Question

Show that the equation 12x² – 10xy + 2y² + 14x – 5y + 2 = 0 represents a pair of straight lines and also find the separate equations of the straight lines.

Answer 1

Comparing 12x² – 10xy + 2y² + 14x – 5y + 2 = 0 with ax² + 2hxy + by² + 2gh + 2fy + c = 0

We get a = 12, 2h = -10, (or) h = -5, b = 2, 2g = 14 (or) g = 7, 2f = -5 (or) f = `- 5/2`, c = 2

Condition for the given equation to represent a pair of straight lines is `|(a,h,g),(h,b,f),(g,f,c)|` = 0

`|(a,h,g),(h,b,f),(g,f,c)| = |(12,-5,7),(-5,2,(-5)/2),(7,(-5)/2,2)|`

`= 1/2 xx 1/2 |(12,-5,7),(-10,4,-5),(14,-5,4)|`   ....`[("R"_2 -> 2"R"_2),("R"_3 -> 2"R"_3)]`

`= 1/4` [12(16 – 25) + 5(-40 + 70) + 7(50 – 56)]
= `1/4` [12(-9) + 5(30) + 7(-6)]
= `1/4` [-108 + 150 – 42]
= `1/4` [0]
= 0
∴ The given equation represents a pair of straight lines.

Consider 12x² – 10xy + 2y² = 2[6x² – 5xy + y²] = 2[(3x – y)(2x – y)] = (6x – 2y)(2x – y)

Let the separate equations be 6x – 2y + l = 0, 2x – y + m = 0

To find l, m

Let 12x² – 10xy + 2y² + 14x – 5y + 2 = (6x – 2y + l) (2x – y + m) ……. (1)

Equating coefficient of y on both sides of (1) we get
2l + 6m = 14 (or) l + 3m = 7 ………… (2)

Equating coefficient of x on both sides of (1) we get

-l – 2m = -5 ……… (3)

(2) + (3) ⇒ m = 2

Using m = 2 in (2) we get

l + 3(2) = 7

l = 7 – 6

l = 1

∴ The separate equations are 6x – 2y + 1 = 0, 2x – y + 2 = 0.