MATHS M.C.Q (1 Marks)

3. $\frac{35}{3\sqrt{34}}$

Solution:

The given lines can be written as

$5\text{x}+3\text{y}-7=0 \ ...(1)$

$5\text{x}+3\text{y}+\frac{14}{3}=0 \ ...(2)$

Let d be the distance between the lines 5x + 3y - 7 = 0 and 15x + 9y + 14 = 0

Then, $\text{d}=\Bigg|\frac{-7-\frac{14}{3}}{\sqrt{5^2+3^2}}\Bigg|$

$\Rightarrow\text{d}=\frac{35}{3\sqrt{34}}$

2. (4, 1)

Solution:

Let A(-1, -6), B(2, -5) and C(7, 2) be the given vertex. Let D(h, k) be the fourth vertex.

The midpoints of AC and BD are (3, -2) and $\Big(\frac{2+\text{h}}{2},\frac{-5+\text{k}}{2}\Big)$ respectively.

We know that the diagonals of a parallelogram bisect each other.

$\therefore3=\frac{2+\text{h}}{2}$ and $-2=\frac{-5+\text{k}}{2}$

$\Rightarrow\text{h}=4$ and $\text{k}=1$

4. Cannot be found.

Solution:

The point which divides the join of (1, 2) and (3, 4) externally in the ratio 1 : 1 is

$\Big(\frac{1\times3-1\times1}{1-1},\frac{1\times4-1\times2}{1-1}\Big)$

which is not defined.

Therefore, it is not possible to externally divide the line joining two points in the ratio 1 : 1

1. (1,1)

Solution:

Let ax + by + c = 0 be the variable line. It is given that the algebraic sum of the distances of the points (1, 1), (2, 0) and (0, 2) from the line is equal to zero

$\therefore \ \frac{\text{a}+\text{b}+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}+\frac{2\text{a}+0+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}+\frac{0+2\text{b}+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}=0$

$\Rightarrow3\text{a}+3\text{b}+3\text{c}=0$

$\Rightarrow\text{a}+\text{b}+\text{c}=0$

Substituting c = -a - b in ax + by + c = 0, we get:

$\text{ax}+\text{by}-\text{a}-\text{b}=0$

$\Rightarrow\text{a}(\text{x}-1)+\text{b}(\text{y}-1)=0$

$\Rightarrow(\text{x}-1)+\frac{\text{a}}{\text{b}}(\text{y}-1)=0$

This line is of the form $\text{L}_1+\lambda\text{L}_2=0$ which passes through the intersection of L₁ =0 and L₂ =0,i.e. x - 1 = 0 and y - 1 = 0.

⇒ x = 1, y = 1

2. $\cos^{-1}\big(\frac{3}{4}\big)$

Solution:

Let the coordinates of the right-angled isosceles triangle be O(0, 0), A(a, 0) and B(0, a).

Here, BD and AE are the medians drawn from the acute angles B and A, respectively.

$\therefore$ Slope of BD = m₁

$=\frac{0-\text{a}}{\frac{\text{a}}{2}-0}$

$=-\frac{1}{2}$

Let $\theta$ be the angle between BD and AE.

$\tan\theta=\Big|\frac{-2+\frac{1}{2}}{1+1}\Big|$

$=\frac{3}{4}$

$\Rightarrow\cos\theta=\frac{4}{\sqrt{3^2+4^2}}$

$\Rightarrow\cos\theta=\frac{4}{5}$

$\Rightarrow\theta=\cos^{-1}\Big(\frac{4}{5}\Big)$

Hence, the acute angle between the medians is $\cos^{-1}\Big(\frac{4}{5}\Big).$

3. 3 : 1

Solution:

Let the points (-3, -4) and (1, -2) be divided by y-axis at (0, t) in the ratio m : n.

