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Given points are A(8,1),B(3,−4) and C(2,k).It is also said that they are collinear and hence the area enclosed by them should be 0. 

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 = \(\frac{1}2\) |8(-4 – k) + 3(k – 1) + 2(1 – (-4))| 

∴ 0 = \(\frac{1}2\) |-32 – 8k + 3k -3 + 10| 

∴ 5k + 25 = 0 

∴ k = -5 

Hence, 

the value of k is -5.

Given points are A(a,2a), B(−2,6) and C(3,1). It is also said that the area enclosed by them is 10 square units. 

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 = 10 

∴ 10 = \(\frac{1}2\) |a(6 – 1) -2 (1 – 2a) + 3(2a - 6)| 

∴ 20 = |5a – 2 + 4a + 6a - 18| 

∴ 20 = | 15a – 20| 

∴ 15a – 20 = ± 20 

Taking positive sign, 

15a – 20 = 20 

a = \(\frac{8}3\)

Taking negative sign, 

15a – 20 = -20 

a = 0 

Hence, 

the value of a are 0 and \(\frac{8}3\)

It is given that A(2, 4), B(2 + \(\sqrt3\), 5) and C(2, 6) are the vertices of the parallelogram ABCD. 

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 □ABCD = 2 × Area of ∆ABC 

Area of ∆ABC = \(\frac{1}2\) |2(5 - 6)+ (2 + \(\sqrt3\))(6 - 4)+2(4 - 5)| 

= \(\frac{1}2\) |-2 + 4 + 2\(\sqrt3\) - 2| 

= \(\frac{1}2\) × 2\(\sqrt3\) = \(\sqrt3\) sq. units 

∴ Area of □ABCD = 2 × \(\sqrt3\) = 2\(\sqrt3\) sq. units 

Hence, 

the area of given parallelogram is 2\(\sqrt3\) sq. units

Let A(1, −3), B(4, p) and C(−9, 7) be the vertices of the ∆ABC. 

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 = 15 

∴ 15 = \(\frac{1}2\) |1(p – 7) +4 (7 – (-3)) - 9(-3 - p)| 

∴ 30 = |p – 7 + 40 + 27 + 9p| 

∴ 30 = | 10p + 60| 

∴ 10p + 60 = ± 30 

Taking positive sign, 

10p + 60 = 30 

p = -3 

Taking negative sign, 

10p + 60 = - 30 

p = -9 

Hence, 

the value of p are -3 and -9

Correct answer is (b) -1

The given ports are A(-3,b) and B(1,b + 4).

Then, (x₁ = -3,y₁ = b) and (x₂ = 1,y₂ = b + 4)

Therefore,

\(x=\frac{[(-3)+1]}{2}\)

= \(\frac{-2}{2}\)

= -1

And

\(y=\frac{[b\,+\,(b\,+\,4)]}{2}\)

= \(\frac{2b\,+\,4}{2}\)

= b + 2

But the midpoint is P(-1,1).

Therefore,

b + 2 = 1

\(\Rightarrow b=-1\)

Correct answer is (a) -8

The point P\((\frac{a}{2},4)\) is the midpoint of the line segment joining the points A(-6,5) and

B(-2,3).

So \(\frac{a}{2}=\frac{-6-2}{2}\)

\(\Rightarrow \frac{a}{2}=-4\)

\(\Rightarrow\) a = -8

Two lines are parallel if their slopes are equal

∴ \(\frac{0-(-8)}{3-(-5)}\) = \(\frac{a-3}{4-6}\) ⇒ \(\frac{8}{8}\) = \(\frac{a-3}{-2}\) ⇒ a – 3 = –2 ⇒ a = 1.

There are two ways to prove it. 

1st way: Area of triangle formed by the given points = 0 if they are collinear.

