In what ratio does the x axis divide the line segment joining the points (-4,-6) and (-1,7)? So, how to find the coordinate of the point of division?

How to answer the following question?rnAmir starts walking from his house to office. Instead of going to the office directly, he goes to the bank first, from there to his daughter's school and then reaches the office. What

In the figure OPQR is a parallelogram.Find a -the coordinates of the vertex Q b-area of the parallelogram

In the figure OPQR is a parallelogram.Find a -the coordinates of the vertex Q b-area of the parallelogram

In the figure OPQR is a parallelogram.Find a -the coordinates of the vertex Q b-area of the parallelogram

Let ABCD be a sqare of side 2a. Find the coordinates of the vertices of this sqare when: 1: A coincide with the origin and AB and coordinayes axes are parallel to the side AB and AD respectively. 2:The centre of the sqare is at the origin amd coordinate axes are parallel to the side AB and AD respectively.

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ABCD is a square. Co-ordinates of A and C are (-1,-1) and (1,1) respectively. Write the coordinates of B and D. Also write the equations of all the sided of the square.

Given, A (-1,-1) and C (1,1)

Then, B (1,-1) and D (-1,1)

Also, equation of sides of square are

AB : y = -1

BC : X = 1

CD: Y=1

DA: X = -1

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Find the coordinates of the points where the line representing the equation cut the x-axis and the y-axis.

Writing in the standared form

On x-axis, y = 0

Point on the x-axis = (4,0)

On y-axis, x = 0

Point on the y-axis = (0,6).

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Shade the triangle formed by the graph of 2x- y = 4, x+ y = 2 and the y-axis. Write the coordinates of vertices of the triangle.

2x -y = 4

X | 0 | 2 | 1 |

Y | -4 | 0 | -2 |

x |
0 |
2 |
1 |

Y |
2 |
0 |
1 |

From the graph, ABC is the required triangle and its verticle s are A (0,2), B (0,-4), C (2,0).

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What you mean by abscissa of a point

The distance of a point from the y-axis is called its x- coordinates, or abscissa.

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Point p is on a x axis and is at a distance of 4 units from y- axis to its left. Write the co-ordintates of the point p.

The p is on x-axis

y = 0

P is at a distance 4 units from y axis to its lft

In second quadrant, the coordinates of the point

p = (-4, 0)

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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

X |
3 |
7 |
-1 |

Y |
3 |
6 |
0 |

and

X |
3 |
7 |
11 |

Y |
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

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The x-co-ordinate of a point P is twice its y- co-ordinate. If P is equidistant from Q (2,-5) and R (-3,6), find the co-ordinates of P.

Let the point P be (2y,y)

PQ = PR

Solving to get y = 8

Hence coordinates of point P are (16,8).

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Find the ratio in which the point (-3, p) the line segment joining the points (-5, -4) and (-2, 3). Hence find the value of P.

Let x (-3, p) divides the join of A (-5,-4) and B (-2,3) in the ratio k :1

The coordinates of P are

But co-ordinates of P are (-3,p)

Substituting k = 2 gives

Ratio of division is 2 : 1 and

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The base of QR of an equilateral triangle PQR lies on x-axis. The coordinates of point Q are (-4, 0) and the origin is the midpoint of the base. Find the coordinates of the point P and R.

Co-ordinates of point R = (4,0)

QR = 8 units

Let the co-ordinates of point P be (0,y)

Since PQ = QR

Coordinates of P are or

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In the figure PQR is an equilateral triangle with co-ordinates of Q and R as (-2,0) and (2,0) respectively. Find the co-ordinates of the vertex P.

From the graph

QR = 2+2 = 4

So, PQ = QR =RP = 4

Now, OR = 2

So from right angle OPR

OP^{2} = PR^{2} - OR^{2}

OP2 = 4^{2} -2^{2 }= 12

So, OP =

Therefore coordinates of P are

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Point P is on the x-axis and is at a distance 4 units from y-axis to its left. write the co-ordinates of the point P.

