Mathematics Chapter 7 Coordinate Geometry

Let the points X (4,k) and Y(1,0)
It is given that the distance XY is 5 units.
By using the distance formula,

 Let A(x₁, y₁)=A(2, 1), B(x₂, y₂)=B(3,−2) and C(x₃, y₃)=C(7/2,y).

Let A(a, a2), B(b, b2) and C(0, 0) be the coordinates of the given points.
We know that the area of a triangle having vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is ∣∣12[x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)]∣∣ square units.
So,
Area of ∆ABC\

Since the area of the triangle formed by the points (a, a²), (b, b²) and (0, 0) is not zero, so the given points are not collinear.

let t line x-y-2=0 divid t line segment in t ratio k : 1 at point C
so coordinates of C are

C lies on x - y - 2 = o

so, subs coordinates of x,y in tis eq.

Given point is P(x,y)

Origin point is O (0,0)

Using distance formula

$$\begin{gathered} PO = \sqrt{(x_2-x_1{)}^2 + (y_2-y_1{)}^2} \\ = \sqrt{(x-0{)}^2 + (y-0{)}^2} \\ = \sqrt{x^2 +y^2} \end{gathered}$$

Suppose the point P(4, m) divides the line segment joining the points A(2, 3) and B(6, -3) in the ratio K : 1

Co-ordinates of the point,

$P = \left(\frac{6K + 2}{K + 1}, \frac{-3K + 3}{K + 1}\right)$

But the co-ordinates of point P are given as (4, m)

$$\begin{gathered} \frac{6K + 2}{K+1} = 4 ....\left(i\right) \\ \frac{-3K + 3}{K+1} = m ..... \left(ii\right) \\ 6K + 2 = 4K +4 \\ 2K =2 \\ K =1 \\ Putting K =1 in equ. \left(2\right) \\ \frac{-3\left(1\right) + 3}{1 + 1} = m \\ \therefore m = 0 \\ Ratio is 1:1 and m = 0 \\ i.e P is the mid point of AB \end{gathered}$$

Let the y-axis divide the line segment joining the points ( -4, -6 ) and ( 10, 12 )in the ratio k:1 and the point of the intersection be ( 0, y ).

