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Mathematics Question on Straight lines

Suppose a line parallel to ax+by=0ax+by=0 (where b≠0b≠0)intersects5x−y+4=0 5x-y+4=0 and 3x+4y−4=03x+4y-4=0 ,respectively at P and Q. If the midpoint of PQ is (1,5)(1,5),then the value of ab\dfrac{a}{b} is

A

1073\dfrac{107}{3}

B

−1073\dfrac{-107}{3}

C

3107\dfrac{3}{107}

D

−3107\dfrac{-3}{107}

E

11

Answer

−1073\dfrac{-107}{3}

Explanation

Solution

Given data

The given lines are:

5x−y+4=0⇒y=5x+45x - y + 4 = 0 ⇒ y = 5x + 4-----------(1)

3x+4y−4=0⇒y=−(3/4)x+13x + 4y - 4 = 0 ⇒ y = -(3/4)x + 1------(2)

From the equations, we can see that the slopes of the two lines are 55 and −34\dfrac{-3}{4}, respectively.

Now, let the coordinates of point P be (p, 5p + 4) (since it lies on the line y = 5x + 4) and the coordinates of point Q be (q,−(34)q+1)(q, -(\dfrac{3}{4})q + 1) (since it lies on the line

y=−3x4+1y = \dfrac{-3x}{4} + 1

Step 2:

Find the midpoint of PQ. The midpoint of two points (x1,y1)(x1, y1) and (x2,y2)(x2, y2) is given by:

Midpoint = ((x1+x2)2,(y1+y2)2)(\dfrac{(x1 + x2)}{2}, \dfrac{(y1 + y2)}{2})

Given that the midpoint of PQ is (1,5)(1, 5), we have: (1,5)=((p+q)/2,((5p+4)−(3/4)q+1)/2)(1, 5) = ((p + q)/2, ((5p + 4) - (3/4)q + 1)/2)

Now, we can write two equations based on the coordinates of the midpoint:

  1. (p+q)/2=1(p + q)/2 = 1
  2. ((5p+4)−(34)q+1)2=5((5p + 4) - \dfrac{(\dfrac{3}{4})q + 1)}{2} = 5

Step 3:

Solve the system of equations to find p and q. From equation 1, we get: p+q=2p + q = 2

From equation 2, we can simplify and get:5p−(3/4)q+5=105p−(3/4)q=5get: 5p - (3/4)q + 5 = 10 5p - (3/4)q = 5

Now, let's multiply the second equation by 4 to get : 20p−3q=2020p - 3q = 20

Now, we can use the first equation (p+q=2)(p + q = 2) to solve for q:q=2−pq: q = 2 - p

Substitute this value of qq into the equation (20p−3q=20):20p−3(2−p)=2020p−6+3p=2023p=26p=2623(20p - 3q = 20): 20p - 3(2 - p) = 20 20p - 6 + 3p = 20 23p = 26 p = \dfrac{26}{23}

Now that we have the value of p, we can find the value of q:q=2−2623q=4623 q = 2 - \dfrac{26}{23} q = \dfrac{46}{23}

Step 4:

Find the equation of the line passing through P and Q. The slope of the line passing through P and Q is given by:

mm =(y2−y1)(x2−x1)= \dfrac{(y2 - y1)}{(x2 - x1)}

Using the coordinates of P(p,5p+4)P (p, 5p + 4) and Q(q,−(34)q+1)Q (q, -(\dfrac{3}{4})q + 1),

we get: m=((−(3/4)q+1)−(5p+4)(q−p)m = (\dfrac{(-(3/4)}q + 1) - \dfrac{(5p + 4)}{(q - p)}

Now, substitute the values of p and q:

m=((−(3/4)(46/23)+1)−(5(26/23)+4))/((46/23)−(26/23)) m= ((-(3/4)(46/23) + 1) - (5(26/23) + 4)) / ((46/23) - (26/23))

m=((−69/92+1)−(130/23+92/23))/(20/23)m= ((-69/92 + 1) - (130/23 + 92/23)) / (20/23)

m=((23/92−222/23)/(20/23)m = ((23/92 - 222/23) / (20/23)

m=((23−5064)/920)/(20/23)m= ((23 - 5064) / 920) / (20/23)

m=(−5041/920)/(20/23)m = (-5041/920) / (20/23)

m=−5041/920∗23/20m = -5041/920 * 23/20

m=−1073m = \dfrac{-107}{3} (_Ans)