Brief Review Of Cartesian System Of Rectangular Co-ordinates

Key points to solve the problem:

• Idea of distance formula- Distance between two points P(x₁,y₁) and Q(x₂,y₂) is given by- PQ =

Given,

Two points P and Q subtends an angle α at the origin as shown in figure:

From figure we can see that points O,P and Q forms a triangle.

Clearly in ΔOPQ we have:

{from cosine formula in a triangle}

⇒ …..equation 1

From distance formula we have-

OP =

As, coordinates of O are (0, 0) ⇒ x₂ = 0 and y₂ = 0

Coordinates of P are (x₁, y₁) ⇒ x₁ = x₁ and y₁ = y₁

=

=

Similarly, OQ =

=

And, PQ =

∴ OP² + OQ² - PQ² =

⇒ OP² + OQ² - PQ² =

Using (a-b)² = a² + b² – 2ab

∴ OP² + OQ² - PQ² = 2x₁ x₂ + 2y₁ y₂ ….equation 2

From equation 1 and 2 we have:

⇒ …Proved.

Key points to solve the problem:

• Idea of distance formula- Distance between two points P(x₁,y₁) and Q(x₂,y₂) is given by- PQ =

Given,

Coordinates of triangle and we need to find cos B which can be easily found using cosine formula.

See the figure:

From cosine formula in ΔABC , We have:

cos B =

using distance formula we have:

AB =

BC =

And, AC =

∴ cos B =

Key points to solve the problem:

• Idea of distance formula- Distance between two points P(x₁,y₁) and Q(x₂,y₂) is given by- PQ =

Ar(ΔPQR) =

Given, coordinates of triangle as shown in figure.

Also,

ar(ΔDBC) =

=

Similarly, ar(ΔABC) =

=

∴

⇒ 24.5 = 28x – 14

⇒ 28x = 38.5

⇒ x = 38.5/28 = 1.375

Key points to solve the problem:

• Idea of distance formula- Distance between two points P(x₁,y₁) and Q(x₂,y₂) is given by- PQ =

• Idea of Rhombus – It is a quadrilateral with all four sides equal.

Given, coordinates of 4 points that form a quadrilateral as shown in fig:

Using distance formula, we have:

AB =

BC =

Clearly, AB ≠ BC ⇒ quad ABCD does not have all 4 sides equal.

∴ ABCD is not a Rhombus

Key points to solve the problem:

• Idea of distance formula- Distance between two points P(x₁,y₁) and Q(x₂,y₂) is given by- PQ =

• Incentre of a triangle - Let A(x_1,y₁) , B(x₂,y₂) and C(x₃,y₃) be the 3 vertices of ΔABC and O be the centre of circle inscribed in ΔABC

O = where a, b and c are length of sides opposite to ∠ A , ∠ B and ∠ C respectively.

Given, coordinates of vertices of triangle as shown in figure:

We need to find the coordinates of O:

Before that we have to find a ,b and c. We will use distance formula to find the same.

As, a = BC =

b = AC =

and c = AB =

∴ coordinates of O =

Key points to solve the problem:

• Idea of distance formula- Distance between two points P(x₁,y₁) and Q(x₂,y₂) is given by- PQ =

• Equilateral triangle- triangle with all 3 sides equal.

• Coordinates of midpoint of a line segment – Let P(x₁,y₁) and Q(x_2,y₂) be the end points of line segment PQ. Then coordinated of midpoint of PQ is given by –

Given, an equilateral triangle with base along y axis and midpoint at (0,0)

∴ coordinates of triangle will be A(0,y₁) B(0,y₂) and C(x,0)

As midpoint is at origin ⇒ y₁+y₂ = 0 ⇒ y₁ = -y₂ …..eqn 1

Also length of each side = 2a (given)

∴ AB = ….eqn 2

∴ from eqn 1 and 2:

y₁ = a and y₂ = -a

∴ 2 coordinates are – A(0,a) and B(0,-a)

See the figure:

Clearly from figure:

DC = x

Also in ΔADC: cos 30° =

∴

Squaring both sides:

∴

∴ Coordinates of C are (√3a,0) or (-√3a,0)

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

Given, P(x₁, y₁) and Q(x₂, y₂) are two points.

i) When PQ is parallel to y-axis

This implies that x – coordinate is constant ⇒ x₂ = x₁

∴ from distance formula:

PQ = = |y₂ - y₁|

ii) When PQ is parallel to x-axis

This implies that y – coordinate is constant ⇒ y₂ = y₁

∴ from distance formula:

PQ = = |x₂ - x₁|

Note: we take modulus because square root gives both positive and negative values but distance is always positive so we make it positive using modulus function.

