(2)     12 units

(1)

Using the mid point formula, the coordinates of mid-point of BC are

Co-ordinates of $D(x,y)=(\frac{{\displaystyle 6+0}}{{\displaystyle 2}},\frac{{\displaystyle 4+0}}{{\displaystyle 2}})=(3,2)$

Now, length of $AD=\sqrt{{(5-3)}^2+{(-6-2)}^2}=\sqrt{4+64}=\sqrt{68}$

(2)

$\text{Here, (}15{\text{)}}^2={(3-x)}^2+{(-5+5)}^2$

$\Rightarrow 225=9-6x+x^2\Rightarrow x^2-6x-216=0$

$\Rightarrow x^2-18x+12x-216=0\Rightarrow x(x-18)+12(x-18)=0$

$\Rightarrow (x-18)(x+12)=0\Rightarrow x=-12,18$

(3)

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

Therefore, midpoint of QS = midpoint of PR. Let the coordinate of S is (x, y)

$\Rightarrow (\frac{{\displaystyle x-2}}{{\displaystyle 2}},\frac{{\displaystyle y+3}}{{\displaystyle 2}})=(\frac{{\displaystyle 3-3}}{{\displaystyle 2}},\frac{{\displaystyle 4-2}}{{\displaystyle 2}})$

$\Rightarrow \frac{x-2}{2}=\frac{3-3}{2} \text{and} \frac{y+3}{2}=\frac{4-2}{2}$

$\Rightarrow x-2=0 \Rightarrow x=2$

$\text{and} y+3=2 \Rightarrow y=-1$

Hence, fourth vertex S are (2, -1).

(3)

$(\frac{{\displaystyle 6+x}}{{\displaystyle 2}},\frac{{\displaystyle 0+y}}{{\displaystyle 2}})=(2,0)\Rightarrow \frac{6+x}{2}=2, \frac{0+y}{2}=0$

⇒ x = –2 and y = 0

(1)

We know that, Length of Diagonals are equal

In Rectangle, ZO = YX

$\Rightarrow Z{O}^2=Y{X}^2\Rightarrow {(x-0)}^2+{(y-0)}^2={(0+3)}^2+{(4-0)}^2$

$\Rightarrow x^2+y^2=25$

$\Rightarrow x^2+y^2=5 \text{Both diagonals are 5 units}$

(2)        (7, 0)

(4)       (8, – 13)

(3)           $(\frac{13}{7},0)$

$x=\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2}=\frac{5\times 1+2\times 4}{5+2}=\frac{5+8}{7}=\frac{13}{7}$

$y=\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}=\frac{5\times 2+2\times (-5)}{5+2}=\frac{10-10}{7}=0$

(2)

Let given end points of diameter be A(-5,4) and B(1,0).

∴ Diameter has length, 

$$\begin{gathered} AB=\sqrt{(1+5{)}^2+(0-4{)}^2} \\ =\sqrt{36+16} \\ =\sqrt{52}=2\sqrt{13} \mathrm{units} \\ \mathrm{Radiusofcircle}=\frac{2\sqrt{13}}{2}=\sqrt{13}\mathrm{units}. \end{gathered}$$

(2)

As we know that distance of the point P(x,y) from y-axis is magnitude of its x-coordinate.
∴ The distance of the point (-4,3) from y-axis is 4 units

(3)

Let given points be A (0,5) and B (-5,0).
Now, 

$$\begin{gathered} AB=\sqrt{(-5-0{)}^2+(0-5{)}^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt{2} \\ \Rightarrow \sqrt{2}AB=\sqrt{2}\times 5\sqrt{2}=5\times 2=10 \mathrm{units} \end{gathered}$$

(3)

Given points be A (m,-n)and B (-m,n).

Now, 

$$\begin{gathered} AB=\sqrt{(m+m{)}^2+(-n-n{)}^2}=\sqrt{4m^2+4n^2}=2\sqrt{m^2+n^2} \\ Required distance=2\sqrt{m^2+n^2} \end{gathered}$$

(4)

Point is on the circle with centre (0,0) and radius 5 units.
So, there are infinitely many points (lying on circumference) in third quadrant.

(1)

Distance between the points A and B :

$$\begin{gathered} p=\sqrt{(sin\theta -cos\theta {)}^2+(cos\theta +sin\theta {)}^2} \\ \Rightarrow p=\sqrt{2{sin}^2\theta +2{cos}^2\theta -2sin\theta cos\theta +2sin\theta cos\theta } \\ \Rightarrow p=\sqrt{2\left({sin}^2\theta +{cos}^2\theta \right)}=\sqrt{2\times 1}=\sqrt{2} \end{gathered}$$

(2)

Let P(x, 0) be a point on the x-axis such that it divides the line segment joining
A(2, –3) and B(6, 7) in the ratio k : 1.