∴ \(\frac{1}{2}\) [\(x\)(2 – (y + 1) ) + 1((y + 1) – 1) + 0(1 – 2)] = 0

⇒ \(\frac{1}{2}\) [2\(x\) – \(x\)y – \(x\) + y + 0] = 0 

⇒ \(x\) + y – \(x\)y = 0 

⇒ \(x\) + y = \(x\)y 

⇒ \(\frac{x}{xy}\) + \(\frac{y}{xy}\) = 1 ⇒ \(\frac{1}{y}\) + \(\frac{1}{x}\) = 1.

2nd way: Slope of the lines formed by joining these points are equal. 

Slope of line joining (\(x\), 1), (1, 2) = \(\frac{(2-1)}{1-x}\)

Slope of line joining (1, 2), (0, y + 1) = \(\frac{y+1-2}{0-1}\)

⇒ \(\frac{1}{1-x}\) = \(\frac{y-1}{-1}\) ⇒ -1 = (1 – x) (y – 1) 

⇒ –1 = y – \(x\)y – 1 + \(x\) 

⇒ x + y = \(x\)y ⇒  \(\frac{1}{x}\) + \(\frac{1}{y}\) = 1.

Slope of AB = \(\frac{2-4}{1-0}\) = -2, Slope of BC = \(\frac{3-2}{3-1}\) = \(\frac{1}{2}\)

Slope of AC = \(\frac{3-4}{3-0}\) = \(-\frac{1}{3}\)

Slope of AB × Slope of BC = -2 x \(\frac{1}{2}\) = -1

∴ AB ⊥ BC, i.e, ∠B = 90º ⇒ ΔABC is a right angled.

Correct option is (B) (2, 2)

Centroid of the triangle whose vertices are (2, 1), (3, 2) and (1, 3) is \((\frac{2+3+1}3,\frac{1+2+3}3)\)

\(=(\frac{6}3,\frac{6}3)=(2,2)\)

Hence, the centroid of the triangle with vertices (2, 1), (3, 2) and (1, 3) is (2, 2).

Correct option is B) (2, 2)

Correct option is (B) (-1, 4)

Mid-point of the line joining the points (4, 5), (- 6, 3) is \(\left(\frac{4+(-6)}2,\frac{5+3}2\right)\)

\(=(\frac{4-6}2,\frac{8}2)\)

\(=(\frac{-2}2,4)=(-1,4)\)

Correct option is B) (-1, 4)

Correct option is (B) 4

Point on X-axis which is nearest to (3, -4) is (3, 0).

\(\therefore\) The distance of point (3, -4) from X-axis

= Distance between points (3, -4) & (3, 0)

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

\(=\sqrt{0+16}\)

= 4 units

Correct option is B) 4

i. L (x₁ , y₁ ) = L (-2, -3) and M (x₂ ,y₂ ) = M (-6, -8) 

Here, x₁ = -2, x₂ = – 6, y₁ = – 3, y₂ = – 8

Slope of line LM = (y₂ - y₁)/ (x₂ - x₁)

= (-8 - (-3))/(-6 -(-2)) = (-8 + 3)/ (-6 + 2)

= -5/-4 = 5/4

∴ The slope of line LM is 5/4

ii. E (x₁ , y₁ ) = E (-4, -2) and F (x₂ , y₂ ) = F (6, 3) 

Here,x₁ = -4, x₂ = 6, y₁ = -2, y₂ = 3

Slope of line EF = (y₂ - y₁)/ (x₂ - x₁)

= (3 - (-2))/(6 - (-4)) = (3+2)/(6 + 4)

= 5/10 = 1/2

The slope of line EF is 1/2.

iii. T (x₁ , y₁ ) = T (0, -3) and S (x₂ , y₂ ) = S (0, 4) 

Here, x₁ = 0, x₂ = 0, y₁ = -3, y₂ = 4

Slope of line TS  = (y₂ - y₁)/(x₂ - x₁)

= (4 - 9(-3))/(0 - 0) = (4 +3)/0 = 7/0

= Not definned

∴ The slope of line TS cannot be determined.

