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Given:

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

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

= 1/2 [x₁(y₂ – y₃) + x₂(y₃ - y₁) + x₃(y₁ – y₂)]

The area of given triangle = 1/2 [2(7 + 1) + 5(-1 - 3) – 3(3 - 7)]

= 1/2 [16 – 20 + 12]

= 1/2 [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).

The new area of triangle = 1/2 [3(4 - (-4)) + 6(-4 - 0) – 2(0 - 4)]

= 1/2 [24 - 24 + 8]

= 1/2 [8]

= 4

Since the area of the triangle before and after the translation after shifting of origin remains same, i.e. 4.

∴ We can say that the area of a triangle is invariant to shifting of origin.

(5, 7) belongs to first quadrant q₁. 

(5, -7) belongs to fourth quadrant q₄. 

(-5, 7) belongs to second quadrant q₂. 

(-5, -7) belongs to third quadrant q₃.

Here both coordinates are negative therefore point C lines in IIIrd quadrant. 

Correct option is (D) 4

The coordinate axes divide the cartesian plane into 4 quadrants.

Correct option is  D) 4

Here x is negative and y is positive therefore point B lies in IInd quadrant. 

i) D 

ii) 3 

iii) F, H 

iv) 0,0 

v) C (3, – 2) 

vi) A (2, 1) 

Vii) Positive Y-axis 

viii) Q₄

We know that, a point lies on the Y-axis, if its x-coordinate is zero. Here, x-coordinate of points C(0, 1), D(0, 0), E(0,-1) and G(0, 5) are zero. So, these points lie on Y-axis. Also, D(0, 0) is the intersection point of both.the axes, so we can consider that it lies on Y-axis as well as on X-axis.

The equation of a line parallel to the X-axis is y = b.

Since, the line is at a distance of 5 units below the Xaxis. 

∴ b = -5 

∴ y = -5 is the equation of the required line.

(i) The point which lies on X and Y-axes both is origin whose coordinates are (0, 0).

(ii) The point whose ordinate is – 4 and which lies on Y-axis, i.e., whose x-coordinate is zero, is (0,-4).

(iii) The point whose abscissa is 5 and which lies on X-axis, i.e., whose y-coordinate is zero, is (5, 0).

we will solve it by taking the coordinates A(x₁,y₁), B(x₂,y₂) and C(x₃,y₃)

 Let the co ordinates of the centroid be G(u, v). 

G(u, v) = \((\frac{x_1+x_2+x_3}3,\frac{y_1+y_2+y_3}3)\)

let the coordinates of P(h, k). 

now we will find L.H.S and R.H.S. separately. 

PA²+PB²+PC² 

= (h - x₁)² +(k - y₁)² +(h - x₂)²+ (k - y₂)² +(h - x₃)² +(k - y₃)² …by distance formula. 

= 3(h²+k²)+(x₁²+x₂²+x₃²)+(y₁²+y₂²+y₃²)-2h(x₁+x₂+x₃)-2k(y₁+y₂+y₃) 

= 3(h²+k²)+(x₁²+x₂²+x₃²)+(y1 2+y₂²+y₃²)-2h(3u)-2k(3v)

GA²+GB²+GC²+3GP² 

= (u-x₁)²+(v-y₁)²+(u-x₂)²+(v-y₂)²+(u-x₃)²+(v-y₃)²+3[(u-h)²+(v-k)²] …by distance formula. 

= 3(u²+v²)+(x₁²+y₁²+x₂²+y₂²+x₃²+y₂³) - 2u(x₁+x₂+x3) -2v(y₁+y₂+y₃) + 3[u²+h²-2uh+v²+k²-2vk] 

= 6(u²+v²)+(x₁²+y₁²+x₂²+y2²+x₃²+y₃²)-2u(3u)-2v(3v) +3(h²+k²) -6uh-6vk 

= (x₁²+x₂²+x₃²)+(y₁²+y₂²+y₃²)+3(h₂+k²)-6uh-6vk 

Hence LHS = RHS 

(The above relation is known as Leibniz Relation) 

Hence Proved.

Let the co-ordinates of any point on the x-axis be (x, 0). Then distance between (x, 0) and (– 4, 8) is 10 units.

⇒ \(\sqrt{(x+4)^2+(0-8)^2}\) = 10 

⇒ x² + 8x + 16 + 64 = 100 

⇒ x² + 8x – 20 = 0 

⇒ (x + 10) (x – 2) = 0 

⇒ x = –10 or 2 

∴ The required points are (– 10, 0) and (2, 0).