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

As, the point is on x-axis so y-coordinate is 0.

Let the coordinate be (x,0)

Given distance of (x,0) from (7,6) and (3,4) is same.

∴ using distance formula we have:

∴ point on x-axis is (7.5,0)

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

How to approach: To find locus of a point we first assume the coordinate of point to be (h, k) and write a mathematical equation as per the conditions mentioned in question and finally replace (h, k) with (x, y) to get the locus of point.

Let the coordinates of point whose locus is to be determined be (h, k)

As we need to maintain a same distance of (h,k) from (2,4) and y-axis.

So we select point (0,k) on y-axis.

From distance formula:

Distance of (h,k) from (2,4) =

Distance of (h,k) from (0,k) =

According to question both distance are same.

∴

Squaring both sides:

⇒

⇒

Replace (h,k) with (x,y)

Thus, locus of point equidistant from (2,4) and y-axis is-

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

How to approach: To find locus of a point we first assume the coordinate of point to be (h, k) and write a mathematical equation as per the conditions mentioned in question and finally replace (h, k) with (x, y) to get the locus of point.

Let the point whose locus is to be determined be (h,k)

Distance of (h,k) from (2,0) =

Distance of (h,k) from (1,3) =

According to question:

Squaring both sides:

⇒

⇒

Replace (h,k) with (x,y)

Thus, the locus of a point which moves such that the ratio of its distance from (2, 0) and (1, 3) is 5 : 4 is –

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

How to approach: To find locus of a point we first assume the coordinate of point to be (h, k) and write a mathematical equation as per the conditions mentioned in question and finally replace (h, k) with (x, y) to get the locus of point.

Let the point whose locus is to be determined be (h,k)

Distance of (h,k) from (ae,0) =

Distance of (h,k) from (-ae,0) =

According to question:

⇒

Squaring both sides:

⇒

⇒

Again squaring both sides:

⇒

⇒

∴

⇒ where b² = a²(e² – 1)

Replace (h,k) with (x,y)

Thus, locus of a point such that difference of its distances from (ae, 0) and (-ae, 0) is 2a:

where b² = a²(e² – 1) ….proved

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

How to approach: To find locus of a point we first assume the coordinate of point to be (h, k) and write a mathematical equation as per the conditions mentioned in question and finally replace (h, k) with (x, y) to get the locus of point.

Let the point whose locus is to be determined be (h,k)

Distance of (h,k) from (0,2) =

Distance of (h,k) from (0,-2) =

According to question:

⇒

Squaring both sides:

⇒

Again squaring both sides:

⇒

⇒

Replace (h,k) with (x,y)

Thus, locus of a point such that sum of its distances from (0,2) and (0,-2) is 6:

….proved

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

How to approach: To find locus of a point we first assume the coordinate of point to be (h, k) and write a mathematical equation as per the conditions mentioned in question and finally replace (h, k) with (x, y) to get the locus of point.

Let the coordinates of point whose locus is to be determined be (h, k)

As we need to maintain a same distance of (h,k) from (2,4) and x-axis.

So we select point (h,0) on x-axis.

From distance formula:

Distance of (h,k) from (1,3) =

Distance of (h,k) from (h,0) =

According to question both distance are same.

∴

Squaring both sides:

⇒

⇒

Replace (h,k) with (x,y)

Thus, locus of point equidistant from (1,3) and x-axis is-

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

How to approach: To find locus of a point we first assume the coordinate of point to be (h, k) and write a mathematical equation as per the conditions mentioned in question and finally replace (h, k) with (x, y) to get the locus of point.

Let the coordinates of point whose locus is to be determined be (h, k)

As we need to maintain a distance of (h,k) from origin such that it is 3 times the distance from x-axis.

So we select point (h,0) on x-axis.

From distance formula:

Distance of (h,k) from (0,0) =

Distance of (h,k) from (h,0) =

According to question both distance are same.

