PAIR OF STRAIGHT LINES SOLVED PROBLEMS

Problem 1 :

The slope of one of the straight lines ax2 + 2hxy + by2 = 0 is twice that of the other, show that 8h2 = 9ab.

Solution :

If the given pair of straight line is in the form ax2 + 2hxy + by2  =  0, then their exists the following relationship.

m1 + m2  =  -2h/b  ------(1)

m1 m2  =  a/b  ------(2)

From the question given, we know that

Slope of one straight line  =  2 (slope of other)

m1  =  2 m2

2m2 + m2  =  -2h/b

3 m=  -2h/b

m2  =  -2h/3b

Now we are going to apply m1  =  2 m2 and m2  =  -2h/3b in the second equation.

2 (-2h/3b)(-2h/3b)  =  a/b

8h2/9b2  =  a/b

8h2  =  9ab

Problem 2 :

The slope of one of the straight lines ax2 + 2hxy + by2 = 0 is three times the other, show that 3h2 = 4ab.

Solution :

If the given pair of straight line is in the form ax2 + 2hxy + by2  =  0, then their exists the following relationship.

m1 + m2  =  -2h/b  ------(1)

m1 m2  =  a/b  ------(2)

From the question given, we know that

Slope of one straight line  =  3 (slope of other)

m1  =  3 m2

3m2 + m2  =  -2h/b

4 m2  =  -2h/b

m2  =  -2h/4b  =  -h/2b

Now we are going to apply m1  =  3 m2 and m2  =  -h/2b in the second equation.

m1 m2  =  a/b

3 m2m2  =  a/b

3 (m2)2  =  a/b

3(-h/2b)2  =  a/b

3h2/4b2  =  a/b

3h2  4ab

Problem 3 :

A ΔOPQ is formed by the pair of straight lines x2 −4xy +y2 = 0 and the line PQ. The equation of PQ is x + y − 2 = 0. Find the equation of the median of the triangle ΔOPQ drawn from the origin O.

Solution :

We cannot find the factors from the pair of straight lines. By solving the given equations, we may get the vertices P and Q.

x2 −4xy + y2 = 0   --------(1)

x + y − 2 = 0   --------(2)

Apply y  =  -x + 2 in the first equation, we get

x2 − 4x (-x + 2) + (-x + 2)2 = 0

x2 + 4x2 - 8x + x2 - 4x + 4  =  0

6x2 - 12x + 4  =  0

3x2 - 6x + 2  =  0

x  =  (-b ± √b2 - 4ac) / 2a

x  =  [6 ± √62 - 4(3)(2)] / 2(3)

x  =  [6 ± √(36 - 24)] / 2(3)

x  =  [6 ± √12] / 6

x  =  [6 ± 2√3] / 6

x  =  1 ± (√3/3)

 x  =  1 + (√3/3)y  =  -1 - (√3/3) + 2  =  (-3 - √3 + 6)/3y  =  (3 - √3)/3P( 1 + (√3/3), 1 - (√3/3)) x  =  1 - (√3/3)y  =  -1 + (√3/3) + 2  =  (-3 + √3 + 6)/3y  =  (3 + √3)/3Q( 1 - (√3/3), 1 + (√3/3))

Midpoint of PQ :

=  (x1 + x2)/2,  (y1 + y2)/2

=  2/2, 2/2

Midpoint of PQ is (1, 1)

Equation of median PQ :

O (0, 0) M (1, 1)

(y - 0)/(1 - 0)  =  (x - 0)/(1 - 0)

y/1  =  x/1

y  =  x

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