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**Question. The lines, x = 2 and y = 3 are **

(A) parallel to each other

(B) perpendicular to each other

(C) neither parallel nor perpendicular to each other

(D) none of these

**Answer**

B

**Question. The lines, x = –2 and y = 3 intersect at the point ________. **

(A) (-2,3)

(B) (2,-3)

(C) (3,-2)

(D) (-3,2)

**Answer**

A

**Question. The point equidistant from vertices of a triangle is called **

(A) circumcentre

(B) incentre

(C) orthocenter

(D) none of these

**Answer**

A

**Question. The slope of the line joining the points (2, k – 3) and (4,-7) is 3. Find k. **

(A) -10

(B) -6

(C) -2

(D) 10

**Answer**

A

**Question. The distance between the points (3, 4) and (6, –3) is. **

(A) √58

(B) √68

(C) √78

(D) √98

**Answer**

A

**Question. If A and B are (1, 4) and (5, 2) respectively, then the co-ordinates of P when AP/PB = 3/4 is **

(A) (19/7 , 22/7)

(B) (20/7 , 21/7)

(C) (21/7 , 22/7)

(D) (21/7 , 23/7)

**Answer**

A

**Question. If the coordinates of the mid-points of the sides of a triangle are (1, 2) (0, –1) and (2, –1). Then the sum of x coordinates of its vertices of the triangle is **

(A) 3

(B) 4

(C) 5

(D) 6

**Answer**

A

**Question. The centre of a circle is C(2,-3) and one end of the diameter AB is A(3,5) . Find the coordinates of the other end B. **

(A) (1,-11)

(B) (5,2)

(C) (1,8)

(D) none of these

**Answer**

A

**Question. Find λ , if the line 3x – λy + 6 = 0 passes through the point (–3,4). **

(A) 3/4

(B) -3/4

(C) 4/3

(D) -4/3

**Answer**

B

**Question. Find the area of the triangle formed by the line 5x – 3y + 15 = 0with coordinate axes. **

(A) 15cm^{2}

(B) 5 cm^{2}

(C) 8 cm^{2}

(D) 15/2 cm^{2}

**Answer**

D

**Question. The point P lying in the fourth quadrant which is at a distance of 4 units from X-axis and 3 units from Y-axis is__________. **

(A) (4,-3)

(B) (4,3)

(C) (3,-4)

(D) (-3,4)

**Answer**

C

**Question. In what ratio does the line 4x+3y -13 = 0 divide the line segment joining the points (2, 1) and (1, 4)? **

(A) 3 : 2 internally

(B) 2 : 3 externally

(C) 2 : 3 internally

(D) 3 : 2 externally

**Answer**

C

**Question. The points on X-axis which are at a distance of 13 units from (-2,3) is ________. **

(A) (0,0), (-2,-3)

(B) (0,0), (-4,0)

(C) (0,0), (2,3)

(D) none of these

**Answer**

B

**Question. If (5, 3), (4, 2) and (1, –2) are the mid points of sides of triangle ABC, then the area of ΔABCis **

(A) 2 sq. units

(B) 3 sq. units

(C) 1 sq. units

(D) 4 sq. units

**Answer**

A

**Question. The points (a,b +c), (b,c +a) and (c,a + b) **

(A) are collinear

(B) form a scalene triangle

(C) form an equilateral triangle

(D) none of these

**Answer**

A

**Coordinate Geometry**

A point in 2-D plane (let’s say XY plane) is represented by its coordinates (x,y)

Here x is the distance of a point from the x-axis and is called y-coordinate.

y is the distance of a point from the and is called its .

**Distance Formula :** the distance between the points P(x_{1},y_{1}) and Q(x_{2},y_{2}) is PQ=

This is called distance formula. F

**Area of a triangle :** the area of ABC is the numerical value of the expression F

Section Formula : the coordinates of the point P(x,y) which divides the line segment joining the points A(x_{1},y_{1}) and B(x_{2},y_{2}) internally, in the ratio m_{1}:m_{2 }are

This is known as section formula. F

Special Case -: if P is mid point then m_{1}:m_{2 }=1:1

Now, the coordinates of mid point

**In order to prove that a given figure is a :**

i) Square, prove that four sides are equal and the diagonals are equal.

ii) Rhombus, prove that the four sides are equal.

iii) Rectangle, prove the opposite sides are equal and the diagonals are also equal.

iv) Parallelogram, prove that the opposite sides are equal.

