MATHS M.C.Q (1 Marks)

1. 90°

Solution:

Let m₁ and m₂ be the slope of the lines 2x - y + 3 = 0 and x + 2y + 3 = 0, respectively.

Let $\theta$ be the angle between them.

Here, m₁ = 2 and $\text{m}_2=-\frac{1}{2}$

$\because\text{m}_1\text{m}_2=-1$

Therefore, the angle between the given lines is 90°.

3. $\frac{1}{2}$

Solution:

Given points: A (6, 4) = (x₁, y₁)

B (2, 12) = (x₂, y₂)

We know that the slope of a line passing through two points (x₁, y₁) and (x₂, y₂) is $\frac{(\text{y}_{2}\text{-y}_{1})}{(\text{x}_{2}-\text{x}_{1})}$

$\text{m} = \frac{(12-4)}{(2-6)} = \frac{8}{-4} = -2.$

We know that the slope of two perpendicular lines m_1.m₂ = -1.

The slope of a line perpendicular to line AB is $\frac{-1}{\text{m}} = \frac{-1}{-2} = \frac{1}{2}$

3. x + 2y - 10 = 0

Solution:

Line 2x - y = 4 has slope 2. Line perpendicular to given line has slope $\frac{-1}{2}$

equation is $(\text{y}-4) = \big(\frac{-1}{2}\big) (\text{x}-2)$

2y - 8 = -x + 2

⇒ x + 2y - 10 = 0

2. x - 3y + 4 = 0

Solution:

The equation of the sides AB, BC and CA of $\triangle \text{ABC}$ are y - x = 2, x + 2y = 1 and 3x + y + 5 = 0, respectively.

Solving the equations of AB and BC, i.e. y - x = 2 and x + 2y = 1, we get:

x = -1, y = 1

So, the coordinates of B are (-1, 1).

The altitude through B is perpendicular to AC.

$\therefore$ Slope of AC = -3

Thus, slope of the altitude through B is 13. Thus, slope of the altitude through B is $\frac{1}{3}.$

Equation of the required altitude is given below:

$\text{y}-1+\frac{1}{3}(\text{x}+1)$

$\Rightarrow\text{x}-3\text{y}+4=0$

2. $\frac{3}{10}$

1. 45°

Solution:

Given, equation is x + y = 2. Reducing the above equation to normal form

$\frac{(\text{x + y})}{\sqrt2} ={\sqrt2}.$

$\text{x} \cos⁡45° + \text{y} \sin 45^\circ = {\sqrt2}$

Angle made with x-axis is 45°.

3. $\sqrt{2}$

Solution:

Let A(0, 6), B(6, 0) and C(6, 6) be the vertices of the given triangle.

Centroid of $\triangle\text{ABC}=\Big(\frac{0+6+6}{3},\frac{6+0+6}{3}\Big)$

$=(4,4)$

Coordinates of $\text{N}=\Big(\frac{6+6}{2},\frac{6+0}{2}\Big)$

$=(6,3)$

Coordinates of $\text{P}=\Big(\frac{0+6}{2},\frac{6+6}{2}\Big)$

$=(3,6)$

Equation of MN is y = 3

Equation of MP is x = 3

As, we know that circumcentre of a triangle is the intersection of the perpendicular bisectors of any two sides.

Therefore, coordinates of circumcentre is (3, 3)

Thus, the coordinates of the circumcentre are (3, 3) and the centroid of the triangle is (4, 4).

Let d be the distance between the circumcentre and the centroid.

$\therefore\text{d}\sqrt{(4-3)^2+(4-3)^2}=\sqrt{2}$

3. 5 units.

Solution:

We know, distance between two points (x₁, y₁) and (x₂, y₂) is $\sqrt{(\text{x}_{1}-\text{x}_{2}) ^{2}+{(\text{y}_{1}-\text{y}_{2})} ^{2}}$

So, distance between (1, 3) and (5, 6) is $\sqrt{{(1-5)}^{2}+(3-6)^{2}} = \sqrt{{(4)}^{2}+(3)^{2}} = 5\text{ units}.$

1. (-1, -14)

Solution:

Let the reflection point be A(h, k)

Now, the mid point of line joining (h, k) and (4, -13) will lie on the line 5x + y + 6 = 0