On x-axis, y coordinate is zero of each point, 

So B(1, 0), D(0, 0) E(-1, 0) and G(4, 0) lie on x-axis.

Let P(1, -1), Q \(\big(\frac{-1}{2},\frac{1}{2}\big)\) and R(1,2) be the vertices of the ΔPQR.

Then, PQ = \(\sqrt{\big(\frac{-1}{2}-1\big)^2+\big(\frac{1}{2}+1\big)^2}\) = \(\sqrt{\frac{9}{4}+\frac{9}{4}}\) = \(\sqrt{\frac{18}{4}}\) = \(\frac{3\sqrt2}{2}\)

QR = \(\sqrt{\big(1+\frac{1}{2}\big)^2+\big(2-\frac{1}{2}\big)^2}\) = \(\sqrt{\frac{9}{4}+\frac{9}{4}}\) = \(\sqrt{\frac{18}{4}}\) = \(\frac{3\sqrt2}{2}\)

PR = \(\sqrt{(1-1)^2+(2+1)^2}\) = \(\sqrt9\) = 3

∵ PQ = QR, the triangle PQR is isosceles.

On x-axis, y coordinate is zero of each point, 

so B(1, 0), D(0, 0) E(-1, 0) and G(4, 0) lie on x-axis.

The three given points are A(a, 1), B(1, −1) and C(11, 4). 

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃) 

= \(\frac{1}2\) |x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)| 

Given that area of ∆ABC = 0 

∴ 0 = \(\frac{1}2\) |a(-1 – 4) + 1(4 - 1) + 11(1 – (-1))| 

∴ 0 = \(\frac{1}2\) |-5a + 3 + 22| 

∴ -5a + 3 + 22 = 0 

a = 5 

Hence the value of a is 5

(i) Axis

(ii) Quadrant

(iii) Origin

Consider the following points A(a,b), B(a₁,b₁), C(a−a₁,b−b₁) 

Since the given points are collinear, we have area(△ABC)=0 

First find the area of area(△ABC) as follows: 

area(△ABC)

= \(\frac{1}2\) |x₁(y₁−y₃)+x₁(y₃−y₁)+x₃(y₁−y₁)| 

= \(\frac{1}2\) |a(b₁−(b−b₁))+a₁((b−b₁)−b)+(a−a₁)(b−b₁)| 

= \(\frac{1}2\) |a(b₁−b+b₁)+a₁(b−b₁−b)+a(b−b₁)−a₁(b−b₁)| 

= \(\frac{1}2\) |−ab−a₁b₁+ab−ab₁+a₁b+a₁b₁| 

= \(\frac{1}2\) |−(ab₁−a₁b)| = (ab₁−a₁b) 

This gives, ab₁−a₁b=0 

∴ ab₁ = a₁b

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 = \(\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)

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) = \(\sqrt{(h - 2)^2 + (k - 4)^2}\)

Distance of (h,k) from (0,k) = \(\sqrt{(h - 0)^2 + (k - k)^2}\)

According to question both distance are same.

\(\therefore\) \(\sqrt{(h - 2)^2 + (k - 4)^2}\) = \(\sqrt{(h - 0)^2 + (k - k)^2}\)

(h - 2)² + (k - 4)² = (h - 0)² + (k - k)²

\(\Rightarrow\) h² + 4 - 4h + k² - 8k + 16 = h² + 0

\(\Rightarrow\) k² - 4h - 8k + 20 = 0

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

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

y^2 _- 4x - 8y + 20 = 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 = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) 

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:

\(\sqrt{(x-7)^2 + (0-6)^2}\) = \(\sqrt{(x - 3)^2 + (0 - 4)^2}\)

Squaring both sides, 

we have:

(x - 7)² + (0 - 6)² = (x -3)² + (0 - 4)²

\(\Rightarrow\) x² + 49 - 14x + 36 = x² + 9 - 6x + 16

\(\Rightarrow\) 8x = 60 \(\Rightarrow\) x = \(\frac{60}{8} = \frac{15}{2} = 7.5\)

∴ point on x-axis is (7.5,0)

Answer is (B) 9 or -3

Distance between P(2, 3) and Q(10, y) = 10

√((10 - 2)² + (y - 3)²) = 10

square both sides

⇒ (8)² + (y – 3)² = 100

⇒ (y – 3)² = 100 – 64

⇒ (y – 3)² = 36

⇒ y – 3 = ±6

Taking the +ve sign

y – 3 = 6

y = 6 + 3 = 9

Taking -ve sign

y – 3 = -6

y = -6 + 3 = -3

So y = 9 or -3

P ≡ (–3, 2), Q ≡ (–5, –5), R ≡ (2, –3), S ≡ (4, 4)