∴

Squaring both sides:

⇒

Replace (h,k) with (x,y)

Thus, locus of point is …

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

Area of a ΔPQR – Let P(x_1,y₁) , Q(x₂,y₂) and R(x₃,y₃) be the 3 vertices of ΔPQR.

Ar(ΔPQR) =

How to approach: To find locus of a point we first assume the coordinate of point to be (h, k) and write a mathematical equation as per the conditions mentioned in question and finally replace (h, k) with (x, y) to get the locus of point.

Let the coordinates of point whose locus is to be determined be (h, k). Name the moving point be C

Given area of ΔABC = 9

According to question:

9 =

⇒

⇒

∴

⇒

Replace (h,k) with (x,y)

Thus, locus of point is …

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

• Pythagoras theorem: In right triangle ΔABC : sum of the square of two sides is equal to square of its hypotenuse.

How to approach: To find locus of a point we first assume the coordinate of point to be (h, k) and write a mathematical equation as per the conditions mentioned in question and finally replace (h, k) with (x, y) to get the locus of point.

Let the coordinates of point whose locus is to be determined be (h, k) and name the moving point be C.

According to question on drawing the figure we get a right triangle Δ ABC.

From Pythagoras theorem we have:

BC² + AC² = AB²

From distance formula:

BC =

AC =

And AB = 4

∴

⇒

⇒

⇒

⇒

Replace (h,k) with (x,y)

Thus, locus of point is

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

Area of a ΔPQR – Let P(x_1,y₁) , Q(x₂,y₂) and R(x₃,y₃) be the 3 vertices of ΔPQR.

Ar(ΔPQR) =

How to approach: To find locus of a point we first assume the coordinate of point to be (h, k) and write a mathematical equation as per the conditions mentioned in question and finally replace (h, k) with (x, y) to get the locus of point.

Let the coordinates of point whose locus is to be determined be (h, k). Name the moving point be C

Given area of ΔABC = 8

According to question:

8 =

⇒

⇒

∴

⇒

Replace (h,k) with (x,y)

Thus, locus of point is …

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

• Idea of section formula- Let two points A(x₁,y₁) and B(x₂,y₂) forms a line segment. If a point C(x,y) divides line segment AB in ratio of m:n internally, then coordinates of C is given as:

C =

How to approach: To find locus of a point we first assume the coordinate of point to be (h, k) and write a mathematical equation as per the conditions mentioned in question and finally replace (h, k) with (x, y) to get the locus of point.

Let the coordinates of point whose locus is to be determined be (h, k). Name the moving point be C

Assume the two perpendicular lines on which rod slides are x and y axis respectively.

Here line segment AB represents the rod of length l also ΔADB formed is a right triangle. Coordinates of A and B are assumed to be (0,b) and (a,0) respectively.

∴ a² + b² = l² …eqn 1

As, (h,k) divides AB in ratio of 1:2

∴ from section formula we have coordinate of point C as-

C = =

As, a and b are assumed parameters so we have to remove it.

∵ h = 2a/3 ⇒ a = 3h/2

And k = b/3 ⇒ b = 3k

From eqn 1:

a² + b² = l²

∴

⇒

Replace (h,k) with (x,y)

Thus, locus of point on rod is:

Key points to solve the problem:

• Idea of distance formula- Distance between two points A(x₁,y₁) and B(x₂,y₂) is given by- AB =

• Idea of section formula- Let two points A(x₁,y₁) and B(x₂,y₂) forms a line segment. If a point C(x,y) divides line segment AB in ratio of m:n internally, then coordinates of C is given as:

C = when m = n =1 , C becomes the midpoint of AB and C is given as C =

How to approach: To find locus of a point we first assume the coordinate of point to be (h, k) and write a mathematical equation as per the conditions mentioned in question and finally replace (h, k) with (x, y) to get the locus of point.

Let the coordinates of point whose locus is to be determined be (h, k). Name the moving point be C

Given that (h,k) is midpoint of line x cos α + y sin α = p intercepted between axes.

So we need to first find the points at which x cos α + y sin α = p cuts the axes after which we will apply the section formula to get the locus.

