# FIND MISSING VALUE FROM THE EQUATION OF PAIR OF STRAIGHT LINES

Example 1 :

Find p and q, if the following equation represents a pair of perpendicular lines

6x2 + 5xy − py2 + 7x + qy − 5 = 0

Solution :

If two lines are perpendicular, then a + b  =  0

6x2 + 5xy − py2 + 7x + qy − 5 = 0

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

a  =  6, b  =  -p

6 - p  =  0

p  =  6

If the given line represent the pair of straight line, then it must satisfy the condition given below.

abc + 2fgh - af2 - bg2 - ch2  =  0

a = 6, b = -6, 2h = 5  ==> h  =  5/2, g  =  7/2, f  =  q/2, c = -5

6(-6)(-5)+2(q/2)(7/2)(5/2)+6(q/2)2+6(7/2)2-(-5)(5/2)2=0

180 + (35q/4) - 6(q2/4) + 6(49/4) + 5(25/4)  =  0

(720 + 35q - 6q2 + 294 + 125) / 4  =  0

1308 + 35q - 6q=  0

6q- 35q - 1139  =  0 6q- 102q + 67q - 1139  =  0

6q (q - 17) + 67 (q - 17)  =  0

(6q + 67) (q - 17)  =  0

6q + 67  =  0             q - 17  =  0

q  =  -67/6                q  =  17

Example 2 :

Find the value of k, if the following equation represents a pair of straight lines. Further, find whether these lines are parallel or intersecting, 12x2 + 7xy − 12y2 − x + 7y + k = 0

Solution :

To check the pair of straight lines is parallel or intersecting, we have to find the separate equation.

=  12x2 + 7xy − 12y2

=  12x2 + 16xy - 9xy − 12y2

=  4x(3x + 4y) - 3y(3x + 4y)

=  (4x - 3y) (3x + 4y)

Separate equations are (4x - 3y + l) (3x + 4y + m)

12x2 + 7xy − 12y2 − x + 7y + k  =  (4x - 3y + l) (3x + 4y + m)

-1  =  4m + 3l  ----(1)

7  =  -3m + 4l ----(2)

(1) ⋅ 3 + (2) ⋅ 4  ==>  12m + 9l - 12m + 16l  =  -3 + 28

25l  =  25  ==>  l  =  1

By applying the value of l in the first equation, we get

4m + 3(1)  =  -1

4m  =  -1 - 3

m  =  -1

(4x - 3y + 1) (3x + 4y - 1)

k  =  lm  =  1(-1)  =  -1

Hence the value of k is -1

Example 3 :

For what value of k does the equation 12x2+2kxy+2y2+11x−5y+2 = 0 represent two straight lines.

Solution :

12x2+2kxy+2y2+11x−5y+2 = 0

ax+ 2hxy + by+ 2gx + 2fy + c = 0

a  =  12, 2h  =  2k ==> h = k, b = 2, g = 11/2, f = -5/2, c = 2

abc + 2fgh - af2 - bg2 - ch2  =  0

12(2)(2) + 2(-5/2)(11/2)k - 12(-5/2)2 - 2(11/2)2 - 2k2  =  0

48 + (-55k/2) - 75 - (121/2) - 2k2  =  0

96 - 55k - 150 - 121 - 4k2  =  0

4k2+ 55k + 175  =  0 k + 5  =  0k  =  -5 4k + 35  =  0k  =  -35/4

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