JKBOSE 10th Class Maths Solutions chapter – 7 Coordinate Geometry

In class IX, you have studied that to locate the position of a point on a plane, we require a pair of coordinate axes. The distance of a point from the y-axis is called its x-coordinate, or abscissa. The distance of a point from the x-axis is called its y-coordinate, or ordinate. The coordinates of a point on the x-axis are of the form (x, 0) and of a point on the y-axis are of the form (0, y).

Also, you have seen that a linear equation in two variables of the form ax + by + c = 0, a, b are not simultaneously zero, when represented graphically, gives a straight line. Further in Chapter 4, you have seen the graph of y = ax² + bx + c (a ‡ 0), is a parabola. In fact, coordinate geometry has been developed as an algebraic tool for studying geometry of figures. It helps us to study geometry using algebra, and understand algebra with the help of geometry. Because of this, coordinate geometry is widely applied in various fields such as physics, engineering, navigation, seismology and art.

Solution. (i) Given points are: (2, 3); (4,1)

Solution. Given points are: A (0, 0) and B (36, 15)

Draw the distinct points A (0, 0) and B (36, 15) as shown in figure.

Hence, required distance between points is 39. Ans.

Solution. Given point are: A (1, 5); B (2, 3) and (-2,- 11).

From above distances, it is clear that sum of any two is not equal to third one.

Hence, given points are not collinear. Ans.

Solution. Given points be A (5,-2); B (6, 4) and C (7,-2).

∴ Given points are vertices of an isosceles triangle. Ans.

Solution. In the given diagram, the vertices of given points are: A (3, 4); B (6, 7); C (9,4) and D (6, 1).

From above discussion, it is clear that

AB = BC = CD = DA = √18 and  AC = BD = 6.

∴ ABCD formed a square and Champa is correct about her thinking. Ans.

Solution. (i) Given points be A (-1,-2); B (1, 0); C (- 1, 2)

and D (-3, 0).

           

From above discussion, it is clear that

(ii) Given points be A (-3, 5); B (3, 1); C (0, 3) and D (-1, – 4)

∴  A, B and C are collinear then A, B, C and D do not form any quadrilateral. Ans.

(iii) Given points are A (4, 5); B (7, 6); C (4, 3) and D (1, 2)

From above discussion, it is clear that

AB = CD and BC = DA and AC ≠ BD.

i.e., opposite sides are equal but their diagonals are not equal.

Hence, given quadrilateral ABCD is a parallelogram. Ans.

Solution. Let required point be P(x, 0)

and given points be A (2,-5) and B (-2,9).

According to question,

Hence, required point be (-7,0) Ans.

Solution. Given points are P (2, -3) and Q (10, y)

Hence,         y = – 9 and 3 Ans.

Solution. Given points Q (0, 1); P (5, -3) and R (x, 6)

According to question,

QP = QR

Solution. Let required points be P (x, y) and given points are A (3, 6) and B (-3, 4)

According to question,

PA = PB

or  3x + y – 5 = 0 is the required relation. Ans.

Solution. Let required point P (x, y) which divides the join of given points A (-1, 7) and B (4,3) in the ratio of 2 : 3.

Hence, required point be (1, 3). Ans.

Solution. Let P (x₁, y₁) and Q (x₂, y₂) be the required points which trisected the line segment joining A (4, -1) and B (– 2, − 3) i.e., P (x₁, y₁) divides AB in ratio 1 : 2 and Q divides AB in ratio 2 : 1.

Solution. In the given figure, we take A as origin.

Taking x-axis along AB and y-axis along AD.

Position of green flag

= distance covered by Niharika

= Niharika runs ¼th distance AD on the 2nd line

= ¼ x 100 = 25 m

∴ Co-ordinates of the green flag are (2, 25) Now, position of red flag

= distance covered by Preet

= Preet runs 1/8 th the distance AD on the 8th line

= 1/5 x 100 = 20 m.

∴ distance between Green and Red flags

Position of blue flag

= mid point of green flag and red flag

Hence, blue flag is in the 5th line and at a distance of 22.5 m along AD. Ans.

Solution. Let point P (- 1, 6) divides the line segment joining the points A (-3, 10) and B (6,- 8) in the ratio K:1

Hence, required ratio is 2: 7. Ans.