Put y = 0

∴ x = p/cos α ⇒ coordinates on x-axis is (p/cos α , 0). Name the point A

Similarly, Put x = 0

∴ y = p/sin α ⇒ coordinates on y-axis is (0, p/sin α ). Name this point B

As C(h,k) is midpoint of AB

∴ coordinate of C is given by:

C =

Thus,

…equation 1

and …equation 2

Squaring and adding equation 1 and 2:

⇒

Replace (h,k) with (x,y)

Thus, locus of point on rod is:

Key points to solve the problem:

• Idea of section formula- Let two points A(x₁,y₁) and B(x₂,y₂) forms a line segment. If a point C(x,y) divides line segment AB in ratio of m:n internally, then coordinates of C is given as:

C = when m = n =1 , C becomes the midpoint of AB and C is given as C =

How to approach: To find locus of a point we first assume the coordinate of point to be (h, k) and write a mathematical equation as per the conditions mentioned in question and finally replace (h, k) with (x, y) to get the locus of point.

Let the coordinates of point whose locus is to be determined be (h, k). Name the moving point be C

As, coordinate of mid point is (h,k) {by our assumption},

Let Q(a,b) be the point such that Q lies on curve y² = x

b² = a ……equation 1

According to question C is midpoint of OQ

∵ C = ⇒ C =

∴

Similarly,

Putting values of a and b in equation 1,we have:

Replace (h,k) with (x,y)

Thus, locus of point is:

Given, equation (x – a²) + (y – b)² = r². For curious readers- this equation represents a circle in the space centered at point (a, b) having a radius of r units.

To find: Transformed equation of given equation when the coordinate axes are transformed parallelly at point (a - c, b).

We know that, when we transform origin from (0, 0) to an arbitrary point (p, q), the new coordinates for the point (x, y) becomes (x + p, y + q), and hence an equation with two variables x and y must be transformed accordingly replacing x with x + p, and y with y + q in original equation.

Since, origin has been shifted from (0, 0) to (a – c, b); therefore any arbitrary point (x, y) will also be converted as (x + (a – c), y + b) or (x + a - c, y + b).

The given equation (x – a)² + (y – b)² = r² will hence be transformed into the new equation by changing x by x – a + c and y by y – b, i.e. substitution of x by x + a and y by y + b.

= ((x + a - c) – a)² + ((y – b ) - b)² = r²

= (x – c)² + y² = r²

= x² + c² – 2cx + y² = r²

= x² + y² = r² - c² + 2cx

Hence, the transformed equation is x² + y² = r² - c² + 2cx.

Given, equation (a – b)(x² + y²) – 2abx = 0

To find: Transformed equation of given equation when the origin (0, 0) is shifted at point (ab/(a – b), 0).

We know that, when we transform origin from (0, 0) to an arbitrary point (p, q), the new coordinates for the point (x, y) becomes (x + p, y + q), and hence an equation with two variables x and y must be transformed accordingly replacing x with x + p, and y with y + q in original equation.

Since, origin has been shifted from (0, 0) to (ab/(a – b), 0); therefore any arbitrary point (x, y) will also be converted as (x + (ab / (a - b)), y + 0) or (x + ab / (a - b), y).

The given equation (a – b)(x² + y²) – 2abx = 0will hence be transformed into new equation by changing x by x + ab/(a-b) and y by y as

Hence, the transformed equation is (a – b)² (x² + y²) = a² b².

To find: Transformed equation of given equation when the origin (0, 0) is shifted at point (ab/(a – b), 0).

We know that, when we transform origin from (0, 0) to an arbitrary point (p, q), the new coordinates for the point (x, y) becomes (x + p, y + q), and hence an equation with two variables x and y must be transformed accordingly replacing x with x + p, and y with y + q in original equation.

Since, origin has been shifted from (0, 0) to (1, 1); therefore any arbitrary point (x, y) will also be converted as (x + 1, y + 1) or (x + 1, y + 1).

(i) x² + xy – 3x – y + 2 = 0

Substituting the value of x by x + 1 and y by y + 1, we have

= (x + 1)² + (x + 1)(y + 1) – 3(x + 1) – (y + 1) + 2 = 0

= x² + 1 + 2x + xy + x + y + 1 – 3x – 3 - y - 1 + 2 = 0

= x² + xy = 0

Hence, the transformed equation is x² + xy = 0.