Using section formula, we have

$$\begin{gathered} \left[\frac{10\mathrm{k}+ ( -4)}{\mathrm{k} + 1} . \frac{12\mathrm{k} + (-6 )}{\mathrm{k} + 1} \right]= 0, \mathrm{y} \\ \therefore \frac{10\mathrm{k} - 4}{\mathrm{k} + 1} = 0 \Rightarrow 10\mathrm{k} -4 = 0 \Rightarrow \mathrm{k} = \frac{4}{10 } =\frac{2}{5} \\ \mathrm{Thus} \mathrm{the} \mathrm{y}-\mathrm{axis} \mathrm{divides} \mathrm{the} \mathrm{line} \mathrm{segment} \mathrm{joining} \mathrm{the} \mathrm{given} \mathrm{points} \mathrm{in} \mathrm{the} \mathrm{ratio} 2:5 \\ \therefore \mathrm{y} = \frac{12\mathrm{k} + (-6)}{\mathrm{k} + 1} = \frac{12 \mathrm{x} {\displaystyle \frac{2}{5}} -6}{{\displaystyle \frac{2}{5}} + 1} =\frac{\left({\displaystyle \frac{24-30}{5}}\right)}{\left({\displaystyle \frac{2+5}{5}}\right)} = -\frac{6}{7} \\ \mathrm{Thus}, \mathrm{the} \mathrm{coordinates} \mathrm{of} \mathrm{the} \mathrm{point} \mathrm{of} \mathrm{division} \mathrm{are} \left(0, -\frac{6}{7}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{The} \mathrm{given} \mathrm{points} \mathrm{are} \mathrm{A}( -2, 3 ) \mathrm{B}( 8, 3 ) \mathrm{and} \mathrm{C}( 6, 7 ). \\ \mathrm{using} \mathrm{distance} \mathrm{firmula}, \mathrm{we} \mathrm{have}: \\ {\mathrm{AB}}^2 =( 8 - ( -2 ){)}^2 +( 3 - 3 {)}^2 \\ \Rightarrow {\mathrm{AB}}^2 = ( 8+ 2 {)}^2 + 0 \\ \Rightarrow {\mathrm{AB}}^2 = {10}^2 \\ \Rightarrow {\mathrm{AB}}^2 = 100 \\ {\mathrm{BC}}^2 = (6-8 {)}^2 + (7-3{)}^2 \\ \Rightarrow {\mathrm{BC}}^2 = ( - 2 {)}^2 + 4^2 \\ \Rightarrow {\mathrm{BC}}^2 = 4+16 \\ \Rightarrow {\mathrm{BC}}^2 = 20 \\ {\mathrm{CA}}^2 = ( -2 -6 {)}^2 + ( 3-7 {)}^2 \\ \Rightarrow {\mathrm{CA}}^2 = (-8{)}^2 +(-4 {)}^2 \\ \Rightarrow {\mathrm{CA}}^2 = 64 + 16 \\ \Rightarrow {\mathrm{CA}}^2 =80 \\ \mathrm{It} \mathrm{can} \mathrm{be} \mathrm{observe} \mathrm{that}: \\ {\mathrm{BC}}^2 + {\mathrm{CA}}^2 = 20 + 80 =100 = {\mathrm{AB}}^2 \\ \mathrm{So}, \mathrm{by} \mathrm{the} \mathrm{converse} \mathrm{of} \mathrm{pythagoras} \mathrm{Theorem}, \\ ∆\mathrm{ABC} \mathrm{is} \mathrm{a} \mathrm{right} \mathrm{angled} \mathrm{triangle} \mathrm{with} \mathrm{right} \mathrm{angled} \mathrm{at} \mathrm{C}. \end{gathered}$$

$$\begin{gathered} \mathrm{The} \mathrm{given} \mathrm{vertises} \mathrm{are} \mathrm{A}(\mathrm{x},\mathrm{y}), \mathrm{B}(1,2) \mathrm{C}(2,1). \\ \mathrm{It} \mathrm{is} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{a} \mathrm{tringle} \mathrm{whose} \mathrm{vertises} \mathrm{are} ({\mathrm{x}}_1,{\mathrm{y}}_1). ({\mathrm{x}}_{2,} {\mathrm{y}}_2 ), \\ \mathrm{and} ({\mathrm{x}}_{3, }{\mathrm{y}}_{3) }\mathrm{is} \mathrm{given} \mathrm{by} \\ \frac{1}{2} \left|{\mathrm{x}}_1 ({\mathrm{y}}_2-{\mathrm{y}}_{3 }) + {\mathrm{x}}_{2 }({\mathrm{y}}_3-{\mathrm{y}}_1) + {\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)\right| \\ \therefore \mathrm{Area} \mathrm{of} ∆\mathrm{ABC} \\ = \frac{1}{2} \left|\mathrm{x}(2-1) + 1(1-\mathrm{y}) +2(\mathrm{y}-2)\right| \\ =\frac{1}{2} \left|\mathrm{x}+1-\mathrm{y}+2\mathrm{y}-4\right| \\ =\frac{1}{2} \left|\mathrm{x}+\mathrm{y}-3\right| \\ \mathrm{The} \mathrm{area} \mathrm{of} ∆\mathrm{ABC} \mathrm{is} \mathrm{given} \mathrm{as} 6 \mathrm{sq}.\mathrm{units}. \\ \Rightarrow \frac{1}{2}\left|\mathrm{x}+\mathrm{y}-3\right| = 6 \\ \Rightarrow \mathrm{x}+\mathrm{y}-3 = 12 \\ \therefore \mathrm{x}+\mathrm{y}=15 \end{gathered}$$

C.