The point of intersection of the medians of a triangle is called its called its centroid. The coordinates of the centroid of the triangle whose are (x_{1},y_{1}) ,(x_{2},y_{2}) and (x_{3},y_{3}) are given by

**For three points to be collinear,**

i) The sum of the distances between two pairs of points is equal to the distance third pair of points.

Three given points A(x_{1},y_{1}) ,B(x_{2},y_{2}) and C(x_{3},y_{3}) are collinear if ABC =0.

**Question. Find the condition that the point (x, y) may lie on the line joining (3, 4) and (-5, – 6). ****Solution.** Since the point P (x, y) lies on the line joining A (3, 4) and B (-5, -6). Therefore, P, A andB are collinear points.

Hence, the point (x, y) lies on the line joining (3,4) and (-5, -6), if 5x – 4y+1 = 0

**Question. Show that the points A(3, 5), B(6, 0), C(1, -3) and D (-2, 2) are the vertices of a square ABCD. ****Solution.** Let A( 3,5), B(6, 0), C(1, -3 ) and D(-2, 2) be the angular points of a quadrilateral ABCD. Join AC and BD

Thus, ABCD is a quadrilateral in which all sides are equal and the diagonals are equal.

Hence, quad. ABCD is a square.

**Question. If P (x, y) is any point on the line joining the points A(a,0) and B(0, b), then show that x/a + y/b = 1 ****Solution.** It is given that the point P (x, y) lies on the line segment joining points A (a, 0) and B (0,

b).Therefore, points P (x, y), A (a, 0) and B (0, b) are collinear points.

**Question. A (4, 2), B (6, 5) and C (1, 4) are the vertices of ΔABC. Solution.**

**i. The median from A meets BC in D. Find the coordinates of the point D.**

i. Median AD of the triangle will divide the side BC in two equal parts. So D is the midpoint of side BC.

**ii. Find the coordinates of point P on AD such that AP : PD = 2:1.**

**iii. Find the coordinates of the points Q and R on medians BE and CP respectively such that BQ : QE = 2 :1 and CR: RF =2: 1.**

Median BE of the triangle will divide the side AC in two equal parts. So E is the midpoint of side AC.

**iv. What do you observe? **

Now we may observe that coordinates of point P, Q are same. So, all these are representing same point on the plane i.e. centroid of the triangle.

**Question. Find the lengths of the medians of a ΓABC whose vertices are A(0, -1) B(2, 1) and C(0,3). ****Solution.** Let D, E, F be the midpoint of the side BC, CA and AB respectively in ΔABC

**Question. The area of triangle formed by the points (p, 2 – 2p), (1, p, 2 p ) and (-4 -p, 6 – 2p) is 70 sq. units. How many integral values of p are possible. Solution.**

**Question. Find the coordinates of the centroid of a triangle whose vertices are (0,6), (8,12) and (8,0). ****Solution.** Coordinates of the centroid of a triangle whose vertices are (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}) are

**Question. Point A is on x-axis, point B is on y-axis and the point P lies on line segment AB, such that P (4, – 5) and AP : PB = 5 : 3. Find the coordinates of point A and B. ****Solution.** Let coordinates of A are (x, 0) and coordinates of B are (0, y)

**Question. If the points A(x, 2), B(- 3, – 4), C(7, – 5) are collinear, then find the value of x. ****Solution.** Since the points are collinear, then,

Area of triangle = 0

**Practice Exercise :**

**Question. Prove that(4,-1),(6,0),(7,2)&(5,1) are the vertices of a rhombus .Is it a square?****Solution.** Proof

**Question. Find the coordinates of a point A, where AB is diameter of a circle whose Centre is(2,-3)and B is(1,4)****Solution.** (3,-10)