(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. Slope of line AB = (y₂ - y₁)/(x₂ - x₁)

= (1-(-1))/(0 - (-1)) = (1 + 1)/(0 + 1) = 2

Slope of line BC = (y₂ - y₁)/(x₂ - x₁)

= (3-1)/(1 - 0) = 2

∴ slope of line AB = slope of line BC 

∴ line AB || line BC 

Also, point B is common to both the lines. 

∴ Both lines are the same. 

∴ Points A, B and C are collinear.

ii. Slope of line DE = (y₂ - y₁)/(x₂ - x₁)

= (0 - (-3))/(1-(-2))

= (0 + 3)/(1 + 2) = 3/3 = 1

Slope of line EF = (y₂ - y₁)/(x₂ - x₁)

= (1 - 0)/(2 - 1) = 1

∴ slope of line DE = slope of line EF 

∴ line DE || line EF 

Also, point E is common to both the lines. 

∴ Both lines are the same. 

∴ Points D, E and F are collinear

i. A (x₁ , y₁ ) = A (2, 3) and B (x₂ , y₂ ) = B (4, 7) 

Here, x₁ = 2, x₂ = 4, y₁ = 3, y₂ = 7

Slope of line AB = (y₂ - y₁)/(x₂ - x₁)

=(7 - 3)/(4 - 2) 

= 4/2 = 2

∴ The slope of line AB is 2.

ii. P (x₁ , y₁ ) = P (-3, 1) and Q (x₂ , y₂ ) = Q (5, -2) 

Here, x₁ = -3, x₂ = 5, y₁ = 1, y₂ = -2

Slope of line PQ = (y₂ - y₁)/(x₂ - x₁)

= (-2 - 1)/(5 - (-3)) 

= -3/(5+3)  = -3/8

The slope of line PQ is -3/8

iii. C (x₁ , y₁ ) = C (5, -2) and D (x₂ , y₂ ) = D (7, 3)

Here, x₁ = 5, x₂ = 7, y₁ = -2, y₂ = 3

Slope of line CD = (y₂ - y₁)/(x₂ - x₁)

= (3 - (-2))/ (7 - 5)

= (3 + 2)/2 = 5/2

∴The slope of line CD is 5/2

Slope of line PQ = (y₂ -y₁)/(x₂ - x₁) = (6 -4)/(3 - 2) = 2

Slope of line RS = (y₂ -y₁)/(x₂ - x₁) = (k - 1)/(5 - 3) = (k -1)/2

But, line PQ || line RS … [Given] 

∴ Slope of line PQ = Slope of line RS 

∴ 2 = (k -1)/2

∴ 4 = k – 1 

∴ k = 4 + 1 

∴ k = 5

(C) 5

Distance of (-3, 4) form origin 

= \(\sqrt {(-3)^2 + (4)^2}\)

= \(\sqrt {9 + 16 }\)

= \(\sqrt {25 }\) = 5

i. Angle made with the positive direction of 

X-axis (θ) = 45° 

Slope of the line (m) = tan θ 

∴ m = tan 45° = 1 

∴ The slope of the line is 1.

ii. Angle made with the positive direction of 

X-axis (θ) = 60° 

Slope of the line (m) = tan θ

∴ m = tan 60° = √3

∴ The slope of the line is √3

iii. Angle made with the positive direction of 

X-axis (θ) = 90° 

Slope of the line (m) = tan θ 

∴ m = tan 90° 

But, the value of tan 90° is not defined. 

∴ The slope of the line cannot be determined.