$\therefore\Big(\frac{\text{m}-3\text{n}}{\text{m}+\text{n}},\frac{-2\text{m}-4\text{n}}{\text{m}+\text{n}}\Big)=(0,\text{t})$

$\Rightarrow0=\frac{\text{m}-3\text{n}}{\text{m}+\text{n}}$

$\Rightarrow\text{m}:\text{n}=3:1$

1. 0

Solution:

$\text{x} - 2\text{y} + 3 = 0 \ ...(\text{i})$

$\lambda\text{x} + 3\text{y} + 1 = 0 \ ...(\text{ii})$

$4\text{x} - \lambda\text{y} + 2 = 0 \ ...(\text{iii})$

It is given that (1), (2) and (3) are concurrent.

$\therefore\begin{vmatrix} 1&-2&3\\\lambda&3&1\\4&-\lambda&2\end{vmatrix}=0$

$\Rightarrow(6+\lambda)+2(2\lambda-4)+3(-\lambda^2-12)=0$

$\Rightarrow6+\lambda+4\lambda-8-3\lambda^2-36=0$

$\Rightarrow5\lambda-3\lambda^2-38=0$

$\Rightarrow3\lambda^2-5\lambda+38=0$

The discriminant of this equation is $25 - 4 \times 3 \times 38 = -431$

Hence, there is no real value of $\lambda$ for which the lines $\text{x} - 2\text{y} + 3 = 0, \lambda\text{x} + 3\text{y} + 1 = 0$ and $4\text{x} - \lambda\text{y} + 2 = 0$ are concurrent.

2. 2x + 5y = 20

Solution:

Let the equation of the line be $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$

The coordinates of the intersection of this line with the coordinate axes are (a, 0) and (0, b).

The midpoint of (a, 0) and (0, b) is $\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big)$

According to the question:

$\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big)=(5,2)$

$\Rightarrow\frac{\text{a}}{2}=5,\frac{\text{b}}{2}=2$

$\Rightarrow\text{a}=10,\text{b}=4$

The equation of the required line is given below:

$\frac{\text{x}}{10}+\frac{\text{y}}{4}=1$

$\Rightarrow2\text{x}+5\text{y}=20$

1. 17

Solution:

Let A be the area of the triangle formed by the points (-4, -1), (1, 2) and (4, -3).

$\therefore\text{A}=\frac{1}{2}\big|\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\}\big|$

$\Rightarrow\text{A}=\frac{1}{2}\big|\{-4(2+3)+1(-3+1)+4(-1-2)\}\big|$

$\Rightarrow\text{A}=17$

1. 9x - 20y + 96 = 0

Solution:

Let the required line intersects the coordinate axis at (a, 0) and (0, b).

The point (−4, 3) divides the required line in the ratio 5 : 3

$\therefore \ -4=\frac{5\times0+3\times\text{a}}{5+3}$ and $3=\frac{5\times\text{b}+3\times0}{5+3}$

$\Rightarrow\text{a}=\frac{ -32}{3}$ and $\text{b}=\frac{ 24}{5}$

Hence, The equation of the required line is given below:

$\frac{\text{x}}{\frac{-32}{3}}+\frac{\text{y}}{\frac{24}{5}}=1$

$\Rightarrow\frac{-3\text{x}}{32}+\frac{5\text{y}}{24}=1$

$\Rightarrow-9\text{x}+20\text{y}=96$

$\Rightarrow9\text{x}-20\text{y}+96=0$

2. (5, 0) or, (1, 4)

Solution:

Let (h, k) be the third vertex of the triangle.

It is given that the area of the triangle with vertices (h, k), (-2, -1) and (3, 2) is 4 square units.

$\frac{1}{2}\big|\text{h}(-1-2)-3(-1-\text{K})-2(2-\text{K})\big|=4$

$\Rightarrow3\text{h}-5\text{k}+1=\pm 8$

Taking positive sign, we get,

3h - 5k + 1 = 8

3h - 5k - 7 = 0 ...(1)

Taking negative sign, we get,

3h - 5k + 9 = 0 ...(2)

The vertex (h, k) lies on the line x + y = 5.

h + k - 5 = 0 ...(3)

On solving (1) and (3), we find (4, 1) to be the coordinates of the third vertex.

Similarly, on solving (2) and (3), we find (2, 3) to be the coordinates of the third vertex.

