Explore topic-wise InterviewSolutions in .

(c) √5 units 

Area of Δ ABC = 0 for collinearity of A, B, C.

⇒ \(\frac{1}{2}\)[1(4 – a) + 2(a – 2) + 3(2 – 4)] = 0 

⇒ 4 – a + 2a – 4 + 6 – 12 = 0 

⇒ a – 6 = 0 ⇒ a = 6. 

∴ Point C ≡ (3, 6)

⇒ BC = \(\sqrt{(3-2)^2+(6-4)^2}\) = \(\sqrt{1+4}\) = √5 units .

(c) (5, 0) or (1, 4) 

Let the third vertex of the triangle be P(a, b). 

Since it lies on the line x + y = 5, a + b = 5      ...(i) 

Also, given area of triangle formed by the points (2, –1), (3, 2) and (a, b) = 4 units

⇒ \(\frac{1}{2}\)[2(2 – b) + 3(b + 1) + a(–1 – 2)] = ± 4 

⇒ [4 – 2b + 3b + 3 – 3a] = ± 8 

⇒ –3a + b = 1          ...(ii) or –3a + b = –15             ...(iii) 

Eqn (i) – Eqn (ii) ⇒ (a + b) – (–3a + b) = 5 – 1 

⇒ 4a = 4 ⇒ a = 1 ⇒ b = 4 

Eqn (i) – Eqn (iii) ⇒ (a + b) – (–3a + b) = 5 + 15

⇒ 4a = 20 ⇒ a = 5 ⇒ b = 0. 

∴ The points are (1, 4) and (5, 0).

(d) 2 : 1 externally

The mid-point of the line joining the points (–10, 8) and (– 6, 12) is

\(\bigg(\frac{-10+(-6)}{2},\frac{8+12}{2}\bigg)\), i.e., (–8, 10).

Let (–8, 10) divide the join of (4, –2) and (–2, 4) in the ratio k : 1. 

Then, –8 = \(\frac{-2k+4}{k+1}\) and 10 = \(\frac{4k-2}{k+1}\)

⇒ – 8k – 8 = – 2k + 4 ⇒ – 6k = 12 ⇒ k = –2 

Since the value of k is negative, it is a case of external division and the ratio is 2 : 1.

Correct option is (C) 1 : 1

Mid point of a line segment divides it into the ratio 1:1.

Correct option is C) 1 : 1

The distance of a point (x,y) from the origin O(0,0) is \(\sqrt{x^2+y^2}\)

Let P(x = -6,y = 8) be the gen point. Then

\(OP=\sqrt{x^2+y^2}\)

= \(\sqrt{(-6)^2+8^2}\)

= \(\sqrt{36+64}\)

= \(\sqrt{100}=10\)

Let coordinates of the point=(x, 0)(given that the point lies on x axis)

x₁=7. y₁=-4

x₂=x. y₂=0

Distance =√(x₂-x₁)²+ (y₂-y₁)²

According to the question,

2√5=√(x-7)²+ (0-(-4))²

Squaring L.H.S and R.H.S

20=x²+49-14x+16

20=x²+65-14x

0=x²-14x+45

0=x²-9x-5x+45

0=x(x-9)-5(x-9)

0=(x-9) (x-5)

x-9 =0. x-5= 0

x=9 or x=5

Therefore, coordinates of points…..(9,0)or(5,0)

(B) 5√2

Distance formula: d² = (x₂ – x₁)² + (y₂ – y₁)²

According to the question,

We have;

x₁ = 0, x₂ = – 5

y₂ = 5, y₂ = 0

d² = (( – 5) – 0)² + (0 – 5)²

d= √(-5-0)²+ (0-5)²

d= √((-5)²+ (-5)²)

d = √(25 + 25)

d= √50= 5√2

So the distance between (0, 5) and ( – 5, 0) = 5√ 2

Answer is (B) 0

Let the points A(1, 2), B(1, x) and C(2, 3) are collinear then area of triangle made by these points will be zero.

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

⇒ 1/2[(x – 3) + (-1)(3 – 2) + 2(2 – x)] = 0

⇒ x – 3 – 1 + 2(2 – x) = 0

⇒ x – 3 – 1 + 4 – 2x = 0

⇒ -x – 4 + 4 = 0

⇒ x = 0

1. a) (2,25)

2. c) (8,20)

3. c) √61

4. a) (5, 22.5)

5. a) (3.5,24)

Correct answer is (a) (2,5)

Let (x,y) be the coordinates of A. then,

\(0=\frac{-2\,+\,x}{2} \Rightarrow x=2\)

\(4=\frac{3\,+\,y}{2}\Rightarrow y=8-3=5\)

Thus, the coordinates of A are (2,5).