**Question. Find the centroid of triangle whose vertices are (3,-7), (-8,6) and (5,10).****Solution.** (0,3)

**Question. If A (-2,4),B(0,0),C(4,2) are the vertices of a ΔABC, then find the length of median through the vertex A.****Solution.** 5 units

**Question. Find the area of the triangle formed by joining the mid points of the sides of the triangle whose vertices are (0,-1),(2,1) and (0,3).Find the ratio of this area to the area of the given triangle.****Solution.** 1:4

**Question. Determine the ratio in which the point P(a,-2) divides the line joining of points(-4,3) and B(2,-4).Also find the value of a.****Solution.** a=2/7

**Question. If the point C(-1,2) divides internally the line segment joining A(2,5)and in the ratio 3:4. Find the Co- ordinates of B.****Solution.** B(-5,-2)

**Question. Show that points (1,1), (4,4) ,(4,8)and (1,5) are the vertices of a parallelogram.****Solution.** Proof

**Question. Find the value of p, for which the points(-1,3),(2,p) & (5,-1)are collinear****Solution.** p=1

**Question. If the points(-1,3),(1,-1)and(5,1)are the vertices of a triangle .Find the length of the median through the first vertex.****Solution.** 5

**Question. Find the center of a circle passing through the points (6,-6), (3,-7)and(3,3).****Solution.** (3,-2)

**Question. If the distance between the points (3,0) and (0,y)is 5 units and y is positive ,what is the value of y?****Solution.** 4

**Question. If the points(x ,y),(-5,-2)and(3,-5)are collinear r, then prove that 3x+8y+31=0.****Solution.** Proof

**Question. A line intersects y–axis and x-axis at the points P and Q respectively. If (2,-5) is the midpoint of PQ, then find the coordinates of P and Q respectively.****Solution.** (0,-10) and(4,0)

**Question. Show that the three points (a,a) ,(-a,-a) &(-a√3 , a√3) are the vertices of an equilateral triangle.****Solution.** Proof

**Question. Find the value of k, if the point P (2, 4) is equidistant from the points (5, k) and (k, 7).****Solution. **K=3

**Question. If the point A(0,2)is equidistant from the points B (3,p) and C(p,5),find p .Also find the length of AB.****Solution.** P=1, AB=√10

**Question. Find the distance between the points A and B in the following : A(a, 0), B(0, a) ****Solution.** A(a, 0), B(0, a)

**Question. Find the perpendicular distance of A(5,12) from the y-axis. ****Solution.** The point on the y-axis is (0,12)

Distance between (5,12) and (0,12)

**Question. Find the distance of the point (- 4, – 7) from the y-axis. ****Solution.** Points are (- 4, – 7) and (0, – 7)

**Question. Find the distance between the points: A(-6, -4) and B(9, -12) ****Solution.** The given points are A(-6, -4) and B(9, -12)

Then, (x_{1} = -6, y_{1} = -4) and (x_{2} = 9, y_{2} = -12)

**Question. Show that four points (0, -1), (6, 7), (-2, 3) and (8, 3) are the vertices of a rectangle. Also,find its area. Solution.**Let A (0 – 1), B (6, 7), C (-2, 3) and D (8, 3) be the given points. Then,

**Question. Find the co-ordinates of the points of trisection of the line segment joining the points A(1,- 2) and B(- ,4). Solution.**

**Question. Find the distance between the points A and B in the following:A(1,-3), B(4, 1) ****Solution.** A(1, -3), B(4, 1)

**Question. If the points A (a, -11), B (5, b), C (2, 15) and D (1, 1) are the vertices of a parallelogram ABCD, find the values of a and b. **

**Solution. **Let A(a, -11), B(5, b), C(2, 15) and D(1, 1) be the given points.

We know that diagonals of parallelogram bisect each other.

Therefore, Coordinates of mid-point of AC = Coordinates of mid-point of BD

Hence value of a and b is equal to 4 and 3 respectively.

**Question. Find the coordinates of the point , where the line x – y = 5 cuts Y-axis. ****Solution.** x – y = 5 is a given line

x – y = 5 cuts Y-axis.