(ii) x² – y² – 2x + 2y = 0

Substituting the value of x and y by x + 1 and y + 1 respectively, we have

= (x + 1)² – (y + 1)² – 2(x + 1) + 2(y + 1) = 0

= x² + 1 + 2x - y² – 1 – 2y – 2x – 2 + 2y + 2 = 0

= x² - y² = 0

Hence, the transformed equation is x² - y² = 0.

(iii) xy – x – y + 1 = 0

Substituting the value of x and y by x + 1 and y + 1 respectively, we have

= (x + 1)(y + 1) – (x + 1) - (y + 1) + 1 = 0

= xy + x + y + 1 – x – 1 – y – 1 + 1 = 0

= xy = 0

Hence, the transformed equation is xy = 0.

(iv) xy – y² – x + y = 0

Substituting the value of x and y by x + 1 and y + 1 respectively, we have

= (x + 1)(y + 1) – (y + 1)² - (x + 1) + (y + 1) = 0

= xy + x + y + 1 – y² – 1 – 2y - x – 1 + y + 1 = 0

= xy - y² = 0

Hence, the transformed equation is xy - y² = 0.

Given, equation x² + xy – 3x + 2 = 0

Let’s assume that the origin is shifted at point (p, q).

To find: The shifted point (p, q) satisfying the question’s conditions.

We know that, when we transform origin from (0, 0) to an arbitrary point (p, q), the new coordinates for the point (x, y) becomes (x + p, y + q), and hence an equation with two variables x and y must be transformed accordingly replacing x with x + p, and y with y + q in original equation.

Since, origin has been shifted from (0, 0) to (p, q); therefore any arbitrary point (x, y) will also be converted as (x + p, y + q).

The New equation hence becomes:

= (x + p)² + (x + p)(y + q) – 3(x + p) + 2 = 0

= x² + p² + 2px + xy + py + qx + pq – 3x – 3p + 2 = 0

= x² + xy + x(2p + q – 3) + y(q – 1) + p² + pq – 3p – q + 2 = 0

For no first degree term, we have 2p + q - 3 = 0 and p – 1 = 0, and for no constant term we have p² + pq – 3p - q + 2 = 0.

Solving these simultaneous equations we have p = 1 and q = 1 from first equation. And, p = 1 and q = 1 satisfies p² + pq – 3p - q + 2 = 0.

Hence, the point to which origin must be shifted is (p, q) = (1, 1).

Given points (2, 3), (5, 7), and (-3, -1).

To show: The area of a triangle is invariant to shifting of origin.

The area of triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is

= [x₁(y₂ – y₃) + x₂(y₃ -y₁) + x₃(y₁ – y₂)]

Hence, the area of given triangle = [2(7+1) + 5(-1-3) – 3(3-7)]

= [16 – 20 + 12]

= [8]

= 4

Origin shifted to point (-1, 3), the new coordinates of the triangle are (3, 0), (6, 4), and (-2, -4) obtained from subtracting a point (-1, 3).

Hence, the new area of triangle = [3(4-(-4)) + 6(-4-0) – 2(0-4)]

= [24-24+8]

= [8]

= 4

Since the area of the triangle before and after the translation after shifting of origin remains same, i.e. 4. Therefore we can say that the area of a triangle is invariant to shifting of origin.

To find: Transformed equations of given equations when the origin (0, 0) is shifted at point (1, 1).

We know that, when we transform origin from (0, 0) to an arbitrary point (p, q), the new coordinates for the point (x, y) becomes (x + p, y + q), and hence an equation with two variables x and y must be transformed accordingly replacing x with x + p, and y with y + q in original equation.

Since, origin has been shifted from (0, 0) to (1, 1); therefore any arbitrary point (x, y) will also be converted as (x + 1, y + 1).

(i) x² + xy – 3y² – y + 2 = 0

Substituting x and y with (x+1) and (y+1) respectively, we have

= (x+1)² + (x+1)(y+1) - 3y² – (y + 1) + 2 = 0

= x² + 1 + 2x + xy + x + y +1 – 3y² – y – 1 + 2 = 0

= x² - 3y² + xy +3x – 6y = 0

Hence, the transformed equation is x² - 3y² + xy +3x – 6y = 0

(ii) xy – y² – x + y = 0

Substituting x and y with (x+1) and (y+1) respectively, we have

= (x+1)(y+1) - y² – (x + 1) + (y + 1) = 0

= xy + x + y +1 – y² – x – 1 – y – 1 = 0

= xy - y² = 0

Hence, the transformed equation is xy - y² = 0

(iii) xy – x – y + 1 = 0

Substituting x and y with (x+1) and (y+1) respectively, we have

= (x+1)(y+1) - (x + 1) – (y + 1) + 1 = 0

= xy + x + y +1 – y – 1 – x – 1 + 1 = 0

= xy = 0

Hence, the transformed equation is xy = 0.

