Question

Choose correct alternatives:

If the equation 3x² – 8xy + qy² + 2x + 14y + p = 1 represents a pair of perpendicular lines, then the values of p and q are respectively ______.

Options

• –3 and –7

• –7 and –3

• 3 and 7

• –7 and 3

Answer 1

If the equation 3x² – 8xy + qy² + 2x + 14y + p = 1 represents a pair of perpendicular lines, then the values of p and q are respectively –7 and –3.

Explanation:

3x² − 8xy + qy² + 2x + 14y + p = 1

Ax² + 2Hxy + By² + 2Gx + 2Fy + C = 0

3x² − 8xy + qy² + 2x + 14y + p = 1

3x² − 8xy + qy² + 2x + 14y + (p − 1) = 0

So the coefficients are:

• A=3

• 2H = −8 ⇒ H = −4

• B = q

• 2G = 2 ⇒ G = 1

• 2F = 14 ⇒ F = 7

• C = p − 1

A + B = 0

3 + q = 0 ⇒ q = −3

`|(A,H,G),(H,B,F),(G,F,C)| = 0`

`|(3,-4,1),(-4,-3,7),(1,7,p-1)| = 0`

`= 3 |(-3,7),(7,p-1)| -(-4)|(-4,7),(1,p-1)| +1|(-4,-3),(1,7)|`

3((−3) (p−1) − (7) (7)) = 3[−3 (p−1) −49] = 3[−3p + 3 − 49] = 3[−3p − 46] = −9p − 138

+4((−4) (p−1) − (7) (1)) = 4[−4(p − 1) −7] = 4[−4p + 4 − 7] = 4[−4p − 3] = −16p − 12

+1((−4) (7) − (−3) (1)) = −28 + 3 = −25

−9p − 138 − 16p − 12 − 25 = −25p − 175 = 0 ⇒ p = −7

p = −7, q = −3