Put x = 0 in the equation of line x- y = 5

⇒ (0) – y = 5

⇒ y = – 5

Therefore , the point is (0,-5)cuts x – y = 5 at Y-axis..

**Question. If the point A(2, –4) is equidistant from P(3, 8) and Q(–10, y) then find the values of y. Also find distance PQ. ****Solution.** According to the question,we are given that,

PA = QA

⇒ PA^{2} = QA^{2}

⇒ (3 – 2)^{2} + (8 + 4)2 = (–10 – 2)^{2} + (y + 4)^{2}

⇒ 12 + 12^{2} = (–12)^{2} + y^{2} + 16 + 8y

⇒ y^{2} + 8y + 16 – 1 = 0

⇒ y^{2} + 8y + 15 = 0

⇒ y^{2} + 5y + 3y + 15 = 0

⇒ y(y + 5) + 3(y + 5) = 0

⇒ (y + 5) (y + 3) = 0

⇒ y + 5 = 0 or y + 3 = 0

⇒ y = –5 or y = –3

So, the co–ordinates are P(3, 8), Q1(–10, –3), Q2(–10, –5).

**Question. Find the value of ‘k’ if the points (7, –2), (5, 1), (3, k) are collinear. ****Solution.** (7, –2), (5, 1), (3, k)

**Question. Find the perimeter of a triangle with vertices (0, 4), (0,0) and (3,0). ****Solution.** Here, A→ (0,4),B→ (0,0),C→ (3,0)

**Question. Find the area of a quadrilateral PQRS whose vertices area P(- 5, 7), Q(- 4, – 5), R (-1, – 6) and S(4, 5). ****Solution. **Area PQRS = ar PQS + ar QRS

Hence, area PQRS = 53 + 19 = 72 sq. units

**Question. The point R divides the line segment AB where A(-4, 0), B(0, 6) are such that AR =3/4 A B. Find the coordinates of R. **

**Solution.** ** **

**Question. Prove that the lines joining the middle points of the opposite sides of a quadrilateral and the join of the middle points of its diagonals meet in a point and bisect one another. ****Solution.** Let OBCD be the quadrilateral P, Q, R, S be the mid-points of OB, CD, OD and BC.

Let the coordinates of O,B, C, D are (0, 0), (x, 0), (x, y) and (0, y)

Coordinates of P are (x/2 ,0)

Coordinates of Q are (x/2 , y)

Coordinates of R are (0,y/2)

Coordinates of S are (x , y/2)

Coordinates of mid-point of PQ are

Since, the coordinates of the mid-point of PQ = coordinates of mid-point of RS.

PQ and RS bisect each other.

**Question. In the given triangle ABC as shown in diagram D, E and F are the mid-points of AB, BC and AC respectively. Find the area of ΔDEF. ****Solution.**

**Question. Find the value of m for which the points with coordinates (3, 5), (m, 6) and (1/2 , 15/2) are collinear. ****Solution.** If points are collinear, then one point divides the other two in the same ratio.

Let point (m, 6) divides the join of (3, 5) and (1/2 , 15/2) in the ratio k: 1.

**Question. If origin is the mid-point of the line segment joined by the points (2, 3) and (x, y) then find the value of (x, y). ****Solution.**

**Question. Find the centroid of the triangle whose vertices are given below: (3, -5), (-7, 4), (10,-2). ****Solution.** The given vertices of triangle are (3, -5), (-7, 4) and (10, -2).

Let (x, y) be the coordinates of the centroid. Then

**Question. Find the number of points on x-axis which are at a distance of 2 units from (2, 4). ****Solution.** Distance of the point (2, 4) from x-axis is 4 units. There is no point on x-axis which is at a distance of 2 units from the given point.

**Question. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: ****(-1, -2), (1, 0), (-1, 2), (-3, 0) (4)****Solution.** (-1, -2), (1, 0), (-1, 2), (-3, 0)

Let A→ (-1, -2), B→ (1, 0)

C→ (-1, 2) and D→ (-3, 0)

Since AB = BC = CD = DA (i.e., all the four sides of the quadrilateral ABCD are equal) and AC = BD (i.e. diagonals of the quadrilateral ABCD are equal). Therefore, ABCD is a square.