Correct option is (A) (0, 0)

The coordinate of origin are (0, 0).

Alternative method :

Since, origin is intersection point of both coordinate axes (X & Y-axes).

The equation of X-axis is y = 0 and the equation of Y-axis is x = 0.

\(\therefore\) Coordinate of intersection point of both coordinate axes are (0, 0).

\(\therefore\) Coordinate of origin are (0, 0).

Correct option is A) (0, 0)

Correct option is (D) non-collinear

Giver points are A(1, 5), B(2, 3) and C(-2, -1).

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

\(=\sqrt{1+4}\)

\(=\sqrt{5}\) units

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

\(=\sqrt{16+16}\)

\(=4\sqrt{2}\) units

\(AC=\sqrt{(-2-1)^2+(-1-5)^2}\)

\(=\sqrt{9+36}\)

\(=3\sqrt{5}\) units

\(\because\sqrt5<\sqrt{32}<\sqrt{45}\)

\(\Rightarrow AB<BC<AC\)

But AB+BC \(=\sqrt5+4\sqrt2\)

\(=2.236+5.657\)

\(=7.893\neq6.708\)

i.e., AB+BC \(\neq\) AC

\(\therefore\) Points A, B and C are not collinear.

Also, they are not vertices of isosceles & right angled triangle.

Hence, given points A, B and C are non-collinear.

Correct option is D) non-collinear

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.

Let point P on the x-axis divides the line segment joining the points A and B the ratio m : n

Consider P lies on x-axis having coordinates (x, 0).

x = (m × 5 + n x2)/ (m + n)

x = (5m + 2n ) / (m + n)

5m + 2n = x(m + n )

(5 – x)m + (2 – x)n = 0 ….(1)

And,

y = 0 = (m × 6 + n(– 3))/ (m+n)

0 = (6m – 3n ) /(m + n)

6m – 3n = 0

6m = 3n

or m/n = 3/6 = 1/2

P divides AB in the ratio 1:2.

=> m = 1 and n = 2

From (2)

(5 – x) + (2 – x)(2) = 0

5 – x + 4 – 2x = 0

3x = 9

x = 3

Hence coordinates are (3,0)

Let P divides the join of A and B in the ratio k : 1, then

Step 1: Find coordinates of P:

6 = (k × 8 + 1×3)/ (k+1)

=> 6k + 6 = 8k + 3

or k = 3/2

P divides the join of A and B in the ratio 3:2

Step 2: Find the value of m

m = (2k-4)/(k+1) = (2×3/2 – 4)/(3/2+1)

= -1/(5/2)

= -2/5

If point P (x, 2) divides the join of A (12, 5) and B (4, -3), then

using section formula, we get

2 = (m x (-3) + n x (5)) / (m + n)

2m + 2n = -3m + 5n

5m = 3n

m/n = m:n = 3:5

The required ratio is 3:5.

Points A (4, 3) and B (x, 5) lie on the circle with centre O(2, 3)

Which means: OA = OB

=> OA² = OB²

(2-4)² + (3-3)² = (2-x)² + (3-5)²

(-2)² + 0² = (2-x)² + (-2)²

(2-x)² = 0

2 - x = 0

x = 2

The value of x is 2.

Correct answer is (b) 26

The given points are A(-6,7) and B(-1,-5). So

AB = \(\sqrt{(-6+1)^2+(7+5)^2}\)

= \(\sqrt{(-5)^2+(12)^2}\)

= \(\sqrt{25+144}\)

= \(\sqrt{169}\)

= 13

Thus, 2AB = 26.

Correct option is (D) (0, 1)

The equation of Y-axis is x = 0.

\(\therefore\) x-coordinate of any point on Y-axis is zero.

\(\therefore\) (0, 1) is a point lying on Y-axis.

Correct option is  D) (0, 1)

Correct option is (B) perpendicular

In the coordinate plane the axes are perpendicular to each other.

B) perpendicular

C. origin

Explanation:

The points at which the two coordinate axes meet is called the origin.

Hence, (C) is the correct option.

(c) \(\bigg(\frac{3+\sqrt3}{2},\frac{5-\sqrt3}{2}\bigg)\)

Let the co-ordinates of vertex C are (x, y). 