R(x₁ , y₁ ) = R (1, -1), S (x₂ , y₂ ) = S (-2, k) 

Here, x₁ = 1, x₂ = -2, y₁ = -1, y₂ = k

Slope of lines RS = (y₂ - y₁)/(x₂ - x₁) 

= (k - (-1))/(-2 - 1) = (k + 1)/-3

But, slope of line RS is -2. … [Given] 

∴ -2 = (k + 1)/-3

∴ k + 1 = 6 

∴ k = 6 – 1 

∴ k = 5

B(x₁ , y₁ ) = B (k, -5), C (x₂ , y₂ ) = C (1, 2) 

Here, x₁ = k, x₂ = 1, y₁ = -5, y₂ = 2

Slope of line BC = (y₂ - y₁)/(x₂ - x₁) = (2 - (-5))/(1 - k)

= (2 +5)/(1 -k) = 7/(1 - k)

But, slope of line BC is 7. …[Given] 

∴ 7 = 7/(1 - k)

∴ 7(1 – k) = 7 

∴ 1 – k = 7/7

∴ 1 – k = 1 

∴ k = 0

The correct answer is : (C) 1/√3

Slope = \(\frac{y_2-y_1}{x_2-x_1}\)

\(\frac{-2+8}{5-4}\)

= 6

False. Since the area of the triangle formed by the points is 4 sq. units, the points are not collinear.

As the x – coordinate of all these points is zero, all points lie on Y – axis.

True. The coordinates of the mid-points of both the diagonals AC and BD are 1/2, 1, i.e., the diagonals bisect each other.

Correct option is (A) \(\frac{-7}{13}\)

Given line is 13x – 7y + 1 = 0

\(\Rightarrow7y=13x+1\)

\(\Rightarrow y=\frac{13}7x+\frac17\)

\(\therefore\) Slope of line \(13x-7y+1=0\) is \(m=\frac{13}7\)

(By comparing given line with \(y=mx+c)\)

\(\therefore\) Slope of line perpendicular to the line \(13x-7y+1=0\)

\(=\frac{-1}{\text{Slope of line }13x-7y+1=0}\)

\(=\frac{-1}{\frac{13}7}\) \(=\frac{-7}{13}\)

Correct option is A) \(\cfrac{-7}{13}\)

Correct option is (B) -2

Given line is 2x+y+4 = 0

\(\Rightarrow y=-2x-4\)

By comparing with y = mx+c, we obtain \(m=-2\)

\(\therefore\) Slope of the line 2x + y + 4 = 0 is m = -2.

\(\therefore\) Slope of line parallel to line 2x+y+4 = 0 = Slope of line 2x+y+4 = 0

= -2

Hence, the slope of the line parallel to the line 2x+y+4 = 0 is -2.

Correct option is B) -2

Let the points are;

A = (x₁, y₁) = (1, 2)

B = (x₂, y₂) = (0, 0)

C = (x₃, y₃) = (a, b)

∵ Area of ∆ABC= ∆ = 1/2

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

∴ ∆ =1/2 [1(0 - b) + 0(b - 2) + a(2 - 0)]

⇒Δ=1/2 ( - b + 0 + 2a)=1/2(2a - b)

As, the points A (1, 2), B (0, 0) and C (a, b) are collinear, then area of ΔABC will be equals to the zero

Area of ΔABC = 0

⇒1/2 (2a - b)

→ 2a - b = 0

→ 2a = b

Hence, the required relation is 2a = b

Given points = A (- 4, – 3) and B (- 8, – 3) 

These two points lie on the line parallel to X – axis. 

∴ Distance between A (- 4, – 3) and B (- 8, – 3) = |x₂ – x₁|

= |-8 – (-4)| 

= |-8 + 4| 

= |-4| = 4 units.

Given points = A (3, 8) and B (6, 8) 

These two points lie on the line parallel to X – axis. 

Distance between A (3, 8) and B (6, 8) = |x₂ – x₁| 

= |6 – 3| = 3 units.

Correct option is (B) Q₃

If x & y coordinate of a point are negative then that point lies in \(3^{rd}\) quadrant.

