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Solve graphically the pair of linear equations: 3x- 4y +3 = 0 and  3x + 4 y -21 = 0.Find the coordinates of the vertices of the triangular region formed  by these lines and x-axis. also calculate the area of this triangle. 

 

3x - 4y +3 = 0

          Rightarrow y = frac{3x+ 3}{4}

 

X 3 7 -1
Y 3 6 0

and  3x +4 y - 21 = 0

                         y = frac{21-3x }{4}

X 3 7 11
3 0 -3

(i) These lines intersects each other at a point (3,3)

Hence  x= 3  and y = 3.

(ii) The vertices of triangular region are (3, 3), (-1,0) and  (7,0)

(iii) Area of a  Delta = frac{1}{2}times 8times 3

Rightarrow Area = 12 sq units

Find the coordinates of the point : 

(i) which the lies on x-axis y-axis both

(ii) Whose abscissa is 5 and ordinate is 6.

(iii) whose ordinates is 6 and which lies on y-axis

(iv) whose ordinates is 3 and abscissa is -7

(V) whose abscissa  is 3 and which lies on x-axis

(VI) whose abscissa  is 4 and ordinates is 4.

 

(i ) A point which lies on x-axies is (0,0) i.,e., origin 

(ii) A point whose abscissa is 5 and ordinates is 6 i.e., x= 5 and y = 6 is (5,6)

(iii) A point whose ordinate is 6 i.e., y = 6 and lies on  y-axis is (0,6)

(iv) A point whose ordinates is 3 and absicssa is -7 ie., y = 3 and x = -7 is (-7,3)

(v) A point whose abscissa is 3 ie., x = 3 and lies on x-axis  is (3,0)

(vi) A point whose abscissa is 4 and ordinates is 4 i.e., x = 4 and y = 4 is (4,4)

If the coordinates of  a point M are (-2,9) which can also be expressed as (1+x,y2) and y >0, then  find in which quadrant do the following points lie :  P (y,x) Q (2,x) , R (x2,y-1), S (2x,-3y)

According to the question, 

(-2,9) = (1+x,y2)

            -2 = 1+x or x = -3

and        9 =y2 or y

                =3, as y>0

So,      P(y,x) = P (3,-3)

Which will be in IV quadrant  Q (2,x) = Q (2,-3)

Which will be in IV quadrant  R (x2,y-1) = R (9, 2 )

Which will be in  I quadrant  S (2x, -3y) = S (-6,-9)

Which will be in III quadrant  

 

 

Find the coordinates of the point:

(i) Which lies on x and the y-axes, both.

(ii) Whose abscissa is 5 and ordinates is 6.

(iii) Whose ordinates is 6 and which lies on the y-axis

(iv) Whose ordinate is 3 and abscissa is -7

(v) Whose abscissa is 3 and which lies on x-axis.

(vi) Whose abscissa is 4 and ordinates is 4.

 

 

 

(i) A point which lies on x and y-axes is (0,0) ie., origin

(ii)A point whose abscissa is 5 and ordinate is 6 ie., x = 5 and y = 6 is (5,6)

(iii) A point whose ordinates is 6 i.e., y = 6 and lies on y-axis is (0,6)

(iv) A point whose ordinates is 3 and abscissa is -7 i.e.,y = 3 and x = -7 is(-7,3).

(v) A point whose abscissa is 3, ie., x=3 and lies on x-axis is (3,0)

(vi) A point whose abscissa is 4  and ordinate is 4 ie., x = 4 and y = 4 is (4,4)

 

Three vertices of a rectangle are (3,2), (-4,2) and (-4,5). Plot these points and find the coordinates of the fourth vertex.

 

Plot the three vertices of the rectangle as A(3,2), B(-4,2), C(-4,5). 

We have to find the coordinates of the fourth vertex D so that that ABCD is a rectangle. Since the opposite sides of a rectangle are equal, so the abscissa of D should be equal to the abscissa of A ie., 3 and the coordinate of D should be equal to the ordinate of C, i.e., 5

Coordinates of D are given by (3,5)

 

 

A circle with centre (3,2) passes through the point (6,3)

a. What is the radius of the circle?

b. Check whether each of the points with coordinates (0,2) (3,6) (0,3) inside, outside or on the circle.

a. Given points (3,2) and (6,3)  (using the distance formula)

             r = sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}

               = sqrt{(6-3)^{2} + (3-2)^{2}} = sqrt{3^{2} + 1^{2}} = sqrt{10}

b. (1) (3,2)  and (0,2) (Using distance formula)

               = sqrt{(x_{2}-x_{1})^{2} +(y_{2}-y_{1})^{2}} = sqrt{(0-3)^{2} + (2-2)^{2}} = sqrt{9} = 3

Here 3 is less than the radius, so the point is inside the circle

(2) (3,2) and (3,6)      (using the distance formula)

sqrt{(3-3)^{2}+(6-2)^{2}} = sqrt{0^{2} + 4^{2} } = sqrt{16} = 4

Here 4 is greater than the radius, so the point be outside the circle.

