Applications Based on Co-ordinate System MCQ Quiz - Objective Question with Answer for Applications Based on Co-ordinate System - Download Free PDF

let perpendicular to the line passing through the points (x₁,y₁) and (x₂,y₂)

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

Slope of the line = (0 - 1) / (-2 - (-5)) = -1/3

∴ Slope of the perpendicular line = -1/(-1/3) = 3

Formula:

The coordinates of the point which divides the line joining (x₁, y₁) and (x₂, y₂) in the ratio m: n

x = (mx₂ + nx₁) / (m + n)

y = (my₂ + ny₁) / (m + n)

Calculation:

The coordinates of the point which divides the line joining (- 1, 7) and (4, - 3) in the ratio 2 : 3

(x, y) = [(2 × 4 + 3 × - 1) / (2 + 3), (2 × - 3 + 3 × 7) / (2 + 3)]

⇒ (x, y) = [(8 – 3)/5, (- 6 + 21)/5]

⇒ (x, y) = (5/5, 15/5)

Since, ABCD is a parallelogram.

Mid-point of diagonal AC will be same as the mid-point of diagonal BD.

Now,

$\frac{-2-3}{2},\frac{a+1}{2}=\frac{b-8}{2},\frac{2a-3-1}{2}$

-5 = b – 8

b = 3

a + 1 = 2a – 4

a = 5

Now,

$=\sqrt{a^2-b^2}$

$=\sqrt{5^2-3^2}$

= 4 units

Let the segment divides y axis in K : 1 ratio.

If ratio m ∶ n is the intersection then formula for co-ordinates dividing is given by,

(mx₂ + nx₁)/(m + n), (my₂ + ny₂)/(m + n)

As, the intersection point is on y axis, the value of x co-ordinate = 0

So,

(-6k + 2)/(k + 1) = 0

⇒ -6k + 2 = 0

⇒ k = 1/3

Given:

Equation of a straight line: 5x + 12y = 60

Concept used:

The intercept form of a straight-line equation: $\frac{x}{a}+\frac{y}{b}=1$

Where a and b are x and y intercepts respectively.

Calculation:

5x + 12y = 60

Dividing by 60 on both sides,

⇒ $\frac{5x}{60}+\frac{12y}{60}=\frac{60}{60}$

⇒ $\frac{x}{12}+\frac{y}{5}=1$

Here,

x-intercept = a = 12

y-intercept = b = 5

Now,

The length of the line segment intercepted between the axes is given by the distance between the intercepts:

Length = $\sqrt{a^2+b^2}$

⇒ Length = $\sqrt{{12}^2+5^2}$

⇒ Length = $\sqrt{144+25}$

⇒ Length = $\sqrt{169}$ = 13 cm

∴ The correct answer is option (2).

Given:-

Vertices of triangle = (1,2), (-4,-3), (4,1)

Formula Used:

Area of triangle = ½ [x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)]

whose vertices are (x1, y1), (x2, y2) and (x3, y3)

Calculation:

⇒ Area of triangle = (1/2) × [1(-3 – 1) + (-4) (1 – 2) + 4{2 – (-3)}]

= (1/2) × {(-4) + 4 + 20}

= 20/2

= 10 sq. units

Let the reflected point be (x, y).

Then the midpoint of (-1, 3) and (x, y) i.e. [(x-1)/2, (y+3)/2] must lie on the line x = -4.

⇒ (x – 1) / 2 = -4

⇒ x = -8 + 1 = -7

Since the line joining the original and the reflected point in perpendicular to x = -4, hence its slope will be equal to 0.

⇒ [y – 3]/[(-7) – (-1)] = 0

⇒ y – 3 = 0

⇒ y = 3

⇒ Section formula for internal division = {[(mx₂ + nx₁)/(m + n)], [(my₂ + ny₁)/(m + n)]}

⇒ Here, x₁, y₁ = (- 1, 0) and x₂, y₂ = (2, 6). m : n = 2 : 1

⇒ [(2 × 2) + (1 × - 1)]/(2 + 1), [(2 × 6) + (1 × 0)]/(2 + 1) = (1, 4)

Given:

PQ is straight line of 13 units length

Co-ordinates of P = (2, 5)

Co-ordinates of Q = (x, -7)

Formula used:

D² = (x₂ - x₁)² + (y₂ - y₁)²

D = distance

(x₁, y₁) = co-ordinates of first point

(x₂, y₂) = co-ordinates of second point

Calculation:

D2 = (x2 - x1)2 + (y2 - y1)2

(13)² = (x - 2)² + (-7 - 5)²

⇒ 169 = (x - 2)² + 144

⇒ (x - 2)² = 25

⇒ x - 2 = 5 

⇒ x = 7

∴ The value of x is 7.