7.5

From the figure, the coordinates of A, B, and C are (1,3), (-1, 0) and (4, 0)

respectively.

$$\begin{gathered} \mathrm{Area} \mathrm{of}∆\mathrm{ABC} \\ = \frac{1}{2} \left|1(0-0) + (-1)(0-3) +(3-0)\right| \\ =\frac{1}{2} \left|0+3+12\right| \\ =\frac{1}{2} \left|15\right| \\ = 7.5 \mathrm{sq}. \mathrm{units} \end{gathered}$$

Point P lies  on X-axis  so it's coordinate is 0   ( Using section formula )

Let the ratio  be k:1 

Let the coordinate of the point be P(x,0)

As given A(3, -3) and B( -2,7)

$$\begin{gathered} \mathrm{y} =\frac{ \left({\mathrm{my}}_2 + {\mathrm{ny}}_1\right)}{\mathrm{m}+\mathrm{n}} \\ 0 = \left(\frac{\mathrm{k} \mathrm{x} 7 +1 \mathrm{x} (-3)}{\mathrm{k} +1}\right) \\ 0 \mathrm{x} (\mathrm{k}+1) = 7\mathrm{k} - 3 \\ 0 = 7\mathrm{k} - 1 \\ 3 = 7\mathrm{k} \\ \mathrm{k} = \frac{3}{7} \\ \mathrm{k}:1 = 3:7 \\ \mathrm{x} = \left(\frac{{\mathrm{mx}}_2 + {\mathrm{nx}}_1}{\mathrm{m}+\mathrm{n}}\right) \\ \mathrm{x} = \frac{ \left[\left(\frac{3}{7} \mathrm{x} (-2)\right)+\left(1 \mathrm{x}3\right)\right]}{\left(\frac{3}{7} + 1\right)} \\ \mathrm{x} = 1.5 \end{gathered}$$

    AC² = ( 5 -2 )² + ( 6 + 1 )² 

          = 9 + 49 

          = 58 sq.unit

BD² = ( 5 - 2 )² + ( -1 -6 )² 

      = 9 + 49

      = 58 sq unit

Diagonals of parallelogram are equal so rectangle.

Let co-ordinate of D(x,y) and D is mid point of BC 

$$\begin{gathered} \mathrm{x} = \frac{\left(3+5\right)}{2} = 4 \\ \mathrm{y} = \frac{\left(2-2\right)}{2} = 0 \end{gathered}$$

$$\begin{gathered} \mathrm{Now} \mathrm{area} \mathrm{of} \mathrm{triangle} \mathrm{ABD} =\frac{1}{2}\left\{4\right(-2-0)+3[(0-(-6\left)\right)]+4[(-6)-\left(-2\right)\left]\right\} \\ =0.5 \mathrm{x} [-8+18-16] \\ =3 \mathrm{sq}. \mathrm{unit} \\ \mathrm{And} \mathrm{area} \mathrm{of} \mathrm{triangle} \mathrm{ACD} = \frac{1}{2}\left[5\left(-6-0\right) + 4\left(0-2\right) + 4\left(2+6\right)\right] \\ = 0.5 \mathrm{x} \left[-30 + \left(-8\right) + 32\right] \\ = 3 \mathrm{sq}. \mathrm{unit} \end{gathered}$$

Hence, the median AD divides triangle ABC into two triangles of equal area.

Since a line is  intersecting  Y-axis at P and X- axis at Q,

Coordinates of P = (0,y) and coordinates of Q = (x,0)

Let R be the midpoint of PQ.

$$\begin{gathered} \mathrm{Then}, \mathrm{co}-\mathrm{ordinates} \mathrm{of} \mathrm{R} = \left(\frac{0 + \mathrm{x}}{2}, \frac{\mathrm{y} + 0}{2}\right) = ( 2,-5) \\ \Rightarrow \left(\frac{\mathrm{x}}{2}, \frac{\mathrm{y}}{2}\right) = (2, -5) \\ \Rightarrow \frac{\mathrm{x}}{2} = 2 \mathrm{and} \frac{\mathrm{y}}{2} = -5 \\ \Rightarrow \mathrm{x} = 4 \mathrm{and} \mathrm{y} = -10 \\ \mathrm{Hence}, \mathrm{co}-\mathrm{ordinates} \mathrm{of} \mathrm{P} \mathrm{are} (0, -10) \mathrm{and} \mathrm{co}-\mathrm{ordinates} \mathrm{of} \mathrm{Q} \mathrm{are} (4,0). \end{gathered}$$

