Graphical Representation of Straight Lines MCQ Quiz - Objective Question with Answer for Graphical Representation of Straight Lines - Download Free PDF

4x + 3y = 0

When x = 0 then y = 0

When x = -3 then y = 4

And, 7x + 5y = 0

When x = 0 then y = 0

When x = 5 then y = -7

The two lines 4x + 3y = 0 and 7x + 5y = 0 will intersect each other at one point only in their graphical representation.

Alternate MethodGiven equation are 4x + 3y = 0 and 7x + 5y = 0.

a₁/a₂ ≠ b₁/b₂ ⇒ 4/7 ≠ 3/5

In this situation, the equation will have a unique solution. If we remember this, there is no need to sketch any graph.

kx + 2y = 7

⇒ 2y = - kx + 7

⇒ y = ( - kx/2) + (7/2)         - - - (1)

3x + y = 1

⇒ y = - 3x + 1         - - - (2)

On comparing the two equations with y = mx + c, we get,

Slope of (1) = m₁ = - k/2

Slope of (2) = m₂ = - 3

For a unique solution,

⇒ m₁ should not be equal to m₂

⇒ - k/2 ≠ - 3

∴ k ≠ 6

For no solution,

⇒ m₁ = m₂

⇒ - k/2 = - 3

∴ k = 6

For a triangle ABC with vertices A(A_X, A_y), B(B_X, B_y) and C(C_X, C_y), the coordinates of the centroid of triangle are obtained as,

(O_X, O_y) = [(A_X + B_X + C_X)/3], [(A_y + B_y + C_y)/3]

Given, A(3, -4), B(0, 5) and O(1, -4)

Computing x-coordinate,

⇒ 1 = (3 + 0 + C_X)/3

⇒ C_X = 3 - 3 = 0

Computing y-coordinate,

⇒ -4 = (-4 + 5 + C_y)/3

⇒ C_y = -12 - 1 = -13

Mark points P and Q on the coordinate system

From the above diagram, ΔOPQ is a right-angled triangle.

OP = 1.5

OQ = 0.5

PQ² = OP² + OQ²

⇒ PQ² = (1.5)² + (0.5)²

⇒ PQ² = 2.25 + 0.25

⇒ PQ² = 2.5

Given:

Equations of the lines: 2x - y + 8 = 0, 8x + 3y - 24 = 0, and y = 0

Formula used:

Area of triangle formed by lines = 1/2 × base × height

Calculation:

Solve for intersection points:

Line 1: 2x - y + 8 = 0

⇒ y = 2x + 8

Line 2: 8x + 3y - 24 = 0

Substitute y from Line 1 into Line 2:

8x + 3(2x + 8) - 24 = 0

⇒ 8x + 6x + 24 - 24 = 0

⇒ 14x = 0

⇒ x = 0

Substitute x = 0 into Line 1:

y = 2(0) + 8 = 8

Intersection points are: (0, 8) and y = 0 intersects these lines at x-values:

For 2x - y + 8 = 0, y = 0:

2x + 8 = 0

⇒ x = -4

For 8x + 3y - 24 = 0, y = 0:

8x - 24 = 0

⇒ x = 3

Base = Distance between (-4, 0) and (3, 0) = 7 units

Height = 8 units

Area = 1/2 × base × height

⇒ Area = 1/2 × 7 × 8

⇒ Area = 28 sq units

∴ The correct answer is option (1).

Given:

The distance between two points (-6, y) and (18, 6) is 26 units.

Formula used:

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

Where, 

The distance between two points (x₁, y₁) and (x₂, y₂) is D units.

Calculation:

The distance between two point = 26 units

The value of the first co-ordinate = (x₁, y₁) = (-6, y)

The value of the second co-ordinate = (x2, y2) = (18, 6)

According to the question,

⇒ D = $\sqrt{(18-(-6){)}^2+(6-y{)}^2}$

⇒ 26 = $\sqrt{(24{)}^2+(6-y{)}^2}$

Squaring on both sides of the equation.

⇒ 676 = 24² + (6 - y)²

⇒ 676 = 576 + (6 - y)2

⇒ 100 = (6 - y)2

⇒ 10² = (6 - y)2

Taking square root on both sides of the equation.

⇒ 10 = 6 - y

⇒ y = -4

∴ The required answer is -4.

Given:

The line joining the points (4, 4) and (6, 8)

Formula used:

Slope of the line passing through the points (x1, y1) and (x2, y2) = (y2 - y1)/(x2 - x1)

Calculation:

We know that,

Slope of the line passing through the points (x₁, y₁) and (x₂, y₂) = (y₂ - y₁)/(x₂ - x₁)

⇒ Slope of the line joining the points (4, 4) and (6, 8)

⇒ (8 - 4)/(6 - 4) = 4/2 = 2

Given:

2x – 3y + 6 = 0

4x + y = 16

y = 0

Formula used:

Area of triangle = $\frac{1}{2}$ ∣x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)∣ 

Calculation:

2x – 3y + 6 = 0 

⇒ 2x – 3y = -6 ....(1)

4x + y = 16  ....(2)

Multiplying equation (2) by 3

⇒ 12x + 3y = 48 ...(3)

Now,

Adding (1) and (3), we get

⇒ (2x + 12x) = (48 – 6)

⇒ 14x = 42

⇒ x = 3

Now, Putting the value of x in equation (1)

⇒ 2 × 3 – 3y = -6

⇒ 6 – 3y = -6

⇒ -3y = -1

⇒ y = (-1 – 3) = 4

So,(x₁, y₁) = (3, 4)