**Question. Find the area of the triangle with vertices (0 ,0) (6 ,0) and (0 ,5). ****Solution.** We have to find the area of the triangle with vertices (0 ,0) (6 ,0) and (0 ,5).

Area of triangle

**Question. Prove that the coordinates of the centroid of a triangle ABC, with vertices A(x _{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) are given by (x_{1} + x_{2} + x_{3} /3 , y_{1} + y_{2} + y_{3} /3) 12**

**Solution.**Let the coordinates of vertices of ΔABC be A(x

_{1}, y

_{1}), B(x

_{2}, y

_{2}) and C(x

_{3}, y

_{3}) respectively. Let D be the midpoint of BC.

Using section formula, coordinates of D will be

Now since centroid G will divide the line joining A and D in the ratio of 2 : 1, therefore again using section formula, coordinates of G will be

**Question. The centre of a circle is (2a, a -7). Find the values of a, if the circle passes through the point (11, -9) and has diameter 10√2 units. ****Solution.** Diameter of a circle = 10√2 units

⇒ Radius of a circle = 5√2 units

Let the centre of a circle be O(2a, a – 7) which passes through the point P(11, -9).

⇒ OP is the radius of the circle.

⇒ OP = 5√2 units

⇒ OP^{2} = (5√2)^{2}

⇒ (11 – 2a)^{2} + (-9 – a + 7)^{2} = 50

⇒ 121 + 4a^{2} – 44a + (-2 – a)^{2} = 50

⇒ 121 + 4a2 – 44a + 4 + a^{2} + 4a = 50

⇒ 5a^{2} – 40a + 125 = 50

⇒ 5a^{2} – 40a + 75 = 0

⇒ a^{2} – 8a + 15 = 0

⇒ a2 – 5a – 3a + 15 = 0

⇒ a(a – 5) -3(a – 5) =0

⇒ (a – 5)(a – 3) = 0

⇒ a – 5 = 0 or a – 3 = 0

⇒ a = 5 or a = 3

**Question. Find the area of the rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order. ****Solution.** Let A (3, 0), B (4, 5), C (-1, 4) and D (-2, -1)

**Question. If the points P (-3, 9), Q (a, b) and R (4, -5) are collinear and a + b = 1, find the values of a and b. ****Solution.** It is given that the points P (-3, 9), Q (a, b) and R(4,-5) are collinear.

**Question. Find the area of a triangle ABC with A(1, – 4) and mid-points of sides through A being (2, -1) and (0, -1). 1Solution.**

**Question. If the points (0, 0), (1, 2) and (x, y) are collinear,then find x. ****Solution.** The points are collinear, then area of triangle = 0

**Question. Find the lengths of the medians of a ΔABC having vertices at A (0, -1), B (2, 1) and C (0, 3). Solution.**

Let A(0, -1), B(2, 1) and C(0, 3) be the given points.

**Question. Find the distance of the point (α ,β) from y-axis. ****Solution.** Distance of the point (α ,β) from y-axis is the positive value of its x-coordinate.

∴ Distance = lαl

**Question. If the centre of circle is (2a, a – 7) then find the values of a if the circle passes through the point (11, –9) and has diameter 10√2 units. Solution.**

Let C(2a, a – 7) be the centre of the circle and it passes through the point P(11, –9).

∴ PQ = 10√2

⇒ CP = 5√2

⇒ CP2 = (5√2)^{2} = 50

⇒ (2a – 11)2 + (a – 7 + 9)2 = 50

⇒ (2a)^{2} + (11)^{2} – 2(2a) (11) + (a + 2)^{2} = 50

⇒ 4a^{2} + 121 – 44a + (a)^{2} + (2)^{2} + 2(a)(2) = 50

⇒ 5a^{2} – 40a + 125 = 50

⇒ a^{2} – 8a + 25 = 10

⇒ a^{2} – 8a + 25 – 10 = 0

⇒ a^{2} – 8a + 15 = 0

⇒ a^{2} – 5a – 3a + 15 = 0

⇒ a(a – 5) –3(a – 5) = 0

⇒ (a – 5) (a – 3) = 0

⇒ a – 5 = 0 or a – 3 = 0

⇒ a = 5 or a = 3

Hence, the required values of a are 5 and 3.