3. $\frac{1}{\text{p}^2}=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$

Solution:

It is given that p is the length of the perpendicular from the origin on the line

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

$\frac{1}{\text{a}}\text{x}+\frac{1}{\text{b}}\text{y}-1=0$

$\therefore\text{p}=\begin{vmatrix}\frac{0+0+1}{\sqrt{\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}}} \end{vmatrix}$

Squaring both sides,

$\Rightarrow\frac{1}{\text{p}^2}=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$

2. 3 : 7

Solution:

Here, in all equations the coefficient of x is same.

It means all the lines have same slope

So, all the lines are parallel.

Now, the distance between the line 3x + 4y + 2 = 0 and 3x + 4y + 5 = 0 is given by

$\frac{|2-5|}{\sqrt{3^2+4^2}}$

$=\frac{3}{\sqrt{25}}=\frac{3}{5}$

Hence, the ratio is given by

$\frac{3}{5}:\frac{7}{5}$

$=3:7$

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

Solution:

The midpoint of the line joining the points (4, -5) and (-2, 9) is (1, 2).

Let $\theta$ be the inclination of the straight line passing through the points (-3, 6) and (1, 2).

Then, $\tan\theta=\frac{ 2-6}{1+3}=-1$

$\Rightarrow\theta=\frac{3\pi}{4}$

4. $\text{a}=\pm\sqrt{2}\text{b}$

Solution:

The midpoints of BC and AC are $\text{D}\Big(\frac{\text{a}}{2},0\Big)$ and $\text{E}\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big)$

Slope of $\text{AD}=\frac{0-\text{b}}{\frac{\text{a}}{2}-0}$

Slope of $\text{BE}=\frac{-\frac{\text{b}}{2}}{\frac{\text{-a}}{2}}$

It is given that the medians are perpendicular to each other.

$\frac{0-\text{b}}{\frac{\text{a}}{2}-0}\times\frac{-\frac{\text{b}}{2}}{-\frac{\text{a}}{2}}=-1$

$\Rightarrow\text{a}=\pm\sqrt{2}\text{b}$

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°.

1. $\frac{11}{8}$

Solution:

The area of a triangle with vertices D(x, 3x), B(-3, 5) and C(4, -2) is given below:

Area of $\triangle\text{DBC}=\frac{1}{2}\{\text{x}(5+2)-3(-2-3\text{x})+4(3\text{x}-5)\}$

⇒ Area of $\triangle\text{DBC}=(14\text{x}-7)\text{sq units}$

Similarly, the area of a triangle with vertices A(6, 3), B(-3, 5) and C(4, -2) is given below:

$\triangle\text{ABC}=\frac{1}{2}\{6(5+2)-3(-2-3)+4(3-5)\}$

$\triangle\text{ABC}=\frac{49}{2}\text{sq units}$

Given:

$\triangle\text{DBC}:\triangle\text{ABC}=1:2$

$\frac{2(14\text{x}-7)}{49}=\frac{1}{2}$

$\Rightarrow8\text{x}-4=7$

$\Rightarrow\text{x}=\frac{11}{8}$

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. $\Big(\frac{2}{3},2\Big).$

Solution:

Given:

a + b + c = 0

Substituting c = -a - b in 3ax + by + 2c = 0, we get:

$3\text{ax}+\text{by}-2\text{a}-2\text{b}=0$

$\Rightarrow\text{a}(3\text{x}-2)+\text{b}(\text{y}-2)=0$

$\Rightarrow(3\text{x}-2)+\frac{\text{b}}{\text{a}}(\text{y}-2)=0$

This line is of the form $\text{L}_1+\lambda\text{L}_2=0,$ which passes through the intersection of the lines L₁ and L₂ i.e. 3x - 2 = 0 and y - 2 = 0.

Solving 3x - 2 = 0 and y - 2 = 0, we get

$\text{x}=\frac{2}{3},\text{y}=2$

Hence, the required fixed point is $\Big(\frac{2}{3},2\Big).$

3. 3

Solution:

The lines x + q = 0, y - 2 = 0 and 3x + 2y + 5 = 0 are concurrent.

