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3.

If the line xa+yb=1\frac{x}{a} + \frac{y}{b} = 1 passes through the point of intersection of the lines 2x−y=12x - y = 1 and 3x−4y+6=03x - 4y + 6 = 0 and is parallel to the line 4x+3y−6=04x + 3y - 6 = 0, then find the values of a and b.

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  1. Point of Intersection:
    The given lines are:
    2x−y=1  ⟹  y=2x−12x - y = 1 \implies y = 2x - 1 ...(i)
    3x−4y+6=03x - 4y + 6 = 0 ...(ii)

    Substituting (i) into (ii):
    3x−4(2x−1)+6=03x - 4(2x - 1) + 6 = 0
    3x−8x+4+6=03x - 8x + 4 + 6 = 0
    −5x+10=0  ⟹  x=2-5x + 10 = 0 \implies x = 2

    From (i), y=2(2)−1=3y = 2(2) - 1 = 3.
    So, the point of intersection is (2,3)(2, 3).

  2. Equation of the Parallel Line:
    The required line is parallel to 4x+3y−6=04x + 3y - 6 = 0.
    Therefore, its equation is of the form:
    4x+3y+k=04x + 3y + k = 0

    Since this line passes through the point (2,3)(2, 3):
    4(2)+3(3)+k=04(2) + 3(3) + k = 0
    8+9+k=0  ⟹  k=−178 + 9 + k = 0 \implies k = -17

    Hence, the equation of the line is:
    4x+3y=174x + 3y = 17

  3. Finding aa and bb:
    Dividing both sides by 17 to put it into the intercept form xa+yb=1\frac{x}{a} + \frac{y}{b} = 1:
    4x17+3y17=1\frac{4x}{17} + \frac{3y}{17} = 1
      ⟹  x174+y173=1\implies \frac{x}{\frac{17}{4}} + \frac{y}{\frac{17}{3}} = 1

    Comparing with xa+yb=1\frac{x}{a} + \frac{y}{b} = 1, we get:
    a=174a = \frac{17}{4}
    b=173b = \frac{17}{3}

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