Solution. Let required point on x-axis is P (x, 0) which divides the line segment joining the points A (1,- 5) and B (- 4, 5) in the ratio K: 1.

Solution. Let points of parallelogram ABCD are A (1, 2); B (4, y) ; C (x, 6) and D (3, 5)

But diagonals of a || gm bisect each other.

Case I. When E is the mid point of A (1, 2) and C (x, 6)

∴ Co-ordinates of E are:

Case II. When E is the mid point B (4, y) and D (3, 5)

∴ Co-ordinates of E are:

But values of E in (1) and (2) are same, so comparing the coordinates, we get

Hence, values of x and y are 6 and 3. Ans. 

Solution. Let, co-ordinates of A be (x, y) But, centre is the mid point of the vertices of the diameter.

Hence, required point A be (3, 10) Ans.

Solution. Let required point P be (x, y)

Solution. Let required points are C, D and E which divide the line segment joining the points A (-2, 2) and B (2, 8) into four equal parts.

Then D is mid point of A and B ; C is the mid point of A and D; E is the mid point of D and B such that

AC = CD= DE = EB

Now, mid point of A and B

(i.e., Coordinates of D)

[Hint: Areas of a rhombus = 1/2 (Product of its diagonals)]

Solution. Let co-ordinates of rhombus ABCD are A (3, 0); B (4, 5); C (-1, 4) and D (-2-1).

Hence, area of rhombus is 24 sq. units. Ans.

Solution. (i) Let vertices of the AABC are A (2, 3); B (-1, 0) and C (2,-4)

Here  x₁ = 2, x₂ = -1, x₃ = 2

y₁ = 3, y₂ = 0, y₃ = -4

(ii) Let vertices of the AABC are A (-5, -1); B (3,-5) and C (5,2)

Here x₁ = -5, x₂ = 3, x₃ = 5

y₁ =-1, y₂ = -5, y₃ = 2

(i) Let given points be A (7, – 2); B (5, 1) and C (3, k)

Here x₁ = 7, x₂ = 5, x₃ = 3

y₁ = -2, y₂ = 1, y₃ = k

Three points are collinear iff

Hence k = 4 Ans.

(ii) Let given points be A (8, 1); B (k, – 4) and C (2,-5)

Here x₁ = 8, x₂ = k, x₃ = 2

y₁ = 1, y₂ = 4, y₃ = -5

Three points are collinear iff

Hence k = 3 Ans.

Let vertices of given triangle ABC are A(0, 1); B (2, 1) and C (0, 3).

Also, D, E, F be the mid points of AB, BC, CA respectively.

Using mid point formula,

∴ Co-ordinates of the vertices of ΔDEF are D (1, 0); E (1, 2); F (0,1).

Here x₁ = 1, x₂ = 1, x₃ = 0

y₁ = 0, y₂ = 2, y₃ = 1

Let co-ordinates of the given quadrilateral ABCD are A (-4,-2); B (-3,-5); C (3,-2) and D (2, 3).

Join AC then Quad. ABCD divides in two triangles

ie. ΔABC and ACDA

Solution. Given that co-ordinates of the vertices of AABC are A(4,-6); B (3,-2) and C (5,2).

Let CD is the median i.e. D is the mid point of AB which divides AABC into two parts i.e.

From above discussion it is clear that area of ΔADC = area of ΔCDB = 3 sq. units.

Hence, a median of a triangle divides it into two triangles of equal areas.

Let line 2x + y -4 = 0 divides the line segment joining the points A (2,-2) and B(3, 7) at C (x, y) in the ratio k: 1

Hence required ratio is 2 : 9 Ans.

Solution. Let given points are A (x, y); B (1, 2) and C (7,0).

or x + 3y – 7 = 0 is the required relation. Ans.

Solution. Let O (x, y) be the required centre of the circle which passes through points P (6, 6); Q (3, -7) and R (3, 3).

∴ radii of circle are equal.