∴ PQ = \(\sqrt{(-5+3)^2+(-5-2)^2}\) = \(\sqrt{4+49}\) = \(\sqrt{53}\)

QR = \(\sqrt{(2+5)^2+(-3+5)^2}\) = \(\sqrt{49+4}\) = \(\sqrt{53}\)

RS = \(\sqrt{(4-2)^2+(4+3)^2}\) = \(\sqrt{4+49}\) = \(\sqrt{53}\)

PS = \(\sqrt{(4+3)^2+(4-2)^2}\) = \(\sqrt{49+4}\) = \(\sqrt{53}\)

∵ PQ = QR = RS = PS, therefore PQRS is a rhombus. For PQRS to be a square, diagonals PR and QS should be equal.

PR = \(\sqrt{(2+3)^2+(-3-2)^2}\) = \(\sqrt{25+25}\) = \(\sqrt{50}\) = \(5\sqrt2\)

QS = \(\sqrt{(4+5)^2+(4+5)^2}\) = \(\sqrt{81+81}\) = \(\sqrt{162}\) = \(9\sqrt2\)

As PR ≠ QS, so PQRS is not a square.

Area of rhombus = \(\frac{1}{2}\) x (Product of length of diagonals) 

= \(\frac{1}{2}\) x \(5\sqrt2\) x \(9\sqrt2\) = 45 sq. units.

(c) Parallelogram

AB = \(\sqrt{(4-7)^2+(5-6)^2}\) = \(\sqrt{9+1}\) = \(\sqrt{10}\)

BC = \(\sqrt{(7-4)^2+(6-3)^2}\) = \(\sqrt{9+9}\) = \(3\sqrt2\)

CD =\(\sqrt{(4-1)^2+(3-2)^2}\) = \(\sqrt{9+1}\) = \(\sqrt{10}\)

AD = \(\sqrt{(4-1)^2+(5-2)^2}\) = \(\sqrt{9+9}\) = \(3\sqrt2\)

AC = \(\sqrt{(4-4)^2+(5-3)^2}\) = \(\sqrt4\) = 2

BD = \(\sqrt{(7-1)^2+(6-2)^2}\) = \(\sqrt{36+16}\) = \(\sqrt{52}\) = \(2\sqrt{13}\)

AB = CD, BC = AD and AC ≠ BD ⇒ opposite sides are equal and diagonals are not equal. 

⇒ ABCD is a parallelogram.

(d) 24 units

AB ⊥ chord PQ ⇒ AB bisects chord PQ ⇒ PQ = 2PB. 

AB = \(\sqrt{(2-5)^2+(-3-1)^2}\) = \(\sqrt{(-3)^2+(-4)^2}\)

= \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5

AP = radius of circle = 13 

∴ By Pythagoras’ Theorem, PB = \(\sqrt{AP^2-AB^2}\)

= \(\sqrt{169-25}\) = \(\sqrt{144}\) = 12 units

∴ PQ = 2 × PB = 24 units.

Mid point of the line segment joining the points A(2a, 4) and B (-2, 3b) is C(1, 2a + 1)

Mid point of AB = ( (2a-2)/2, (4+3b)/2 ) ….(1)

Mid point of AB = (1, 2a + 1) ….(2) (given)

Now, from (1) and (2)

1 = (2a-2)/2

=> a = 2

and 2a + 1 = (4+3b)/2

10-4 = 3b

or b = 2

Answer: a = 2 and b = 2

The line segment joining the points A(-2, 9) and B(6, 3) is a diameter of a circle with centre C.

Which means C is the midpoint of AB.

let (x, y) be the coordinates of C, then

x = (-2 + 6)/2 = 2 and

y = (9+3)/2 = 6

So, coordinates of C are (2, 6).

Point P(x, y) is equidistant from the points A(5, 1) and B(– 1, 5), means PA = PB or PA² = PB²

(5 – x)² + (1 – y)² = (– 1 – x)² + (5 – y)²

(25 + x² – 10x) + (1 + y² – 2y) = (1 + x² + 2x + 25 + y² – 10y)

26 + x² – 10x + y² – 2y = (26 + x² + 2x + y² – 10y)

12x = 8y

3x = 2y

Hence proved.

