Free Linear Equations in Two Variables 02 Practice Test - 9th Grade
Question 1
The geometrical representation of a linear equation is a
Straight line
curve
circle
parabola
SOLUTION
Solution : A
A linear relation between two variables is geometrically represented by a straight line on the Cartesian plane.
Question 2
A linear equation in two variables
Has no solution
Has one solution
Has two solutions
Has infinitely many solutions
SOLUTION
Solution : D
A linear equation in two variables is of the form x+by+c=0
⇒x=−c−by
For every unique value of x, we get a unique value of y satisfying the above equation. Therefore, a linear equation in two variables can have infinitely many solutions.
Question 3
Select the equation whose graph is given alongside:
y = 2x + 2
3y = 6x - 15
y - 2x = 4
y = 2x - 4
SOLUTION
Solution : D
By looking at the graph we can easily say that the line passes through the points (2,0) and (0,-4).
We can identify which line passes through these points by substituting the points in the equation of the line.
Plugging x = 2, y = 0 in equation y = 2x - 4, we get0 = 2(2) - 4 = 0
Plugging x = 0, y = -4 in equation y = 2x - 4, we get
-4 = 2(0) - 0 = -4
The equation y = 2x - 4 is satisfied by both the points. So, given graph belongs to y = 2x - 4.
Question 4
Identify the solutions for the given equation 4x + y = 16.
(0, 16)
(4, 0)
(0, 8)
(3, 2)
SOLUTION
Solution : A and B
The given equation is 4x + y = 16.
(i) (0, 16)
Putting x = 0 and y = 16 in the given equation:
LHS = (4 × 0) + 16 = 16 (RHS)
Thus, (0, 16) is a solution of the given equation.
(ii) (4, 0)
Putting x = 4 and y = 0 in the given equation:
LHS = (4 × 4) + 0 = 16 (RHS)
Thus, (4, 0) is a solution of the given equation.
(iii) (0, 8)
If we put x = 0 and y = 8 in the given equation:
4(0)+8=8≠16(RHS), and thus (0, 8) is not a soltuion of the given equation.
(iv) (3, 2)
Similarly, putting x = 3 and y = 2 in the given equation, we get 4(3)+2=14≠16(RHS), and thus (3, 2) is also not a soltuion of the given equation.
Question 5
Find the value of k, when a=−1 and b=4 is the solution of the equation −2a=−5b+k.
SOLUTION
Solution : D
If a=−1,b=4 is the solution of equation −2a=−5b+k, then they will satisfy the given equation.
Putting values of a & b in the given equation.
So,
−2(−1)=−5(4)+k2=−20+k⇒k=20+2⇒k=22.
Therefore, the value of k is 22.
Question 6
Co-ordinates of P and Q are _______ respectively:
(1, 1) & (2, 0)
(1, 1) & (0, 2)
(1, 2) & (0, 1)
(2, 0) & (1,2)
SOLUTION
Solution : B
Coordinate P is (1, 1) and Q is (0, 2)
Question 7
The price of 1 kg oranges is thrice the price of 1 kg apples. Which of the following linear equation represents the given statement ?
(Assume the the price of 1 kg oranges to be x and of 1 kg apples to be y)
SOLUTION
Solution : A
Given, the price of 1 kg of oranges is x and that of 1 kg of apples is y.
According to the given condition, the price of the 1 kg oranges is three times the price of the 1 kg apple.
So, x = 3y.
Therefore, x - 3y = 0
Question 8
If the graph of the equation 4x + 3y = 12 cuts the coordinate axes at A and B, then hypotenuse of right angle triangle AOB is of length ___. (O is origin)
SOLUTION
Solution : C
4x+3y=12Put y=0 then,we get4x+3(0)=12⇒x=3∴A(3,0) cuts the x−axisPut x=0 then, we get4(0)+3y=12⇒y=4∴B(0,4) cuts the y−axis
In △AOB,AB is the hypotenuse of the given triangle.The length OA=3 units and OB=4 unitsBy pythogoras theorem,AB=√OA2+OB2 =√32+42=√25=5 units
Question 9
A point on the line x + y = 0 can be represented as (a,-a).
True
False
SOLUTION
Solution : A
x + y = 0
⇒ x = - y or y = - x
If x = a, y = -a
∴ Any point on the line x + y = 0 can be represented as (a,-a)
Question 10
The straight line x = a (where a is a constant) is parallel to the
SOLUTION
Solution :
Any straight line that is parallel to the Y-axis will always pass through one constant point on the X-axis. Hence, all the straight lines that are parallel to the Y-axis can be represented as x = a.
