Advertisement Remove all ads

Advertisement Remove all ads

Advertisement Remove all ads

Find ‘k' if the sum of slopes of lines represented by equation x^{2}+ kxy - 3y^{2} = 0 is twice their product.

Advertisement Remove all ads

#### Solution

Consider the given equation of the straight lines

2 2 x +kxy −3y =0

Sum of the slopes is given by the formula=-2H/B

Comparing the given equation with the

standard equation,

Ax^{2}+ 2Hxy + By^{2}= 0, we have,

H=k/2 ,A=1, B=- 3

Thus, sum of the slopes= `(-2(k/2))/-3=k/3`

Product of the slopes = `A/B=1/-3`

Also given that, sum of the slopes is twice their product

`k/3=2(1/-3)`

`k=-2`

Concept: Acute Angle Between the Lines

Is there an error in this question or solution?