Given, P(x,y) is equidistant from  A(5,1)  and B(-1,5)

Now,  AP = BP

$$\begin{gathered} \Rightarrow \sqrt{{\left(5-\mathrm{x}\right)}^2 + {\left(1-\mathrm{y}\right)}^2} = \sqrt{{\left(-1-\mathrm{x}\right)}^2 + {\left(5-\mathrm{y}\right)}^2} \\ \Rightarrow {\left(5-\mathrm{x}\right)}^2 + {\left(1-\mathrm{y}\right)}^2 = {\left(-1-\mathrm{x}\right)}^{2 }+ {\left(5-\mathrm{y}\right)}^2 \\ \Rightarrow \left(25 + {\mathrm{x}}^2 - 10\mathrm{x}\right) + \left(1 + {\mathrm{y}}^2 -2\mathrm{y}\right) = \left(1 + {\mathrm{x}}^2 + 2\mathrm{x}\right) + \left(25 + {\mathrm{y}}^2 -10\mathrm{y}\right) \\ \Rightarrow {\mathrm{x}}^2 + {\mathrm{y}}^2 -10\mathrm{x} -2\mathrm{y} +26 = {\mathrm{x}}^2 +{\mathrm{y}}^2 +2\mathrm{x} -10\mathrm{y} +26 \\ \Rightarrow -10\mathrm{x} -2\mathrm{y} = -10\mathrm{y} + 2\mathrm{y} \\ \Rightarrow -12\mathrm{x} = -8\mathrm{y} \\ \Rightarrow 3\mathrm{x} = 2\mathrm{y} .........(\mathrm{Dividing} \mathrm{throughout} \mathrm{by} -4) \end{gathered}$$

 $$\begin{gathered} \mathrm{Suppose} \mathrm{the} \mathrm{point} \mathrm{A}\left(\frac{24}{11}, \mathrm{y}\right) \mathrm{divides} \mathrm{the} \mathrm{line} \mathrm{segment} \mathrm{joining} \mathrm{points} \\ \mathrm{P}\left(2, -2\right) \mathrm{and} \mathrm{Q}\left(3,7\right) \mathrm{in} \mathrm{the} \mathrm{ratio} \mathrm{k} : 1 \\ \mathrm{Then}, \mathrm{the} \mathrm{co}-\mathrm{ordinates} \mathrm{of} \mathrm{A} \mathrm{are} \left(\frac{3\mathrm{k} + 2}{\mathrm{k} + 1}, \frac{7\mathrm{k} - 2}{\mathrm{k} + 1}\right) \\ \mathrm{But}, \mathrm{the} \mathrm{co}-\mathrm{ordinates} \mathrm{of} \mathrm{A} \mathrm{are} \mathrm{given} \mathrm{as} \left(\frac{24}{11}, \mathrm{y}\right). \\ \Rightarrow \frac{3\mathrm{k} + 2}{\mathrm{k} + 1} = \frac{24}{11} \\ \Rightarrow 33\mathrm{k} + 22 = 24\mathrm{k} + 24 \\ \Rightarrow 9\mathrm{k} = 2 \\ \Rightarrow \mathrm{k} = \frac{2}{9} \\ \mathrm{Hence}, \mathrm{the} \mathrm{ratio} \mathrm{is} 2 : 9. \\ \mathrm{Also}, \frac{7\mathrm{k} - 2}{\mathrm{k} + 1} = \mathrm{y} \\ \Rightarrow \frac{7 \mathrm{x} {\displaystyle \frac{2}{9}} - 2}{{\displaystyle \frac{2}{9}} + 1} = \mathrm{y} \\ \Rightarrow \mathrm{y}= \frac{{\displaystyle \frac{14-18}{9}}}{{\displaystyle \frac{2+9}{9}}} = \frac{-4}{9} \mathrm{x} \frac{9}{11} \\ \Rightarrow \mathrm{y}= -\frac{4}{11} \end{gathered}$$

B.