(3) (3,2) and (0,2)      (Using the distance formula)

sqrt{(0-3)^{2} + (3-2)^{2}} = sqrt{9+1} = sqrt{10}

Here sqrt{10} is equal to the radius, so the point be on the circle.

A line of slope 2 passes through the point A (1,3)

a. Check whether B (3,7) is a point on this line.

b. Write down the equation of this line.

c. Find the coordinates of a point C on the line such that BC = 2AB.

a. Slope of A(1,3) and (3,7)

frac{7-3}{3-1}=frac{4}{2}=2

Since the slope of these points is same point B (3,7) is on the line.

b. Equation of line

frac{x-1}{3-1} = frac{y-3}{7-3}

frac{(x-1)}{2} = frac{y-3}{4}

4 (x-1) = 2 (y-) 

4x - 4 = 2y - 6

4x - 2y = -6 + 4

4x - 2y = -2

4x - 2y +2 = 0

c. Coordinate of point C = left ( frac{x_{1}+x_{2}}{2}, frac{y_{1}+y_{2}}{2} right )

                                    = left ( frac{3+1}{2}, frac{7+3}{2} right )

                                    = left ( frac{4}{2}, frac{10}{2} right ) = (2,5)

Equation of a line is y = 2x.

a. A is a point on the line. If the x coordinates of A is -2, find its y coordinate.

b. Verify whether a circle of radius 5 centred at A passes through the point B (5,5)

c. The radius of a circle passing through B is 5 and its centre is on the above-mentioned line. Find the coordinates o its centre.

a. y = 2x

     x = -2

     y = 2 x -2 

        = -4

y coordinate of A is -4

 A (-2,-4)

b. Since y = 2x

here x = 5

y = 2 x 5 = 10

Hence the point (5,5) is not on the line y = 2x

c. If y = 2x

Let (x,2x) be the points on the line 

sqrt{(x-5)^{2} + (2x-5)^{2}} = 5

x^{2} - 10x + 25 +4x^{2} - 20x + 25 = 25

x^{2} -6x +5=0

x = frac{6pm sqrt{36-20}}{2} = frac{6pm 4}{2}

   = 50 

Coordinates of centre (5,10) or (1,2)

In the picture midpoint of the sides of the quadrilateral ABCD are joined to draw PQRS.

a. Find coordinates of all vertices of R.

b. Write coordinates of all vertices of quadrilateral ABCD.

a. The quadrilateral got by joining the midpoint of the sides of a quadrilateral is a parallelogram.PE is drawn parallel to      the x axis and QE parallel to Y axis.

    PE =8-4=4

    QE=8-3=5

    If SF is drawn parallel to the x axis and RF parallel to the y axis.Then SF=4,RF=5.

b. P is the midpoint of AB. Distance between A and the horizontal line through P=3-1=2

    Coordiantes of B=(4+3,3+2)=(7,5)

    Coordinates of C=(8+1,8+3)=(9,11)

    Coordinates of D=(6-3,9-2)=(3,7)

    Coordinates of A =(1,1)

A circle with centre (3,2) passes through the point (6,3).

a. What is the radius of the circle?

b. Check whether each of the points with coordinates (0,2),(3,6),(0,3) is inside ,outside or on the circle.

a. If the centre of the circle is A(3,2) and B(6,3), a point on the circle.

Radius of the circle =sqrt{(6-3)^2 +(3-2)^2 }

                              =sqrt{9+1}=sqrt{10}units

b. Distance between the centre of the circle and (0,2)=sqrt{(3-0)^2 +(2-2)^2 }=sqrt{9+0}=sqrt{9}=3

Since this is less than the radius, the point (0,2) is inside the circle.

Distance between the centre of the circle and (3,6) =sqrt{(3-3)^2 +(6-2)^2 }=sqrt{0+16}=sqrt{16}=4

Since this is greater than the radius, the point (3,6) in outside the circle.

Distance between the centre of the circle and (0,3) =sqrt{(3-0)^2 +(3-2)^2 }=sqrt{9+1}=sqrt{10}

Since this is equal to the radius, the point (0,3) is on the circle.