$\therefore\begin{vmatrix}1&0&\text{q}\\0&1&-2\\3&2&5 \end{vmatrix}=0$

$\Rightarrow1(5+4)-0+\text{q}(0-3)=0$

$\Rightarrow3\text{q}=9$

$\Rightarrow\text{q}=3$

3. (0, 0)

Solution:

Let the coordiantes of the point be (a, b)

Now, the distance of the point (a, b) from 4x + 3y + 10 = 0 is given by

$\Bigg|\frac{4\text{a}+3\text{b}+10}{\sqrt{4^2+3^2}}\Bigg|$

$=\Bigg|\frac{4\text{a}+3\text{b}+10}{5}\Bigg|$

Again, the distance of the point (a, b) from 5x - 12y + 26 = 0 is given by

$\Bigg|\frac{5\text{a}-12\text{b}+26}{\sqrt{5^2+(-12)^2}}\Bigg|$

$=\Bigg|\frac{5\text{a}-12\text{b}+26}{13}\Bigg|$

Again, the distance of the point (a, b) from 7x + 24y - 50 = 0 is is given by

$\Bigg|\frac{7\text{a}+24\text{b}-50}{\sqrt{7^2+(24)^2}}\Bigg|$

$=\Bigg|\frac{7\text{a}+24\text{b}-50}{25}\Bigg|$

Now,

$\Bigg|\frac{4\text{a}+3\text{b}+10}{5}\Bigg|=\Bigg|\frac{5\text{a}-12\text{b}+26}{13}\Bigg|=\Bigg|\frac{7\text{a}+24\text{b}-50}{25}\Bigg|$

Only a = 0 and b = 0 is satisfying the above equation

Hence, the correct answer is option (c).

4. None of these.

Solution:

The given points are {(a + 3)(a + 4), a + 3), ((a + 2)(a + 3), (a + 2)) and ((a + 1)(a + 2), (a + 1)}

Let A be the area of the triangle formed by these points.

Then, $\text{A}=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$

$\Rightarrow\text{A}=\frac{1}{2}[(\text{a}+3)(\text{a}+4)(\text{a}+2-\text{a}-1)+(\text{a}+2)(\text{a}+3)\$\text{a}+1-\text{a}-3)+(\text{a}+1)(\text{a}+2)(\text{a}+3-\text{a}-2)]$

$\Rightarrow\text{A}=\frac{1}{2}[(\text{a}+3)(\text{a}+4)-2(\text{a}+2)(\text{a}+3)+(\text{a}+1)(\text{a}+2)]$

$\Rightarrow\text{A}=\frac{1}{2}[\text{a}^2+7\text{a}+12-2\text{a}^2-10\text{a}-12+\text{a}^2+3\text{a}+2]$

$\Rightarrow\text{A}=1$

2. 3x + 2y - 2 = 0

Solution:

Given:

4x + 3y - 7 = 0 ...(1)

8x + 5y - 1 = 0 ...(2)

The equation of the line with slope $-\frac{3}{2}$ is given below:

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

$\Rightarrow\frac{3}{2}\text{x}+\text{y}-\text{C}=0 \ ...(3)$

The lines (1), (2) and (3) are concurrent.

$\therefore\begin{vmatrix}4&3&-7\\8&5&-1\\\frac{3}{2}&1&\text{-c} \end{vmatrix}=0$

$\Rightarrow4(-5\text{c}+1)-3\Big(-8\text{c}+\frac{3}{2}\Big)-7\Big(8-\frac{15}{2}\Big)=0$

$\Rightarrow-20\text{c}+4+24\text{c}-\frac{9}{2}-56+\frac{105}{2}=0$

$\Rightarrow\frac{-40\text{c}+8+48\text{c}-9-112+105}{2}=0$

$\Rightarrow8\text{c}=8$

$\Rightarrow\text{c}=1$

On substituting c = 1 in $\text{y}=-\frac{3}{2}\text{x}+\text{c},$ we get:

$\text{y}=-\frac{3}{2}\text{x}+1,$

$\Rightarrow3\text{x}+2\text{y}-2=0$

3. a rhombus.