Since the incentre and centroid of an equilateral triangle coincide, co-ordinates of centroid of

ΔABC = \(\bigg(\frac{9+\sqrt3}{6},\frac{15-\sqrt3}{6}\bigg)\)

∴ \(\frac{x+1+2}{3}\) = \(\frac{9+\sqrt3}{6}\) and \(\frac{2+3+y}{3}\) = \(\frac{15-\sqrt3}{6}\)

⇒ 2(x + 3) = 9 + √3 and 2(5 + y) = 15 – √3 

⇒ 2x = 3 + √3 and 10 + 2y = 15 – √3

⇒ x = \(\frac{3+\sqrt3}{2}\) and 2y = 5 - √5 ⇒ y = \(\frac{5-\sqrt3}{2}\)

∴ Required co-ordinates are \(\bigg(\frac{3+\sqrt3}{2},\frac{5-\sqrt3}{2}\bigg)\).

Distance of point (-5, 4) from x – axis = 4 unit.

Correct option is (A) (1,0)

OX is positive x-axis.

Thus, any point on line OX has y-coordinate zero.

\(\therefore\) (1, 0) lies on OX in graph.

Correct option is  A) (1,0)

(c) 25x + 25y – 4 = 0 

The point of intersection of the given lines can be obtained by solving the equations of the two lines simultaneously. 

100x + 50y = 1                    ...(i) 

75x + 25y = –3                  ...(ii) 

Eqn (i) – 2 × Eqn (ii) 

⇒ (100x + 50y) – (150x + 50y) = 1 – (–6) 

⇒ – 50x = 7 ⇒ x = \(\frac{-7}{50}\)

100 x \(\frac{-7}{50}\) + 50y  = 1 ⇒ 50y = 1 + 14 = 15 ⇒ y = \(\frac{15}{50}\) = \(\frac{3}{10}\)

Let the required line make x-intercept = y-intercept = a. 

Then eqn of required line is \(\frac{x}{a}\) + \(\frac{y}{a}\) = 1 ⇒ x + y = a

Since it passes through \(\bigg(\)\(\frac{-7}{50}\), \(\frac{3}{10}\)\(\bigg)\), therefore

⇒ \(\frac{-7}{50}\) + \(\frac{3}{10}\) = a ⇒ \(\frac{-7+15}{50}\) = a ⇒ a = \(\frac{8}{50}\) = \(\frac{4}{25}\)

∴ Eqn of required line: x + y = \(\frac{4}{25}\) ⇒ 25 + 25y – 4 = 0.

(2) (0, 4)

Points in y-axis have abscissa a zero

(3) (3, 6)

x + 2 = 5 ⇒ x = 3; 

y – 2 = 4 ⇒ y = 6

(4) √2

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

(3) on x-axis

(i) False 

(ii) True 

(iii) True 

(iv) True 

(v) True

(i) I quadrant 

(ii) III quadrant 

(iii) II quadrant 

(iv) IV quadrant.

(3) 3

Points in y-axis have ordinate zero. 

∴ a – 3 = 0 ⇒ a = 3

(3) II and IV quadrant respectively

(-, +) lies II^nd quadrant and (+, -) lies in IV^th quadrant

(2) II

(-, +) lies in II^nd quadrant

Correct option  (b) 12 Sq.units

Explanation:

2x + 3y ≤ 6, x ≥ 0, y ≥ 0

2x - 3y ≤ 6, x ≥ 0,  y ≤ 0

- 2x + 3y ≤ 6, x ≥ 0,  y ≤ 0

- 2x - 3y ≤ 6,  x ≤ 0, y ≤ 0

Form a rhombus with diagonals 4 & 6.

Area = 1/2 x 4 x 6 = 12 sq. units.

Given points are A(2, 0), B(9, 1), C(11, 6) and D(4, 4).

Coordinates of mid-point of AC are (11+2/ 2, 6+0/ 2) = (13/2, 3)

Coordinates of mid-point of BD are (9+4/ 2, 1+4/ 2) = (13/2, 5/2)

As the coordinates of the mid-point of AC ≠ coordinates of mid-point of BD, ABCD is not even a parallelogram.

Therefore, ABCD cannot be a rhombus too.

