Question

Direction :

Answer the questions based on the following case.

1. Mohit went for shopping in the vening by metro with his father who is an expert in mathematics. He told Mohit that path of metro A is given by the equation x + 2y = 4 and path of metro B is given by the equation x + 2y = 6. Help Mohit to solve the questions.

(a)(4,0), (0,2)

i) Equation x + 2y = 4 intersect the X-axis and Y-axis respectively at (d) (0,4), (0,2)

(b)(0,4),(2,0) (c)(4,0),(2,0)

ii)Equation x + 2y = 9 intersects the

(a)(6,0), (0,8) (b)(0,6), (0,8)

X-axis and Y-axis respectively at (d)(0,6), (0,3)

(c)(6,0), (0, 3)

iii) Coordinates of point of intersection

(a)(1,2)

of two given equations are (c)(3,7)

(b)(2,4)

(d) does not exist​

Answer 1

Answer:

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Mohit went for shopping in the vening by metro with his father who is an expert in mathematics. He told Mohit that path of metro A is given by the equation x + 2y = 4 and path of metro B is given by the equation x + 2y = 6. Help Mohit to solve the questions.

MCQ :-

❶ Equation x + 2y = 4 intersect the X-axis and Y-axis respectively at :-

⦿ (a) (4,0), (0,2)

⦿ (b) (0,4),(2,0)

⦿ (c) (4,0),(2,0)

⦿ (d) (0,4), (0,2)

Correct options is (a)(4,0), (0,2)

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❷ Equation x + 2y = 6 intersects the :

⦿ (a) (6,0), (0,8)

⦿ (b) (0,6), (0,8)

⦿ (c) (6,0), (0, 3)

⦿ (d) (0,6), (0,3)

Correct options is (c)(6,0), (0, 3)

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❸ Coordinates of point of intersection :

⦿ (a) (1,2)

⦿ (b) (2,4)

⦿ (c) (3,7)

⦿ (d) does not exist

Correct options is (d) does not exist

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Answer 2

Answer:

(1) Equation$x+2 y=4$  intersect the $X$-axis and $Y$-axis respectively at

(a)$(4,0), (0,2)$.

(2) Equation $x + 2y = 6$ intersects the points (c) $(6,0), (0, 3)$.

(3)  Coordinates of point of intersection is (d) does not exist.

Step-by-step explanation:

(1)

To find :  Point of intersection of x+2y= 4 with the X-axis and Y-axis

Step 1

For the intersection  $x+2y = 4$ with the $X$-axis, put $y= 0$ in the equation since every y coordinate on $X$-axis $= 0$.

$x+ 2y =4$

$\Rightarrow x+2 \times 0=4$

$\Rightarrow x=4$.

Hence, the coordinates of intersection with the $X$-axis is $(4,0)$.

Step 2

For the intersection  $x+2y = 4$ with the $Y$ axis, put$x= 0$in the equation since every $x$coordinate on $Y$axis$= 0$.

$x+ 2y =4$

⇒ $0 + 2y= 4$

⇒ $y = 2$.

Hence, the coordinates of intersection with the $Y$-axis is $(0,2)$.

Therefore, Equation $x+2 y=4$intersect the $X$-axis and $Y$-axis

respectively at (a) $(4,0),(0,2)$.

The correct option is (a) (4,0), (0,2).

(2)

To find :  Point of intersection of $x+2y= 6$ with the $X$axis and $Y$axis.

Step 1

For the intersection of $x+2y = 6$ with the $X$axis, put $y= 0$ in the equation since every $y$coordinate on $X$ axis $= 0$.

$x+ 2y =6$

⇒ $x+2 \times 0=4$

⇒ $x = 6$.

Hence, the coordinates of the intersection with $X$-axis is $(6,0)$.

Step 2

For the intersection of $x+2y = 6$ with the $Y$ axis, put $x= 0$ in the equation since every $x$coordinate on $Y$axis $= 0$.

$x+ 2y =6$

⇒ $0 + 2y= 6$

⇒ $y = 3$

Hence, the coordinates of intersection with $Y$axis is $(0,3)$.

Therefore, the Equation $x + 2y = 6$ intersects the points (c) $(6,0), (0, 3)$.

The correct option is  (c) $(6,0), (0, 3)$.

(3)

To find: Coordinates of intersection of the two given equations.

Step 1

Let$x+2y-4=0$ be $ax+by+c= 0$ and

$x+2y-6= 0$be $px+qy+r= 0$.

Here,

$a/p = 1/1$

$= 1$.

$b/q= 2/2$

$= 1$.

$c/r = -4/-6$

$= 2/3$.

Step 2

Here, $a/p = b/q$ which is not equal to $c/r$ which is a critical requirement for no answer to a pair of linear equations.

Therefore, coordinates of the point of intersection - option (d) does not exist.