The P is on x-axis

y= 0

P is at distance of 4 units from y- axis to its left

In second quadrant the coordinates of the point

p = (-4,0)

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Mention the coordinates of a point which is 7 units away from the x-axis and lies on the negative direction of the y-axis

(0,-7)

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Write the coordinates of A, B, C and D from the following figure :-

coordinates of A = (5,0)

coordinates of B = (5,3)

coordinates of C = (-2,4)

coordinates of D = (0,-3)

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A point lies on x-axis at a distance of 9 units from y-axis. What are its coordinates what will be the coordinates of a point if it lies on y-axis at a distance of 9 units from x axis

(i) Since the point lies on x-axis at a distance of 9 units from y-axis. Hence its coordinates are (9,0)

(ii) According to the question, the required coordinates are (0,-9)

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From the given figure, write the following :

A. The coordinates of P

B. The abscissa of the point Q

C. The ordinates of the point R.

D. The points whose abcissa is 0.

A. (-3,-2)

B. 0

C. 0

D. Q and T.

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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)

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write the coordinates of the vertices of a rectangle whose length and breadth are 6 and 3 units respectively one vertex at the origin the longer side lies on the x-axis and one of the vertices lies in the III quadrant.

The vertices of the rectangle OABC are O (0,0), A (-6,0), B (-6,-3), C(0,-3)

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Observe the points plotted in the figure and find the following

(i) The coordinates of E

(ii) The point which the coordinates (-4,-1)

(iii) The abscissa of A- abscissa of B

(iv) The ordinates of C+ ordinate of F.

(i), E (-1,2)

(ii) D (-4, -1)

(iii) Coordinates of A = (2,-2)

The abscissa of A - abscissa of B = 2-4 =-2

(iv) Coordinates of C = (3,3)

Coordinates of F = (-1,-4)

The ordinate of C + ordinates of F = 3+ (-4) = -1

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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 (x^{2},y-1), S (2x,-3y)

^{2},y-1), S (2x,-3y)

According to the question,

(-2,9) = (1+x,y^{2)}

-2 = 1+x or x = -3

and 9 =y^{2} 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 (x ^{2},y-1) = R (9, 2 )**

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

**Which will be in III quadrant **

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Locate and the write the coordinates of a point:

(A) Above x-axis lying on the y-axis at a distance of 5 units from the origin.

(B) Below x-axis lying on the y-axis at a distance of 3 units from the origin.

(C) lying on the x-axis to the right of origin at a distance of 5 units

(D) lying on the x-axis to the left of origin at a distance of 2 units

(A) P (0,5)

(B) Q (0,-3)

(C) R (5,0)

(D) S (-2,0)

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A point lies on the x-axis at a distance of 9 units from the y-axis. What are its coordinates? what will be the coordinates of a point if it lies on the y-axis at a distance of -9 units from x-axis?

(i) Since the point lies on the x-axis at a distance of 9 units from the y-axis. Hence its coordinates are (9,0).

(ii) According to the question, the required coordinates are (0,-9)

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From the given figure, write the following:

(A) The coordinates of P.

(B) The abscissa of the point Q

(C) The ordinates of the point R.

(D) The points whose abscissa is 0.

(A) (-3,-2)

(B) 0

(C) 0

(D)Q and T

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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)

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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:

According to the question,

and

= 3 as y >0

So,

Which will be in IV quadrant

Q(2,x) = Q (2,-3)

Which will be in IV quadrant

Which will be in I quadrant

Which will be in III quadrant

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From the figure given below find the coordinates of point Q.

The coordinates of point Q = (-3,-3.5)

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Plot the point P(2,-6) on a graph paper and from it draw PM and PN perpendiculars to the x-axis and y-axis respectively. Write the coordinates of the points M and N.

Plotting of P(2,-6)

Coordinates of M (2,0)

Coordinates of N (0,-6)

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In the coordinate plane, draw a square of side 3 units, taking origin as one vertex. Also write the coordinates of its vertices.

Vertices are (0,0) (3,0) (3,3) and (0,3)

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In figure are equilateral triangles. Find the coordinates of point C and D.

OB = a, OA = a

AB = 2a

AC = 2a

Using Pythagoras theorem in

So, coordinates of C are (0,)

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In fig: PQR is an equilateral triangle in the coordinates of Q and R as (0,4) and (0,-4). Find the coordinates of the vertex P.