Given,

The given equation is 2x = 5 – 3y

Concept:/Formula:

If equation cuts the x-axis, then y = 0

Calculation:

⇒ 2x = 5 – 3y

If equation cuts the x-axis at P (α, β), then y = 0

⇒ 2x = 5 – 0

⇒ x = 5/2

⇒ α = 5/2 and β = 0

Now,

⇒ (2 α + β)

⇒ (2 × 5/2 + 0)

⇒ 5

Given:

(x1 ,y1) =  (2, 4),

(x2,y₂) = (-3, -1) 

(x3 ,y3)= (5, 3)

Formula Used:

Area of the triangle = [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]/2

Calculation:

Area of the triangle = [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]/2

Substitute the values in the formula

⇒ [2(-1 – 3) + (-3)(3 – 4) + 5(4 + 1)]/2

⇒ [-8 + 3 + 25]/2 = 10 sq. units

The given points form a right angle triangle with base 6 units and height 8 units.

∴ Area of triangle = (Base × Height) /2 = (6 × 8) /2 = 24 sq. units.

Alternate Method Formula used:

Area of the triangle having vertices on the points (x₁, y₁), (x₂, y₂) and (x₃, y₃) = (1/2) × [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)]

Calculation:

Area of the triangle with vertices (3, 4), (-3, 4) and (-3, -4)

⇒ (1/2) × [3(4 - (-4)) + (-3)(-4 - 4) + (-3)(4 - 4)]

⇒ (1/2) × [3(8) + (-3)(-8) + 0]

⇒ (1/2) × [24 +24]

⇒ 24

∴ The correct answer is 24.

Given:

The line joining the two points (5, 4) and (-3, k) is parallel to the line kx – y + 11 = 0

Calculation:

Equation of line joining two points:

$\left(y-4\right)=\frac{\left(k-4\right)}{-3-5}\times \left(x-5\right)$

Slope of above line = $m_1=\frac{4-k}{8}$

Another line:

kx – y + 11 = 0

y = kx + 11

Slope of above line = k

Both the lines are parallel:

m₁ = m₂

⇒ (4 – k)/8 = k

⇒ 4 – k = 8k

⇒ 4 = 9k

⇒ k = 4/9

Given:

Quadrilateral formed with the vertices (-3, 2), (5, 4),(7, -6) and (-5, -4).

Concept:

Area of quadrilateral ABCD = (1/2) ⋅ [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)]

Calculations:

Let A(-3, 2), B(5, 4), C(7, -6) and D(-5, -4) be the vertices of a quadrilateral ABCD.

Thus,

A(-3, 2) = (x1, y1)

B(5, 4) = (x2, y2) 

C(7, -6) = (x3, y3)

D(-5, -4) = (x4, y4)

We know that, 

Area of quadrilateral ABCD = (1/2) ⋅ [(x1y2 + x2y3 + x3y4 + x4y1) – (x2y1 + x3y2 + x4y3 + x1y4)]

Substituting the values,

= ($\frac{1}{2}$). {[-3(4) + 5(-6) + 7(-4) + (-5)2] – {[5(2) + 7(4) + (-5)(-6) + (-3)(-4)]}

= ($\frac{1}{2}$).[(-12 – 30 – 28 – 10) – (10 + 28 + 30 + 12)]

= ($\frac{1}{2}$) [-80 – 80]

= 160/2 {since area cannot be negative}

= 80

$\therefore$The area of the quadrilateral formed with the given vertices is 80 sq. units.

Given: 

Coordinates = (a, - b) and (- a, - b)

Formula used:

Distance between two points (x1, y1) and (x2, y2) is √{(x2 - x1)2 + (y2 - y1)2}

Calculation:

Let the two coordinates A and B be (a, -b) and (-a, -b) respectively

Coordinate, A = (a, - b)

Where, x1 = a, y1 = - b

Coordinate, B = (- a, - b)

Where, x2 = - a, y2 = - b

We know that,

Distance between AB = √{(x2 - x1)2 + (y2 - y1)2}

⇒ √{(- a - a)2 + (- b + b)²}

⇒ √(- 2a)2 

⇒ √(4a²)

⇒ 2a

∴ The distance between (a, - b) and (- a, - b) is 2a.