The coordinates of the vertices of a triangle are A(1,1), B(5,5), C(2,5).

a. Write the coordinates of the midpoint D of AB.

b. What is the length of CD ?

c. What are the coordiaates of the point dividing the line CD in the ratio 2:1 ?

a. A(1,1), b(5,5), c(2,5)

Coordinates of D=left ( frac{1+5}{2},frac{1+5}{2} right )=(3,3)

b. Cd=sqrt{(2-3)^2 +(5-3)^2 }=sqrt{1+4}=sqrt{5}

c. x coordinates  of the point dividing the line joining c(2,5) and D(3,3) in the ratio 2:1

x_{1}+frac{p}{p+q}(x_{2}-x_{1})=2+frac{2}{3}(3-2)

                          =2+frac{2}{3}times 1=2+frac{2}{3}=frac{6}{3}+frac{2}{3}=frac{8}{3}

y coordinates =y_{1}+frac{p}{p+q}(y_{2}-y_{1})

                      =5+frac{2}{3}(3-5)=5+frac{2}{3}times -2

                      =5-frac{4}{3}=frac{15}{3}-frac{4}{3}=frac{11}{3}

Coordinates of this point =left ( frac{8}{3} ,frac{11}{3}right )

a. Find the centre of the circle with the line joining te points (3,-1), (13,-9) as diameter. Find the equation of this circle.

b. Show taht there is no point on the circle whose x and y coordiantes are equal.

a. Centre of the circle = midpoint of the line joining the points (3,-1) and (13,-9)=left ( frac{3+13}{2},frac{-1+-9}{2} right )=(8,-5)

Radius of teh circle = distance between the point (3,-1) and (8,-5)=sqrt{(3-8)^2 +(-1--5)^2 }=sqrt{5^2 +4^2 }=sqrt{25+16}=sqrt{41}

Eqaution of the circle with centre (8,-5) and radius sqrt{41} is (x-8)^2 +(y--5)^2 =left ( sqrt{41} right )^2

(x-8)^2 (y+5)^2 =41

b. Ket x=y

(x-8)^2 +(x+5)^2 =41

x^2 -16x+64+x^2 +10x+25-41=0

2x^2 -6x+48=0,x^2 -3x+24=0

sqrt{b^2 -4ac}=sqrt{9-96}=sqrt{-87}

Since negative numbers has no squareroot, there is no solution to the equation x^2 -3x+24=0.

So there cannot be any point on the circle where x and y coordinates are equal.

In the figure the radius of the circle centred at 0 is 6 units. Line Ab touches the circle at P and angle OAB=30^0.

a. Find the coordinates of the point A and P.

b. Find teh equation ao AB.

a. In Delta OPA, angle A=30^0 ,angle P=90^0 and angle O=60^0 .

    Since the angles are 30^0 ,60^ 0 and 90^0 , sides are in the  ratio 1:sqrt{3}:2

    Since OP =6

    PA =6sqrt{3}, OA=12

   Since OA =12, coordinates of A =(12,0)

   Draw PQ perpendicular to OA.

   Angles of Delta OPQ are 30^0 ,60^0 and 90^0.

   Since OP=6,  OQ=3,  PQ=3sqrt{3}

   Coordinates of P=(3,3sqrt{3})

b. Slope of AB=frac{y_{2}-y_{1}}{x_{2}-x_{1}}=frac{3sqrt{3}-0}{3-12}

                     =frac{3sqrt{3}}{-9}=frac{sqrt{3}}{-3}=frac{sqrt{3}}{-sqrt{3}times sqrt{3}}

                     =frac{1}{sqrt{3}}=frac{-1}{sqrt{3}}

   Let (x,y) be a point on the line AB.

   Then slope of the line joining (x,y) and (12,0) will also be frac{-1}{sqrt{3}}

    frac{0-y}{12-x}=frac{-1}{sqrt{3}},-1(12-x)=sqrt{3}times -y

    -12+x=-sqrt{3}y

    x+sqrt{3}y-12=0

    x+sqrt{3}y=12.This is the equation of the line.

In the figure, equation of the line joining the points A and B is x+2y=10. P and Q are points on this line.

a. Find the coordinates of the points A and B.

b. Find the coordinates of the points P that divides the line AB in the ratio 2:3.

c. If AQ: BQ =2:3, find the coordinates of the point Q.

a. Equation of the line AB is x+2y=10

    Since the y coordinates of the point A is zero, x+2times 0=10,x=10

    Coordinates of A = (10,0)

    Since the x coordinates of the point B is sero, 0+2y=10,y=5

    Coordinates of B =(0,5)

b. AP : PB =2:3

   x coordinates of P =x_{1}+frac{p}{(p+q)}(x_{2}-x_{1})

                                =10+frac{2}{5}(0-10)=10+frac{2}{5}times -10

                                =10+-6

    y coordinates of p =y_{1}+frac{p}{p+q}(y_{2}-y_{1})

    =0+frac{2}{5}(5-0)=0+2=2

    Coordinates of P=(6,2)

c. AQ: BQ=2:3

   Then BA: AQ=1:2

   if x is the x coordinates of Q.

   0+frac{1}{3}(x-0)=10

   frac{1}{3}(x-0)=10

   x-0=10times 3=3 0,x=30

   If y is the y coordinates of Q.

   5+frac{1}{3}(y-5)=0,5+frac{(y-5)}{3}=0

  3times 5+frac{3(y-5)}{3}=3times 0

  15+y-5=0,y=5-15=-10

  Coordinates of Q =(30,-10)