Now,

We put y = 0 in equation (2)

⇒ 4x + 0 = 16

⇒ 4x = 16

⇒ x = 4

So,(x₂, y₂) = (4, 0)

Again,

We put y = 0 in equation (1)

⇒ 2x – 0 = -6

⇒ 2x = -6

⇒ x = -3

So,(x₃, y₃) = (-3, 0)

Now,

Area of triangle = $\frac{1}{2}$ ∣x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)∣ 

⇒ $\frac{1}{2}$∣3(0 – 0) + 4(0 – 4) + (-3)(4 – 0)∣ 

⇒ $\frac{1}{2}$∣0 + (-16) + (-12)∣ 

⇒ $\frac{1}{2}$∣-16 – 12∣ 

⇒ $\frac{1}{2}$∣-28∣ 

⇒ 14  [Negative sign is always treated as positive in modulus]

∴ The required value is 14.

Alternate Method

 When we find the vertices from the given equations

Given:

2x – 3y + 6 = 0

4x + y = 16

y = 0

Area of triangle = $\frac{1}{2}$ × Base × Height

Area of triangle = $\frac{1}{2}$ × 7 × 4 = 14

Given: 

 According to the question, :

⇒ kx - 3y = 6 

⇒ k (3) - 3 x 2 = 6 

⇒ 3k – 6 = 6 

⇒ k = $\frac{12}{3}$ = 4

Given:

8x + 3y = 24, 2x + 8 = y and the x-axis

Formula used:

Area of triangle = 1/2 × base × height

The Y co-ordinate of point of intersection of the two lines 8x + 3y = 24, 2x + 8 = y gives the height of the triangle.

The difference between the x intercepts of the lines 8x + 3y = 24, 2x + 8 = y gives the base of the triangle

Calculation:

For 8x + 3y = 24, 2x + 8 = y to find the point of intersection put one equation in other

⇒ 8x + 3(2x + 8) = 24

⇒ 14x = 0

⇒ x = 0

⇒ y = 2×0 + 8 = 8

⇒ Height of the triangle = 8

For 8x + 3y = 24, 2x + 8 = y to find the X – intercepts put y = 0

⇒ 8x + 3×0 = 24

⇒ x₁ = 3

⇒ 2x + 8 = 0

⇒ x₂ = -4

Their difference is the base = 3 – (-4) = 7

∴ The area of triangle = 8 × 7/2 = 28

Given:

Equation of line p: x + y = 5

Equation of line q: x - y = 3

Formula Used:

To find the coordinates of the intersection point, solve the system of equations:

1. x + y = 5

2. x - y = 3

Calculation:

Adding equations 1 and 2:

x + y + x - y = 5 + 3

⇒ 2x = 8

⇒ x = 8 / 2

⇒ x = 4

Substituting x = 4 into equation 1:

4 + y = 5

⇒ y = 5 - 4

⇒ y = 1

Coordinates of the point common to both lines:

(4, 1)

GIVEN:

CONCEPT:

Here we need to understand that the line x = - 1 is parallel to y axis does there will be no change in the value of the point at y axis.

FORMULAE USED:

Reflected point = (-x + 2a, y)

SOLUTION:

Reflected point = (2 + 2 × (-1), 6)

∴ Reflected point = (0, 6)

Given:

$4x+\frac{1}{3}y=\frac{8}{3}$ and 

$\frac{1}{2}x+\frac{3}{4}y+\frac{5}{2}=0$

Calculation:

$4x+\frac{1}{3}y=\frac{8}{3}$ ....(1)

$\frac{1}{2}x+\frac{3}{4}y+\frac{5}{2}=0$

⇒ $\frac{1}{2}x+\frac{3}{4}y=-\frac{5}{2}$ .....(2)

Now,

Multiplying the number by 3 in equation (1), we get

⇒ 12x + y = 8 ....(3)

Again, multiplying the number by 24 in equation (2), we get

⇒ 12x + 18y = -60 ...(4)

Now,

From equation (3) and (4), we get

⇒ (18y – y) = (-60 – 8)

⇒ 17y = -68

⇒ y = -4

Now, Putting the value of y in equation (3), we get

⇒ 12x – 4 = 8

⇒ 12x = 12

⇒ x = 1

So, P(1, -4)

Now,

According to given option

x + 2y – 5 = 0

⇒ (1 – 8 – 5) = 0

⇒ -12 is not equal to 0

So, It is not satisfying the point P

Again,

3x – y – 7 = 0

⇒ (3 + 4 – 7) = 0

⇒ (7 – 7) = 0

⇒ 0 = 0

It is satisfying the point P

Again,

x – 3y – 12 = 0

⇒ (1 + 12 – 12) = 0

⇒ (13 – 12) = 0

⇒ 1 is not equal to 0

So, It is not satisfying the point P

Again,

4x – y + 7 = 0

⇒ (4 + 4 + 7) = 0

⇒ 15 is not equal to 0

So, It is not satisfying the point P

∴  The point P also lies on the graph of the equation is 3x – y – 7 = 0

According to the given information,

Slope of line passing through the points (2, 1) and (6, 3) = $\frac{3-1}{6-2}$ = 1/2

We know that, parallel lines have the same slopes.

Given:

ax + 5y = 8

Calculation:

ax + 5y = 8

⇒ y = -ax/5 + 8/5

Comparing it with y = mx + c, where m = Slope

So, m = -a/5

⇒ Slope = -a/5 = -4/3 (given)

⇒ a/5 = 4/3