Let four vertices of quadrilateral be A (1, 2) and B (−5, 6) and C (7, −4) and D (k, −2) 

Area of □ ABCD = Area of ∆ABC + Area of ∆ACD = 0 sq. unit 

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃)

= \(\frac{1}2\) |x₁(y₂ - y₃)+x₂(y₃ - y₁)+x₃(y₁ - y₂)| 

Area of ∆ABC

= \(\frac{1}2\) |1(6 – (-4)) - 5(-4 -2) + 7(2 – 6)| 

= \(\frac{1}2\) |10 + 30 -28| 

= 6 sq. units 

Area of ∆ACD 

= \(\frac{1}2\) |1(-2 – (-4)) + k(-4 -2) + 7(2 – (-2))| 

= \(\frac{1}2\) |2 - 6k + 30| 

= \(\frac{1}2\) (3k -15) sq. units 

Area of ∆ABC + Area of ∆ACD = 0 sq. unit 

∴ 6 + 3k -15 =0 

3k -9 = 0 

∴ k =3 

Hence, 

the value of k is 3 

(d) \(\frac{23}{\sqrt{17}}\)

The given lines are:

L : \(\frac{x}{5}+\frac{y}{b}=1\)            ....(i)

K : \(\frac{x}{c}+\frac{y}{3}=1.\)          ...(ii)

Since line L passes through (13, 32),

\(\frac{13}{5}\) + \(\frac{32}{b}\) = 1 ⇒ \(\frac{32}{b}\) = 1 - \(\frac{13}{5}\) = \(\frac{-8}{5}\) ⇒ b = \(\frac{32\times5}{-8}\) = -20.

∴ Line L is \(\frac{x}{5}\) - \(\frac{y}{20}\) = 1, i.e., y = 4x – 20

⇒ Slope of line L = 4 

As L || K, slope of line K = Slope of line L. 

Eqn of line K can be written as y = \(\frac{-3x}{c}+3\)

∴ Slope of K = \(-\frac{3}{c}.\)

given, \(-\frac{3}{c}\) = 4 ⇒ c = \(-\frac{3}{4}\)

∴ Equation of line K : \(\big(\frac{-4}{3}\big)x\) + \(\frac{y}{3}\) = 1 ⇒ 4x – y + 3 = 0.

Distance between the lines 

L ≡ 4x – y – 20 = 0 and K ≡ 4x – y + 3 = 0 is

d = \(\bigg|\frac{3-(-20)}{\sqrt{4^2+(-1)^2}}\bigg|\) = \(\frac{23}{\sqrt{17}}\)
\(\bigg(\)Distance between parallel lines ax + by + c₁ = 0 and ax + by c₂ = 0 is d = \(\frac{|c_2-c_1|}{\sqrt{a^2+b^2}}\bigg)\)

(i) θ = 90°

m = tan θ = tan 90° = ∝ (undefined)

(ii) m = tan θ = tan 0° = 0

(i) Slope = 0

tan θ = 0

tan 0 = 0

∴ θ = 0°

(ii) Slope = 1

tan θ = 1

tan 45° = 1

∴ θ = 45°

Angle of inclination is 45°

Given:

The equation, (x – a) ² + (y – b) ² = r² 

The given equation (x – a)² + (y – b)² = r² can 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 -2cx = r² – c² 

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

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.

i. The equation of a line parallel to the Y-axis is x = a. 

∴ The line x = 3 is parallel to the Y-axis. 

ii. y – 2 = 0 

∴ y = 2 

The equation of a line parallel to the X-axis is y = b. 

∴ The line y – 2 = 0 is parallel to the X-axis. 

iii. x + 6 = 0 

∴ x = -6 

The equation of a line parallel to the Y-axis is x = a. 

∴ The line x + 6 = 0 is parallel to the Y-axis. 

iv. The equation of a line parallel to the X-axis is y = b. 

∴ The line y = – 5 is parallel to the X-axis.

i. Q(-2, 2) and R(4, -1) 

ii. T(0, -1) and M(3, 0) 

iii. Point S lies in the third quadrant. 

iv. The x and y co-ordinates of point O are equal.

(C) The answer is (a, 0)

i. Quadrant I 

ii. Quadrant III

iii. Quadrant IV 

iv. Quadrant II

The equation of a line parallel to the Yaxis is x = a. 

Since, the line is at a distance of 7 units to the left of Y-axis, 

∴ a = -7 

∴ x = -1 is the equation of the required line.