( 3, 5 )

It is given that the point P divides AB in the ratio  2 : 1.

Using the section formula, the coordinates of the point P are

$$\begin{gathered} \left( \frac{\left(1\mathrm{x}1\right) + \left(2\mathrm{x}4\right)}{2 + 1} , \frac{\left(1\mathrm{x}3\right) + \left(2\mathrm{x}6\right)}{2 + 1}\right) \\ = \left(\frac{1 + 8}{3}, \frac{3 + 12}{3}\right) = (3,5) \end{gathered}$$

Hence, the coordinates of the point P are (3,5).

It is given that the point  A(0, 2) is equidistant from the points B(3, p) and C(p, 5).

So, AB = AC $\Rightarrow$ AB² = AC²

Using distance formula, we have

$\Rightarrow$ ( 0 - 3 )² + (2 - p )² = ( 0 - p )² = ( 2 - 5 )²

$\Rightarrow$ 9 + 4 + p² - 4p = p² + 9

$\Rightarrow$4 - 4p = 0

$\Rightarrow$ p = 1

Hence, the value of p = 1.

Let the co-ordinates of point P be (x, y)

Then using the section formula co-ordinates of P are.

$\mathrm{x} = \frac{-4\mathrm{k} + 3}{\mathrm{k} + 1} \mathrm{y} = \frac{8\mathrm{k} - 5}{\mathrm{k} + 1}$

Since P lies on  x + y = 0

$$\begin{gathered} \therefore \frac{-4\mathrm{k} + 3}{\mathrm{k} + 1} + \frac{8\mathrm{k} - 5}{\mathrm{k} + 1} = 0 \\ \Rightarrow 4\mathrm{k} - 2 = 0 \\ \Rightarrow \mathrm{k} = \frac{2}{4} \\ \Rightarrow \mathrm{k} = \frac{1}{2} \\ \mathrm{Hence} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{k} = \frac{1}{2}. \end{gathered}$$

The area of the triangle, whose vertises are (x₁, y₁),  (x₂, y₂)  and (x_3, y₃) is

$\mathrm{Area} \mathrm{of} \mathrm{a} ∆ = \frac{1}{2}$[ x₁ (y₂ - y₃) + x₂ ( y₃ -  y₁ ) + x₃ ( y₁ - y₂ ) ]

Substituting the given coordinates

$\mathrm{Area} \mathrm{of} ∆ = \frac{1}{2}\left[{\mathrm{x}}_1 ({\mathrm{y}}_2 - {\mathrm{y}}_3 ) + {\mathrm{x}}_2 ({\mathrm{y}}_3 - {\mathrm{y}}_1) + {\mathrm{x}}_3 ( {\mathrm{y}}_1 - {\mathrm{y}}_2)\right]$

$\Rightarrow \frac{1}{2}\left[\left(\mathrm{p} - 7\right) + 40 + 27 + 9\mathrm{p}\right] = 15$

$\Rightarrow$$<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mstyle\; mathsize="14px"><mfrac><mn> </mn></mfrac></mstyle></math> 10p\; +\; 60$  = $\pm$30

$\Rightarrow$$<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mstyle\; mathsize="14px"><mfrac><mn> </mn></mfrac></mstyle></math> 10p\; =\; 30 \; or\; 10p\; =\; -90$

$\Rightarrow$$<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mstyle\; mathsize="14px"><mfrac><mn> </mn></mfrac></mstyle></math> p\; =\; -3 \; \; or \; \; p\; =\; -9$