x}_1+24\text{y}_1-50}{25}\Big|$

We see that putting x₁ = 0 and y₁ = 0, the above relation is satisfied i.e.,

$=\frac{10}{5}=\frac{26}{13}=\frac{50}{25}=2$

Hence, the correct option is (c).

3. $\frac{\text{b}^2-\text{a}^2}{2\text{ab}}$

Solution:

Intercepts of line are a and -b; i.e., line passes through the points (a, 0), (0, -b).

$\therefore$ Slope of line, $\text{m}_1=\frac{-\text{b}-0}{0-\text{a}}=\frac{\text{b}}{\text{a}}$

Intercepts of line are b, -a; i.e., line passes through the points (b, 0), (0, -a).

$\therefore$ Slope of line, $\text{m}_2=\frac{-\text{a}-0}{0-\text{b}}=\frac{\text{a}}{\text{b}}$

If $\theta$ is the angle between the lines, then

$\tan=\theta=\frac{\frac{\text{b}}{\text{a}}-\frac{\text{a}}{\text{b}}}{1+\frac{\text{a}}{\text{b}}\times\frac{\text{b}}{\text{a}}}=\frac{\frac{\text{b}^2-\text{a}^2}{\text{ab}}}{2}=\frac{\text{b}^2-\text{a}^2}{2\text{ab}}$

4. (-1, -1)

Solution:

Equation of line passing through the points (2, -3) and (4, -5) is

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

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

$\Rightarrow \text{y}+3=-(\text{x}-2)$

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

$\Rightarrow \text{x}+\text{y}=-1$

$\Rightarrow \frac{\text{x}}{-1}+\frac{\text{y}}{-1}=1$ (intercept from)

$\therefore \text{a}=-1,\text{b}=-1$

Hence, the correct option is (d).