Answer:

(i) For collinear points, area of triangle formed by them is zero.
Therefore, for points (7, -2) (5, 1), and (3, k), area = 0
1/2 [7 { 1- k} + 5(k-(-2)) + 3{(-2) + 1}] = 0
7 - 7k + 5k +10 -9 = 0
-2k + 8 = 0
k = 4

(ii) For collinear points, area of triangle formed by them is zero.
Therefore, for points (8, 1), (k, - 4), and (2, - 5), area = 0
1/2 [8 { -4- (-5)} + k{(-5)-(1)} + 2{1 -(-4)}] = 0
8 - 6k + 10 = 0
6k = 18
k = 3

Let A = (x₁,y₁) = (a, c + a), B = (x₂, y₂) = (a, c) and C = (x₃, y₃) = (- a, c – a) be the given points

Then,

The area of ∆ABC is given by

= \(\frac{1}{2}\)[a ( – {c – a}) + a(c – a – (c + a)) +( – a)(c + a – a)]

= \(\frac{1}{2}\) [a(c – c + a) + a(c – a – c – a) – a(c + a – c)]

= \(\frac{1}{2}\)[a × a + ax( – 2a) – a × a]

= \(\frac{1}{2}\)[a² – 2a² – a²]

= \(\frac{1}{2}\)×(-2a)²

= – a²

Point P(2, 4) is equidistant from the points A(5, K) and B(K, 7). 

So PA = PB

⇒ PA² = PB²

⇒ (5 – 2)² + (k – 4)² = (k – 2)² + (7 – 4)²

⇒ 9 + K² + 16 – 8K = K² + 4 – 4K + 9

⇒ K² – 8K + 16 – K² – 4 + 4K = 0

⇒ -4K + 12 = 0

⇒ -4K = -12

⇒ K = -12/-4 = 3

Hence, K = 3

Answer is (A) Abscissa

Let co-ordinate of B is (x, y)

Given : co-ordinate of mid point of (3, 4) and (x, y)

4 = (3 + x)/2 

⇒ 3 + x = 8

⇒ x = 8 – 3 = 5

and 5 = (y + 4)/2

⇒ y + 4 = 10

⇒ y = 10 – 4 = 6

Hence, co-ordinate of B = (5, 6)

Answer is (A) (4, 6)

Let co-ordinate of mid point of (6, 8) and (2, 4) is (x, y), then :

x = (6 + 2)/2 = 8/2 = 4

And y = (8 + 4)/2 = 12/2 = 6

Hence, co-ordinate of mid point = (4, 6)

Co-ordinate of points A,B, C are (6, -1), (1, 3) and (x, 8) respectively.

According to question

AB = BC

√((6 - 1)² + (-1 - 3)²) = √((1 - x)² + (3 - 8)²)

Squaring both sides

25 + 16 = (1 – x)² + 25

(1 – x)² = 16

(1 – x) = ±4

taking +ve sign = 1 – x = 4

x = 1 – 4 = -3

taking -ve sign 1 – x = -4

x = 1 + 4 = 5

So, x = -3 or 5

Answer is (B) Origin

Let the point P(x,y) be equidistant from the points A(7,1) and B(3,5)

Then,

PA = PB

\(\Rightarrow\) PA² = PB²

\(\Rightarrow\) (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²

\(\Rightarrow\) x² + y² - 14x - 2y + 50 = x² + y² - 6x - 10y + 34

\(\Rightarrow\) 8x - 8y = 16

\(\Rightarrow\) x - y = 2

Coordinates of the points are A(3, 6) and B(-3, 4) 

Let the point P(x, y) is equidistant from A and B 

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

⇒ PA = PB

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

On squaring both sides, we get

(x - 3)² + (y - 6)² = (x + 3)² + (y - 4)²

x² - 6x + 9 + y² - 12y + 36 = x² + 6x + 9 + y² - 8y + 16

3x + y = 5

Correct option is (C) (0, c)

x-coordinate of any point on Y-axis is zero.

Put x = 0 in line y = mx+c, we get

y = c

\(\therefore\) Intersection point of line y = mx+c & Y-axis is (0, c).

Hence, the line y = mx+c meets Y-axis at the point (0, c).

Correct option is C) (0, c)

Correct option is (B) 2 : 1

The centroid divides the median of a triangle in the ratio 2:1.

Correct option is C) 2 : 1

3\(x\) – 2y + 5 = 0 ⇒ –2y = –3\(x\) – 5 ⇒ y = \(\frac{3}{2}\)\(x\) + \(\frac{5}{2}\)

On comparing with y = m\(x\) + c, we see that slope of given line = \(\frac{3}{2}\)

As the required line is perpendicular to the given line, 

Slope of required line = \(\frac{-3}{2}\)

∴ Equation of required line: (y – 5) = \(\frac{-3}{2}\) (\(x\) - 4)

⇒ 3(y – 5) = – 2\(x\) + 8 ⇒ 3y – 15 = –2\(x\) + 8 ⇒ 3y + 2\(x\) – 23 = 0