Solution:

 The given lines can be written separately in the following manner:

ax + by + c = 0 ...(1)

ax + by - c = 0 ...(2)

ax - by - c = 0 ...(3)

ax - by - c = 0 ...(4)

Graph of the given lines is given below:

Clearly, $\text{AB}=\text{BC}=\text{CD}=\text{DA}=\sqrt{\frac{\text{a}^2}{\text{c}^2}+\frac{\text{b}^2}{\text{c}^2}}=\frac{\sqrt{\text{a}^2+\text{b}^2}}{|\text{c}|}$

Thus, the region formed by the given lines is ABCD, which is a rhombus. 

4. 4 : 9

Solution:

Let the line segment joining the points (1, 2) and (−2, 1) be divided by the line 3x + 4y = 7 in the ratio m:n.

Then, the coordinates of this point will be $\Big(\frac{-2\text{m}+\text{n}}{\text{m}+\text{n}},\frac{\text{m}+2\text{n}}{\text{m}+\text{n}}\Big)$ that lie on the line.

$3\text{x}+4\text{y}=7$

$3\times\frac{-2\text{m}+\text{n}}{\text{m}+\text{n}}+4\times\frac{\text{m}+2\text{n}}{\text{m}+\text{n}}=7$

$\Rightarrow-2\text{m}+11\text{n}=7\text{m}+7\text{n}$

$\Rightarrow-9\text{m}=-4\text{n}$

$\Rightarrow\text{m}:\text{n}=4:9$

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

Solution:

The equation of the line perpendicular to 3x + y = 3 is given below:

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

This line passes through (2, 2).

$2-6+\lambda=0$

$\Rightarrow\lambda=4$

So, the equation of the line will be

x - 3y + 4 = 0

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

Hence, the y-intercept is $\frac{4}{3}.$

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.

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

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. (-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.

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

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

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

$\Rightarrow4\lambda=4$

$\Rightarrow\lambda=1$ 

3. A.P.

Solution:

The given lines are

ax + 12y + 1 = 0 ...(1)

bx + 13y + 1 = 0 ...(2)

cx + 14y + 1 = 0 ...(3)

It is given that (1), (2) and (3) are concurrent.

$\begin{vmatrix} \text{a}&12&1\\\text{b}&13&1\\\text{c}&14&1\end{vmatrix}=0$

$\Rightarrow\text{a}(13-14)-12(\text{b}-\text{c})+14\text{b}-13\text{c}=0$

$\Rightarrow-\text{a}-12\text{b}+12\text{c}+14\text{b}-13\text{c}=0$

$\Rightarrow-\text{a}+2\text{b}-\text{c}=0$

$\Rightarrow2\text{b}=\text{a}+\text{c}$

Hence, a, b and c are in AP.

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).

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.

1. $4\text{p}_1^2 + \text{p}_2^2 = \text{a}^2$

Solution:

The given lines are

$\text{x} \sec \theta + \text{y} \text{cosec}\theta = \text{a} \ ...(1)$

$\text{x} \cos \theta - \text{y} \sin \theta = \text{a} \cos 2 \theta \ ...(2)$

p₁ and p₂ are the perpendiculars from the origin upon the lines (1) and (2), respectively.

$\text{p}_1=\Big|\frac{-\text{a}}{\sqrt{\sec^2\theta+\text{cosec}^2}\theta}\Big|$ and $\text{p}_2=\Big|\frac{-\text{a}\cos2\theta}{\sqrt{\cos^2\theta+\sin^2}\theta}\Big|$

$\Rightarrow\text{p}_1=\Big|\frac{-\text{a}\sin\theta\cos\theta}{\sqrt{\sin^2\theta+\cos^2}\theta}\Big|$ and $\text{p}_2=\big|-\text{a}\cos2\theta\big|$

$\Rightarrow\text{p}_1=\frac{1}{2}\big|-\text{a}\times2\sin\theta\cos\theta\big|$ and $\text{p}_2=\big|-\text{a}\cos2\theta\big|$

$\Rightarrow\text{p}_1=\frac{1}{2}\big|-\text{a}\times\sin2\theta\big|$ and $\text{p}_2=\big|-\text{a}\cos2\theta\big|$

$\Rightarrow4\text{p}_1^2+\text{p}_2^2=\text{a}^2(\sin^22\theta+\cos^22\theta)$

$=\text{a}^2$