$\therefore5\Big(\frac{\text{h}+4}{2}\Big)+\frac{\text{k}-13}{2}+6=0$

$\Rightarrow5\text{h}+20+\text{k}-13+12=0$

$\Rightarrow5\text{h}+\text{k}+19=0 \ ...(1)$

Now, the slope of the line joining points (h, k) and (4, -13) are perpendicular to the line 5x + y + 6 = 0.

slope of the line = -5

slope of line joining by points (h, k) and (4, -13)

$\frac{\text{k}+13}{\text{h}-4}$

$\therefore\frac{\text{k}+13}{\text{h}-4}(-5)=-1$

$\Rightarrow5\text{k}-\text{h}+60=0 \ ...(2)$

Solving (1) and (2), we get

h = -1 and k = -14

2. 1

Solution:

 It is given that the lines $3\text{x} + 4\text{y} = 5, 5\text{x} + 4\text{y} =4$ and $\lambda\text{x} + 4\text{y} = 6$ meet at a point. In other words, the given lines are concurrent.

$\begin{vmatrix}3&4&-5\\5&4&-4\\\lambda&4&-6 \end{vmatrix}=0$

$\Rightarrow3(-24+16)-4(-30+4\lambda)-5(20-4\lambda)=0$

$\Rightarrow-24+120-16\lambda-100+20\lambda=0$

$\Rightarrow4\lambda=4$

$\Rightarrow\lambda=1$ 

1. 5x + 3y - 20 = 0

Solution:

A line perpendicular to 3x - 5y + 7 = 0 is given by

$5\text{x}+3\text{y}+\lambda=0$

This line passes through (1, 5)

$5+15+\lambda=0$

$\Rightarrow\lambda=-20$

Therefore, the equation of the required line is 5x + 3y - 20 = 0.

1. x = 3

Solution:

Let general equation of line be y = m (x - d)

$\Rightarrow\text{x} =\frac{ \text{y}}{\text{m + d}}$ Since line is parallel to y-axis so, $\text{m}=\frac{1}{0} $ or $\frac{1}{\text{m}} =0$

⇒ x = d

⇒ x = 3 by substituting the point (3, 4)

2. (4, 7)

Solution:

Let A(4, 8) and B(-2, 6) be the given vertex. Let C(h, k) be the third vertex.

The centroid of $\triangle\text{ABC}$ is $\Big(\frac{4-2+\text{h}}{3},\frac{8+6+\text{k}}{3}\Big)$

It is given that the centroid of triangle ABC is (2, 7).

$\therefore\frac{4-2+\text{h}}{3}=2,\frac{8+6+\text{k}}{3}=7$

$\Rightarrow\text{h}=4,\text{h}=7$

Thus, the third vertex is (4, 7).

2. $3$

Solution:

We know, slope of line joining two points (x₁, y₁) and (x₂, y₂) is given by $\frac{{\text{y}}_{2}-{\text{y}}_{1}}{{\text{x}}_{2}-{\text{x}}_{1}}$

So, slope of line joining (1, 2) and (4, 11) is $\frac{11-2}{4-1} = \frac{9}{3} = 3$

3. 0

Solution:

Let A(1, 2), B(2, 1) and C $\Big(\frac{3+\sqrt{3}}{2},\frac{3+\sqrt{3}}{2}\Big)$ be the given points.

$\therefore \ \text{AB}=\sqrt{(2-1)^2+(1-2)^2}$

$=\sqrt{2}$

$\text{BC}=\sqrt{\Big(\frac{3+\sqrt{3}}{2}-2\Big)^2,\Big(\frac{3+\sqrt{3}}{2}-1\Big)^2}$

$=\sqrt{2}$

$\text{AC}=\sqrt{\Big(\frac{3+\sqrt{3}}{2}-1\Big)^2,\Big(\frac{3+\sqrt{3}}{2}-2\Big)^2}$

$=\sqrt{2}$

Thus, ABC is an equilateral triangle.

We know that the orthocentre and the circumcentre of an equilateral triangle are same.

So, the distance between the the orthocentre and the circumcentre of the triangle with vertices (1, 2), (2, 1) and $\Big(\frac{3+\sqrt{3}}{2},\frac{3+\sqrt{3}}{2}\Big)$ is 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).

1. $10$

Solution:

Given, equation is y = 5x + 10. Y-intercept means value of y when x is zero. y = 0 + 10

⇒ y = 10

2. $\text{k} = \underline{+} 3 \sqrt{5}$