\(\therefore\) (-2, -8) lies in \(3^{rd}\) quadrant, i.e., in \(Q_3\) quadrant.

Correct option is  B) Q₃ 

Correct option is (C) abscissa

The x-coordinate of a point is called abscissa.

Correct option is  C) abscissa

Correct option is (D) (0,4)

\(\because\) Equation of X-axis is y = 0.

\(\therefore\) Any point on X-axis has y-coordinate zero.

Since, y-coordinate of point (0, 4) is \(4\neq0.\)

Therefore, it does not lie on X-axis.

Correct option is  D)(0,4)

Correct option is (A) (-2,0)

\(\because\) Equation of Y-axis is x = 0.

\(\therefore\) Any point on Y-axis has x-coordinate zero.

Since, x-coordinate of point (-2, 0) is \(-2\neq0.\)

\(\therefore\) (-2, 0) does not lie on Y-axis.

Correct option is  A) (-2,0)

Correct option is (D) (0,0)

Origin is intersection point of both axes.

\(\therefore\) The point which lies on both axes is (0, 0).

Correct option is  D) (0,0)

Correct option is (B) x = 0

Y-axis is denoted by x = 0.

Correct option is  B) x = 0

Correct option is (D) (3, 8)

The distance of point from Y-axis is x = 3 and the distance of point from X-axis is y = 8.

\(\therefore\) Point is (3, 8).

Correct option is  D) (3, 8)

Correct option is (A) y = 0

X-axis is denoted by y = 0.

Correct option is  A) y = 0

(d) 2a = b

Points A(1, 2), B(0, 0) and C(a, b) are collinear if area of ΔABC = 0.

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

⇒ –b + 2a = 0 ⇒ 2a = b.

Correct option is (B) (0, 3)

Any point that has x-coordinate zero will lie on Y-axis.

\(\because\) Point (0, 3) has x-coordinate zero.

\(\therefore\) Point (0, 3) lies on Y-axis.

Correct option is B) (0, 3)

Correct option is (C) (4, -3)

We know that any point in IV quadrant has x-coordinate positive and y-coordinate negative.

\(\because4>0\;\&\;-3<0\)

\(\therefore\) (4, -3) lies in IV quadrant.

Correct option is C) (4, -3)

Correct option is (C) (2, 0)

Any point on X-axis has y-coordinate zero.

Or any point that has y-coordinate zero, will lie on X-axis.

\(\because\) y-coordinate of point (2, 0) is O.

\(\therefore\) Point (2, 0) lies on X-axis.

Correct option is C) (2, 0)

Correct option is B) Φ

Correct option is (B) \(\sqrt{x^2+y^2}\)

The distance between two points \((x_1,y_1)\;\&\;(x_2,y_2)\) is given by \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

\(\therefore\) Distance of point P(x, y) from origin O(0, 0) is

\(d=\sqrt{(0-x)^2+(0-y)^2}\)

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

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

Correct option is B) \(\sqrt{x^2+y^2}\)

Correct option is (B) X – axis

 In a coordinate system the horizontal line is called X-axis.

Correct option is B) X – axis

Correct option is (A) origin

The coordinates of a point are (0, 0) then the point is origin.

Correct option is A) origin

Correct option is (A) Not defined

\(\because\) The equation of Y-axis is x = 0.

We know that the slope of Y-axis is not defined.

i.e., the slope of line x = 0 is not defined.

\(\because\) Line x = 9 is parallel to  line x = 0 and parallel lines have same slope.

\(\therefore\) The slope of line x = 9 is same as the slope of line x = 0.

\(\therefore\) The slope of line x = 9 is not defined.

Correct option is A) Not defined

Correct option is (A) Y – axis

In a coordinate system the vertical line is called Y-axis.

Correct option is A) Y – axis

Correct option is (D) (0, 0)

The point which lies on both of the axis is origin (0, 0).

Correct option is D) (0, 0)