In

The coordinates of P are

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Draw the line passing through (2,3) and (3,2) Find the coordinates of the points at which this lin meets the x-axis and y-axis.

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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)

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Find the coordinates of the given vertices of a rectangle placed in III quadrants in the Cartesian plane with length 'p' units on x-axis and breadth 'q' units on y-axis.

From figure the coordinate of vertices of a rectangle are :

O(0,0)

A (-p,0)

B (-p,-q)

C (0,-q)

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In the figure, OABC is a rectangle and its breadth is 3. Write the coordinates of the vertices B and C.

Given OABC is a rectangle

Breadth = 3 cm

A (6,0)

From the figure AB = 3 cm (breadth)

Hence, B = (6,3)

C = (0,3) [OC = 3 cm breadth]

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PQR are the midpoints of the side of the triangle ABC.

a. What type of the quadrilateral is PQCR?

b. Write the Co-ordinates of the vertices A and C.

a. From the figure, PQCT be a parallelogram.

b. From the figure, We can see that

The coordinates of A = (0,0)

The coordinates of C = (2,4)

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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)

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

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

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

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

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

Here is equal to the radius, so the point be on the circle.

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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 coordinates of the point dividing the line CD in the ratio 2:1?

Given, the vertices of the,

A (1,1) B (5,5) C(2,5)

a. Mid-point of D of AB

(Formula)

Points A (1,1) B (5,5)

ie., D =

b. Length of CD points C(2,5) D (3,3) [Using distance formula]

c. Given CD = 2:1 (ie., m:n)

(Formula)

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In the figure of the radius of the circle centred at C is 5. The circle passes through the point A(8,0). If PC is perpendicular to x-axis find the coordinates of the points P, B and C.

Coordinates of P (4,0), B (0,8) C(4,4)

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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)

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

b. Equation of line

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 =

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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

= 50

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

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Find the slope od line joining (2,4) and (4,7). Write the coordiantes of another pointb on the line.Check whether (5,8) is on this line.

Slope of the line joining (2,4) and (4,7)=

For any point on this line if the cahnge in y is 3, the change in x will be 2.

Coordinates of another point on this line =(4+2,7+3)=(6,10)

Slope of the line joining the points (2,4) and (5,8)=

Since this slope is not equal to the slope of the first line the point (5,8) is not on the first line.

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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)

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Find the length of the line joining A(-2,-3) and B(4,5). Write the equation of circle whose diameter is AB.

=

=

Radius of the circle =

Coordinates of the centre of the circle =

=

Equation of the circle:

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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 =

=

b. Distance between the centre of the circle and (0,2)=

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) =

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) =

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

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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=

b. Cd=

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

y coordinates =

Coordinates of this point =

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Calculate the coordinates of the point at which the circle of centre (2,4) and radius 4 units cuts the y axis.

Centre of the circle =(2,4). radius =4

Equation of the circle =

The coordiantes of the point where the circle cuts the y axis is zero.

Coordinates of the point where the circle cuts the y axis =

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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)=

Radius of teh circle = distance between the point (3,-1) and (8,-5)=

Eqaution of the circle with centre (8,-5) and radius is

b. Ket x=y

Since negative numbers has no squareroot, there is no solution to the equation .

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

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In the figure the radius of the circle centred at 0 is 6 units. Line Ab touches the circle at P and .

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

b. Find teh equation ao AB.

a. In and

Since the angles are and sides are in the ratio

Since OP =6

PA =

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

Draw PQ perpendicular to OA.

Angles of are and .

Since OP=6, OQ=3, PQ=

Coordinates of P=

b. Slope of AB=

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

Then slope of the line joining (x,y) and (12,0) will also be

.This is the equation of the line.

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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,

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 =

y coordinates of p =

Coordinates of P=(6,2)

c. AQ: BQ=2:3

Then BA: AQ=1:2

if x is the x coordinates of Q.

If y is the y coordinates of Q.

15+y-5=0,y=5-15=-10

Coordinates of Q =(30,-10)