Algebra
- The area in sq. unit. of the triangle formed by the graphs of x = 4, y = 3 and 3x + 4y = 12 is
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x = 4, a straight line parallel to y – axis.
y = 3, a straight line parallel to x – axis.
Putting x = 0 in 3x + 4y = 12, 3 × 0 + 4y = 12,⇒ 4y = 12 ⇒ y = 12 = 4 3
∴ Point of intersection on y – axis = (0, 4)
Again, putting y = 0 in 3x + 4y = 12, 3x + 4 × 0 = 12⇒ 3x = 12 ⇒ x = 12 = 4 3
∴ Point of intersection onx - axis = (4, 0)
Area of ∎ OACB = OA × OB = 4 × 3 = 12 sq. unitsArea of ∆ OAB = 1 × OA × OB = 1 × 4 × 3 = 6 sq. units 2 2
∴ Area of ∆ABC = 12 – 6 = 6 sq. unitsCorrect Option: D
x = 4, a straight line parallel to y – axis.
y = 3, a straight line parallel to x – axis.
Putting x = 0 in 3x + 4y = 12, 3 × 0 + 4y = 12,⇒ 4y = 12 ⇒ y = 12 = 4 3
∴ Point of intersection on y – axis = (0, 4)
Again, putting y = 0 in 3x + 4y = 12, 3x + 4 × 0 = 12⇒ 3x = 12 ⇒ x = 12 = 4 3
∴ Point of intersection onx - axis = (4, 0)
Area of ∎ OACB = OA × OB = 4 × 3 = 12 sq. unitsArea of ∆ OAB = 1 × OA × OB = 1 × 4 × 3 = 6 sq. units 2 2
∴ Area of ∆ABC = 12 – 6 = 6 sq. units
- The straight line 2x + 3y = 12 passes through :
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Putting y = 0 in 4x + 3y = 12,
we get x = 3
Putting x = 0 in 4x + 3y = 12, we get, y = 4
Correct Option: B
Putting y = 0 in 4x + 3y = 12,
we get x = 3
Putting x = 0 in 4x + 3y = 12, we get, y = 4
- The graphs of x = a and y = b intersect at
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Point of intersection = (a, b)
Correct Option: A
Point of intersection = (a, b)
- The graph of 3x + 4y – 24 = 0 forms a triangle OAB with the coordinate axes, where O is the origin. Also the graph of x + y+4 =0 forms a triangle OCD with the coordinate axes. Then the area of ∆OCD is equal to
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On putting x = 0 in the equation 3x + 4y = 244y = 24 ⇒ y = 24 = 6 4
∴ Co-ordinates of B = (0, 6)
Again, putting y = 0 in the equation 3x + 4y = 24,
3x = 24 Þ x = 8
∴ Co-ordinates of A = (8,0)
Similarly, for x + y = – 4
Co-ordinates of C = (–4, 0)
Co-ordinates of D = (0, –4)∴ Area of ∆OAB = 1 × OA × OB = 1 × 8 × 6 = 24 sq. units 2 2 Area of ∆OCD = 1 × OC × OD = 1 × 4 × 4 = 8 sq. units 2 2
Clearly,∆ OCD ≡ 1 ∆OAB 3 Correct Option: B
On putting x = 0 in the equation 3x + 4y = 244y = 24 ⇒ y = 24 = 6 4
∴ Co-ordinates of B = (0, 6)
Again, putting y = 0 in the equation 3x + 4y = 24,
3x = 24 Þ x = 8
∴ Co-ordinates of A = (8,0)
Similarly, for x + y = – 4
Co-ordinates of C = (–4, 0)
Co-ordinates of D = (0, –4)∴ Area of ∆OAB = 1 × OA × OB = 1 × 8 × 6 = 24 sq. units 2 2 Area of ∆OCD = 1 × OC × OD = 1 × 4 × 4 = 8 sq. units 2 2
Clearly,∆ OCD ≡ 1 ∆OAB 3
- The equations 3x + 4y = 10 – x + 2y = 0 have the solution (a,b). The value of a + b is
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3x + 4y = 10 ---(i)
– x + 2y = 0
⇒ x = 2y
∴ From equation (i),
3 × 2y + 4y = 10 Þ 10y = 10⇒ y = 10 = 1 10
∴ x = 2
∴ (a, b) = (2, 1)
∴ a + b = 2 + 1 = 3Correct Option: C
3x + 4y = 10 ---(i)
– x + 2y = 0
⇒ x = 2y
∴ From equation (i),
3 × 2y + 4y = 10 Þ 10y = 10⇒ y = 10 = 1 10
∴ x = 2
∴ (a, b) = (2, 1)
∴ a + b = 2 + 1 = 3