(iv) x² – y² – 2x + 2y = 0

Substituting x and y with (x+1) and (y+1) respectively, we have

= (x+1)² - (y + 1)² – 2(x + 1) + 2(y + 1) = 0

= x² + 1 + 2x - y² – 1 – 2y – 2x – 2 + 2y + 2 = 0

= x² - y² = 0

Hence, the transformed equation is x² - y²= 0.

To find: The point to which origin has to be shifted such that there are no first-degree terms, i.e. there are no terms with (variable)¹

We know that, when we transform origin from (0, 0) to an arbitrary point (p, q), the new coordinates for the point (x, y) becomes (x + p, y + q), and hence an equation with two variables x and y must be transformed accordingly replacing x with x + p, and y with y + q in original equation.

In following subproblems, we assume that origin has been shifted from (0, 0) to (p, q); therefore any arbitrary point (x, y) will also be converted as (x + p, y + q).

(i) y² + x² – 4x – 8y + 3 = 0

Substituting x and y with (x+p) and (y+q) respectively, we have

= (x+p)² + (y + q)² – 4(x + p) - 8(y + q) + 3 = 0

= x² + p² + 2px - y² – q² – 2qy – 4x – 4p – 8y – 8q + 3 = 0

= x² + y² + x(2p – 4) + y(2q – 8) + p² + q² – 4p – 8q + 3 = 0

For first degree term to be zero we have,

2p – 4 = 0 and 2q – 8 = 0

Giving us, p = 2 and q = 4.

Hence, the shifted point is (p, q) = (2, 4).

(ii) x² + y² – 5x + 2y – 5 = 0

Substituting x and y with (x+p) and (y+q) respectively, we have

= (x+p)² + (y + q)² – 5(x + p) + 2(y + q) - 5 = 0

= x² + p² + 2px - y² – q² – 2qy – 5x – 5p + 2y + 2q - 5 = 0

= x² + y² + x(2p – 5) + y(2q + 2) + p² + q² – 5p + 2q - 5 = 0

For first degree term to be zero we have,

2p – 5 = 0 and 2q + 2 = 0

Giving us, p = 5/2 and q = 1.

Hence, the shifted point is (p, q) = (5/2, 1).

(iii) x² – 12x + 4 = 0

Substituting x and y with (x+p) and (y+q) respectively, we have

= (x+p)² – 12(x + p) + 4 = 0

= x² + p² + 2px – 12x – 12p + 4 = 0

= x² + x(2p – 12) + p² – 12p + 4 = 0

For first degree term to be zero we have,

2p – 12 = 0.

Giving us, p = 2.

Hence, the shifted point is (p, q) = (2, q), where q can be any real number.

Given points (4, 6), (7, 10), and (1, -2).

To show: The area of a triangle is invariant to shifting of origin.

The area of triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is

= [x₁(y₂ – y₃) + x₂(y₃ -y₁) + x₃(y₁ – y₂)]

Hence, the area of given triangle = [4(10-(-2)) + 7(-2-6) + 1(6-10)]

= [48 – 56 - 4]

= [-12]

= -6

we takes modulus value of -6 i.e. 6 since the area cannot be negative.

Origin shifted to point (-2, 1), the new coordinates of the triangle are (6, 5), (9, 9), and (3, -3) obtained from subtracting a point (-2, 1).

Hence, the new area of triangle = [6(9-(-3)) + 9(-3-5) + 3(5-9)]

= [72 – 72 + (-12)]

= [-12]

= -6

we takes modulus value of -6 i.e. 6 sq. sq. units since the area cannot be negative.

Since the area of the triangle before and after the translation after shifting of origin remains same, i.e. 6 sq. units, therefore we can say that the area of a triangle is invariant to shifting of origin.