**Question. If the centre and radius of circle is (3, 4) and 7 units respectively, then what is the position of the point A(5,8) with respect to circle? ****Solution.** Distance of the point, from the centre

**Question. Find the coordinates of the centre of the circle passing through the points (0, 0), (-2, 1) and (-3, 2). Also, find its radius. **

**Solution. **Let P (x, y) be the centre of the circle passing through the points O (0, 0), A (-2, 1) and B (-3, 2). Then,

**Question. If the coordinates of the mid-points of the sides of a triangle are (1, 1), (2, -3) and (3,4), find the vertices of the triangle. ****Solution. **Let A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) be the vertices of

Let D(1, 1), E(2, -3) and F(3, 4) be the mid-points of sides BC, CA and AB respectively.

Since, D is the mid-point of BC

Similarly E and F are the mid-points of CA and AB respectively.

⇒ x1 + x_{2} = 6 and y_{1} + y_{2} = 8 …(iii)

From (i), (ii) and (iii) we get

x_{2} + x_{3} + x_{1} + x_{3} + x_{1} + x_{2} = 2 + 4 + 6

and, y_{2} + y_{3} + y1 + y_{3} + y_{1} + y_{2} = 2 +(-6) + 8

⇒ 2(x_{1} + x_{2} + x_{3}) = 12 and 2(y_{1} + y_{2} + y_{3}) = 4

⇒ x_{1} + x_{2} + x_{3} = 6 and y_{1} + y_{2} + y_{3} = 2 …(iv)

From (i) and (iv) we get

x_{1} + 2 = 6 and y_{1} + 2 = 2

⇒ x_{1} = 6 – 2 y_{1}= 2 – 2

⇒ x_{1} = 4 y_{1} = 0

So the coordinates of A are (4, 0)

From (ii) and (iv) we get

x_{2} + 4 = 6 and y_{2} + (-6) = 2

⇒ x_{2} = 2 ⇒ y_{2} – 6 = 2

⇒ y_{2} = 8

So the coordinates of B are (2, 8)

From (iii) and (iv) we get

6 + x_{3} = 6 and 8 + y_{3} = 2

⇒ x_{3} = 6 – 6 ⇒ y_{3} = 2 – 8

⇒ x_{3} = 0 ⇒ y_{3} = -6

So the coordinates of c are (0, -6)

Hence, the vertices of triangle ABC are:

A(4, 0), B(2, 8) and C(0, -6)

**Question. Find the distance between the points P (-4, 7) and Q(2, -5). ****Solution.** The given points are P (-4, 7) and Q(2, -5).

Then, x_{1} = -4, y_{1} = 7 and x_{2} = 2, y_{2} = -5.

**Question. Find the point on the X-axis which is equidistant from the points (-1,0) and (5,0) ****Solution.** Let A(x,o) be any point on the X-axis , which is equidistant from points (-1,0) and (5,0).

⇒ (x + 1)^{2} = (x – 5)^{2}

A(x,o) be any point on the X-axis , which is equidistant from points (-1,0) and (5,0).

⇒ x^{2} + 2x + 1 = x^{2} – 10x + 25

⇒ 2x + 1 = -10x + 25

⇒ 2x + 10x = 25 – 1

⇒ 12x = 24

**Question. Find value(s) of y for which the distance between the points P (2, –3) and Q (10, y) is 10 units.****Solution.** Let P (2, –3) and Q (10, y) be the two given points such that PQ = 10.

∴ PQ = √(10 – 2)^{2} + (y + 3)^{2} = 10 (given)

⇒ PQ^{2} = 64 + y^{2} + 6 y + 9 = 100

⇒ y^{2} + 6y – 27 = 0 ⇒ y2 + 9y – 3y – 27 = 0

⇒ y (y + 9) – 3 (y + 9) = 0 ⇒ (y + 9) (y – 3) = 0

⇒ y = – 9 or y = 3 Ans.