∴ OP = OQ = OR

or (OP)² = (OQ)² = (OR)²

Now, (OP)² = (0Q)²

(x-6)² + (y + 6)² = (x-3)² + (y+7)²

or x² + 36 – 12x + y² + 36 + 12y

= x² + 9 – 6x + y² + 49 + 14y

or       – 12x + 12y + 72 = − 6x + 14y + 58

or       – 6x – 2y + 14 = 0

or      3x + y – 7 = 0                …(1)

Also, (OQ)² = (OR)²

or   (x-3)² + (y + 7)² = (x-3)² + (y – 3)²

or   (y+7)² = (y – 3)²

or    y² + 49 + 14y = y² + 9 – 6y

or    20y = – 40

Substitute this value of y in (1), we get

3x – 2 – 7 = 0

or      3x – 9 = 0

or      3x = 9

∴ Required centre is (3,-2) Ans.

Solution. Let two opposite vertices of a square ACBD are A (-1, 2) and B (3, 2) and coordinates of Care (x, y)

∵ Length of each sides of square are equal.

∴ AC = BC

or (AC)² = (BC)²

or (x + 1)² + (y-2)² = (x-3)² + (y-2)²

Now, in rt ∠d ΔACB,

Using Pythagoras Theorem,

(AC)² + (BC)² = (AC)²

(x + 1)² + (y-2)² + (x − 3)² + (y-2)² = (3+1)² + (2-2)²

or x² + 1 + 2x + y² + 4 – 4y + x² + 9 – 6x + y² + 4 – 4y = 16

or 2x² + 2y² – 4x – 8y + 2 = 0

or x² + y² – 2x – 4y + 1 = 0

Putting the value of x = 1 from (1), we get

(1)² + y² – 2 (1) – 4y + 1 = 0

or y² – 4y = 0

or y (y – 4) = 0

Either y = 0 or   y – 4 = 0

Either y = 0 or   y = 4

∴ y = 0, 4

∴ Required points are (1, 0) and (4, 0). Ans.

When taking A as origin then AD is X-axis and AB is Y-axis.

∴ Coordinates of triangular grassy Lawn PQR are P (4, 6); Q (3, 2) and R( 6, 5).

Here x₁ = 4, x₂ = 3, x₃ = 6

y₁ = 6, y₂ = 2, y₃ = 5

When taking C as origin then CB is X-axis and CD is Y-axis.

∴ Coordinates of triangular grassy lawn PQR are P (12, 2); Q (13, 6) and R (10, 3)

Here x₁ = 12, x₂ = 13, x₃ = 10

y₁ = 2, y₂ = 6, y₃ = 3

From above two cases, it is clear that area of triangular grassy lawn is same. Ans.

Solution. The vertices of ΔABC are

A (4, 6); B (1, 5) and C (7, 2)

A line is drawn to intersect sides AB and AC at D (x₁, y₁) and E (x₂, y₂) respectively such that

∴ D and E divides AB and AC in the ratio 1:3

∴ Coordinates of D are

Solution. Given that vertices of ΔABC are A (4, 2); B (6, 5) and C (1, 4).

(i) AD is the median from the vertex A.

∴ D is the mid point of BC.

(ii) Let P (x, y) be point on AD such that AP : PD = 2:1

(iii) Let BE and CF are the medians of ΔABC to AC and AB respectively.

∴ E and F are mid points of AC and AB respectively.

(iv) From above discussion, it is clear that coordinates of P, Q and R are same and coincide at a point, is known as centroid of triangle, which divides each median in the ratio 2 : 1.

(v) The vertices of given ΔABC are

A (x_₁, y_₁); B (x₂, y₂) and C (x₂, Y₃).

Let AD is mediam of ΔABC.

∴ D is the mid point of BC then coordinates of D are

Now, G be the centroid of AABC, which divides the median AD in the ratio 2 : 1

∴ Coordinates of G are             [using (iv)]

Given. The vertices of given rectangle ABCD are

A(-1,-1); B (- 1, 4); C (5, 4) and D (5,- 1).

∵ P is the mid point of AB.

∴ Coordinates of P are

∵ Q is the mid point of BC.

∴ Co-ordinates of Q are

∵ R is the mid point of CD

∴  Coordinates R are

∵ S is the mid point of AD

∴ Coordinates of S are

From above discussion, it is clear that

PQ = RS and QR = SP

Also, PR ≠ QS

⇒ Opposite sides of Quad. PQRS are equal but their diagonals are not equal.