Using the section formula, we get:

$$\begin{gathered} y=\frac{k{y}_2+y_1}{k+1} \\ 0=\frac{k\times 7+1\times (-3)}{k+1} \\ 0=7k-3\Rightarrow k=\frac{3}{7} \\ So, the ratio is: \\ \frac{3}{7}:1 i.e., 3:7 \end{gathered}$$

(1)

Since P(a/8, 4) is the mid-point of the line segment joining points A(–5, 2) and B(4, 6),

$$\begin{gathered} \frac{a}{8}=\frac{-5+4}{2} \\ \Rightarrow \frac{a}{8}=\frac{-1}{2} \\ \Rightarrow a=\frac{-8}{2} \\ \Rightarrow a=-4 \end{gathered}$$

(3)

We know that centre of a circle is the mid-point of the diameter.

Coordinates of the centre are:

$(\frac{-6+6}{2},\frac{3+4}{2})=(0,\frac{7}{2})$

(1)

We have,

$\frac{AP}{AB}=\frac{AQ}{AC}=\frac{3}{4}$

$\Rightarrow \frac{AP}{AP+PB}=\frac{AQ}{AQ+QC}=\frac{3}{4}$

$\Rightarrow \frac{AP+PB}{AP}=\frac{4}{3}\mathrm{and}\frac{AQ+QC}{AQ}=\frac{4}{3}(\mathrm{Take} \mathrm{reciprocal})$

$\Rightarrow 1+\frac{PB}{AP}=\frac{4}{3}\mathrm{and}1+\frac{QC}{AQ}=\frac{4}{3}$

$\Rightarrow \frac{PB}{AP}=\frac{1}{3}\mathrm{and}\frac{QC}{AQ}=\frac{1}{3}$

Again taking reciprocal, we get:

$\frac{AP}{PB}=\frac{3}{1}\mathrm{and}\frac{AQ}{QC}=\frac{3}{1}$

So, P and Q divide AB and AC, respectively, in the same ratio 3 : 1. Thus, the coordinates of P and Q are

$$\begin{gathered} P\left(\frac{3\times 1+1\times 5}{3+1}, p\frac{3\times 5+1\times 5}{3+1}\right)=\left(\frac{3+5}{4}, \frac{15+5}{4}\right)=\left(2,5\right) \\ and \end{gathered}$$

$Q\left(\frac{3\times 9+1\times 5}{3+1}, \frac{3\times 1+1\times 5}{3+1}\right)=\left(\frac{27+5}{4}, \frac{3+5}{4}\right)=\left(8,2\right)$

Now, $PQ=\sqrt{(2-8{)}^2+(5-2{)}^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt{5} \mathrm{units}.$

(1)

$$\begin{gathered} Let A = (x₁, y₁), B = (x₂, y₂), C = (x₃, y₃) \\ be the vertices of \Delta ABC. \\ Let the midpoints be: \\ D(1, 2) on BC, E(0, -1) on CA, F(2, -1) on AB \\ Since D is the midpoint of BC: \\ \frac{x_2+x_3}{2}=1 \mathrm{and} \frac{y_2+y_3}{2}=2 \\ \Rightarrow x_2+x_3=2, y_2+y_3=4...\left(i\right) \\ Since E is the midpoint of AC: \\ \frac{x_1+x_3}{2}=0\mathrm{and}\frac{y_1+y_3}{2}=-1 \\ \Rightarrow x_1+x_3=0, y_1+y_3=-2...\left(ii\right) \\ Since F is the midpoint of AB: \\ \frac{x_1+x_2}{2}=2\mathrm{and}\frac{y_1+y_2}{2}=-1 \\ \Rightarrow x_1+x_2=4, y_1+y_2=-2...\left(iii\right) \\ Adding \left(i\right), \left(ii\right) and \left(iii\right): \\ (x_2+x_3)+(x_1+x_3)+(x_1+x_2)=2+0+4 \\ \Rightarrow x_1+x_2+x_3=3 \\ Similarly for y-values: \\ (y_2+y_3)+(y_1+y_3)+(y_1+y_2)=4-2-2 \\ \Rightarrow y_1+y_2+y_3=0...\left(iv\right) \\ Finding coordinates of A \end{gathered}$$