Correct option is (A) -7

The abscissa of the point (- 7, 2) = x-coordinate of point (-7, 2)

= -7

Correct option is  A) -7

Correct option is (B) Q and R

Correct option is (C) Third

D. do not lie in the same quadrant

Explanation:

Points (1, – 1), (2, – 2), (4, – 5) lie in IV quadrant and (– 3, – 4) lie in III quadrant.

Hence, (D) is the correct option.

i) (- 2, 3) – Q₂ (second quadrant)

ii) (5, – 3) – Q₄ (fourth quadrant)

iii) (4, 2) ) – Q₁ (first quadrant)

iv) (- 7, – 6) – Q₃ (third quadrant)

v) (0, 8) – on Y-axis

vi) (3, 0) – on X-axis

vii) (-4,0) – on X’ – axis

viii) (0, – 6) – on Y’: axis

Equation of Y-axis is x = 0. 

Since, ‘a’ is a real number, there are two possibilities. 

Case I: a > 0 

Case II: a < 0 

∴ Distance between the Y-axis and the line x = a = a-0 = a 

Since, |a| = a, a > 0

= – a, a < 0 

∴ Distance between the Y-axis and the line x = a is |a|

In ∆ABC, ∠B = 900

∴ (AB)² + (BC)² = [(Ac)² …(i) … [Pythagoras theorem]

seg CB || X-axis 

∴ y co-ordinate of B = 2 

seg BA || Y-axis 

∴ x co-ordinate of B = 2 

∴ co-ordinate of B is (2, 2) = (x₁, y₁) 

co-ordinate of A is (2, 3) = (x₂, y₂) 

Since, AB || to Y-axis, 

d(A, B) = Y₂ – Y₁ 

d(A,B) = 3 – 2 = 1 

co-ordinate of C is (-2, 2) = (x₁, y₁) 

co-ordinate of B is (2, 2) = (x₂, y₂) 

Since, BC || to X-axis, 

d(B, C) = x₂ – x₁ 

d(B,C) = 2 – -2 = 4 

∴ AC² = 12 + 42 …[From (i)] 

= 1 + 16 = 17 

∴ AC = √17 units …[Taking square root of both sides].

(i) The coordinates are:

P = (3,2)

R = (3,0)

Q = (3,-1)

(ii) Since, all the points on the line have the same abscissa = 3.

The difference in abscissa of L and M = 0.

Correct answer is

(D) (1,-3)

Since, seg AB || Y-axis.

∴ x co-ordinate of all points on seg AB

will be the same,

x co-ordinate of A (1, 3) = 1 

x co-ordinate of B (1, – 3) = 1

i. Distance of line LM from the Y-axis is 3 units.

ii. P(3, 2), Q (3, -1), R(3, 0) 

iii. x co-ordinate of point L = 3 

x co-ordinate of point M = 3 

∴ Difference between the x co-ordinates of the pointsL and M = 3 –3

=0

Correct option is (D) √5 units

Given points are \(A\,(log_{10}\,1000,tan\,45^\circ)\)

\(=(log_{10}\,10^3,1)\) \(=(3,1)\)

And \(B\,(cosec\,30^\circ,log_7\,343)\)

\(=(2,log_7\,7^3)=(2,3)\)

i.e., \(A(3,1)\;\&\;B(2,3)\)

\(\therefore\) Distance between points A & B is

\(AB=\sqrt{(2-3)^2+(3-1)^2}\)

\(=\sqrt{(-1)^2+2^2}\)

\(=\sqrt{1+4}\) \(=\sqrt{5}\) units

Correct option is D) √5 units

Correct option is (A) (2, 2)

\(\because\) \(log_2\,8=log_2\,2^3\)

\(=3\,log_2\,2=3\)         \((\because log_a\,a=1)\)

\(log_5\,25=log_5\,5^2\)

\(=2\,log_5\,5=2\)

\(log_{10}\,10=1\)

and \(log_{10}\,100=log_{10}\,10^2\)

\(=2\,log_{10}\,10=2\)

\(\therefore\) \(A=(log_2\,8,log_5\,25)=(3,2)\)

& \(B=(log_{10}\,10,log_{10}\,100)=(1,2)\)

\(\therefore\) Mid-point of AB is mid-point of (3, 2) & (1, 2)

\(=(\frac{3+1}2,\frac{2+2}2)\)

\(=(\frac{4}2,\frac{4}2)=(2,2)\)

Correct option is A) (2, 2)