**Question. Find the ratio in which the join of (–3, 10) and (6, –8) is divided by (–1, 6). ****Solution.** Let the given points be A(–3, 10), B(6, –8) and C(–1, 6).

**Question. Prove that the points (2, –2), (–3, 8) and (–1, 4) are collinear. ****Solution.** Let Δ be the area of the triangle formed by given three points A(2, –2), B(–3, 8) and C(–1, 4).

**Question. Show that the points (1, –1), (5, 2) and (9, 5) are collinear. ****Solution.** Let A(1, –1), B(5, 2) and C(9, 5) be the given points.

**Question. Find the coordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, –2) and (2, –2). Also, find its circum-radius.****Solution.** We know circumcentre of a triangle is equidistant from the vertices of a triangle.

Let A(8, 6), B(8, –2) and C(2, –2) be the vertices of a given triangle and let P(x, y) be the circumcentre of this triangle. Then,

PA = PB = PC

⇒ PA^{2} = PB^{2} = PC^{2}

Now, PA^{2} = PB^{2}

⇒ (x– 8)^{2} + (y – 6)^{2} = (x – 8)2 + (y + 2)^{2}

⇒ x^{2} + y^{2} – 16x – 12 y + 100 = x^{2} + y^{2} – 16x + 4y + 68

⇒ 16y = 32 ⇒ y = 2.

and PB^{2} = PC^{2}

⇒ (x – 8)^{2} + (y + 2)^{2} = (x – 2)^{2} + (y + 2)^{2}

⇒ x^{2} + y^{2} – 16x + 4y + 68 = x^{2}+ y^{2} – 4x + 4y + 8

⇒ 12x = 60 ⇒ x = 5

So, the coordinates of the circumcentre P are (5, 2).

Also, circum-radius = PA = PB = PC

√(5 – 8)^{2} + (2 – 6)^{2} = √(-3)^{2} + (-4)^{2}

√9 + 16 = √25 = 5 units Ans.

**Question. Find the distance between the points : ****(i) A (0, 0), B (–5, 12)(ii) A (5, –8), B(–7, –3)**

**(iii) P (cos α, –sin α), Q( – cos α, sin α)**

**(iv) P (a + b, a – b), Q (a – b, a + b)**

**Solution.**(i) AB = √(-5 – 0)2 + (12 – 0)

^{2}

**Question. Find the area of the quadrilateral whose vertices taken in order are (–4, –2), (–3, –5), (3, –2), (2, 3). ****Solution.** Let A(–4, –2), B(–3, –5), C(3, –2) and D (2, 3) be the given points.

Now, area of quad. ABCD = ar (ΔABD) + ar (ΔBCD)

For ar (ΔABD) : A(–4, –2), B(–3, –5), D(2, 3)

**Question. Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle. ****Solution.** Let A(5, –2), B(6, 4) and C(7, –2) be the vertices of a ΔABC.

AB = √(6 – 5)2 + (4 + 2)2 = √12 + 62 = √ 1 + 36 = √37

**Question. If A(5, –1), B(–3, –2) and C(–1, 8) are the vertices of ΔABC, find the length of medians through A and the coordinates of the centroid. ****Solution.** Let AD be the median through the vertex A of ΔABC. Then, D is the mid-point of BC. So, the coordinates of D are (-3/2 , -2 + 8/2) i.e. (–2, 3).

**Question. Show that (1, –2), (3, 0), (1, 2) and (–1, 0) are the vertices of a square. ****Solution.** Let A(1, –2), B(3, 0), C(1, 2) and D(–1, 0) be the given points.

**Question. Find the area of the triangle whose vertices are (–5, –1), (3, –5), (5, 2). ****Solution.** Let A(–5, –1), B(3, –5), C(5, 2) be the given points.

**Question. Find the co-ordinates of the points of trisection of the line joining the points (4, –1) and (–2, –3). ****Solution.** Let C and D be the points of trisection of AB

## Assignments for Class 10 Mathematics Coordinate Geometry as per CBSE NCERT pattern

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