TS Inter 1st Year Maths 1B Pair of Straight Lines Formulas

Learning these TS Inter 1st Year Maths 1B Formulas Chapter 4 Pair of Straight Lines will help students to solve mathematical problems quickly.

TS Inter 1st Year Maths 1B Pair of Straight Lines Formulas

→ If a b and h are not all zero then the equation H ≡ ax² + 2hxy + by² = 0 represents a pair of straight lines if and only if h² ≥ ab.

→ If ax² + 2hxy + by² = 0 represent a pair of lines passing through the origin then the sum of the slopes of lines is \(\frac{-2h}{b}\) and product of the slopes is \(\frac{a}{b}\).
i.e.., if ax² + 2hxy + by² = (y – m₁x) (y – m₂x) then m₁ + m₂ = \(\frac{-2h}{b}\) and m₁ m₂ = \(\frac{a}{b}\).

→ If θ is the angle between the lines represented by ax² + 2hxy + ² = 0 then
cos θ = \(\frac{a+b}{\sqrt{(a-b)^2+4 h^2}}\) and tan θ = \(\frac{2 \sqrt{\mathrm{h}^2-a b}}{a+b}\)

• If h² = ab then ax² + 2hxy + by² = 0 represents coincident or parallel lines.
• ax² + 2hxy + by² = 0 represents a pair of perpendicular lines ⇔ a + b = 0 i.e., coefficient of x² + coefficient of y² = 0.

→ (i) The equation of pair < >f lines passing! Iirough origin and perpendicular to ax² + 2hxy + by² = 0 is bx² – 2hxy + ay² = 0,
(ii) The equation of pair of lines passing through (x₁, y₁) and perpendicular to ax² + 2hxy + by² = 0 is b(x – x₁)² – 2h (x – x₁) (y – y₁) – a(y – y₁)² = 0.
(iii) The equation of pair of lines passing through (x₁, y₁) and parallel to ax² + 2hxy + by² = 0 is a(x – x₁)²+ 2h (x – x₁) (y – y₁) + b(y – y₁)² = 0.

→ The equation of bisectors of angles between the lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 is \(\frac{a_1 x+b_1 y+c_1}{\sqrt{a_1^2+b_1^2}}\) = \(\frac{(a_2 x+b_2 y+c_2)}{\sqrt{a_2^2+b_2^2}}\)

→ The equation to the pair of bisectors of angles between the pair of lines ax² + 2hxy + by² = 0 is h(x² – y²) – (a – b)xy

→ The area of the triangle formed by ax² + 2hxy + by² = 0 and lx + my + n = 0 is \(\frac{n^2 \sqrt{h^2-a b}}{\left|a m^2-2 h l m+b l^2\right|}\)

→ The product of the perpendiculars from (α, β) to the pair of lines ax² + 2hxy + by² = 0 is \(\frac{\left|a \alpha^2+2 h \alpha \beta+b \beta^2\right|}{\sqrt{(a-b)^2+4 h^2}}\)

→ The line ax + by + c – 0 and pair of lines (ax + by)² – 3(bx – ay)² = 0 form an equilateral triangle and the area is \(\frac{c^2}{\sqrt{3}\left(a^2+b^2\right)}\) units

→ If S = ax² + 2hxy + by² + 2gx + 2fy + c = 0 represent the equation of pair of lines then

• Δ = abc + 2fgh – af² – bg² – ch² = 0
• h² ≥ ab, g² ≥ ac, f² ≥ be

→ The point of intersection of the pair of lines S ≡ 0 is \(\left(\frac{h f-b g}{a b-h^2}, \frac{g h-a f}{a b-h^2}\right)\)

→ If S ≡ ax² + 2hxy + by² + 2gx + 2fy + c = 0 represent a pair of parallel lines then

• h² = ab
• bg² = af²
• distance between them is 2\(\sqrt{\frac{g^2-a c}{a(a+b)}}\) (or) 2\(\sqrt{\frac{f^2-b c}{a(a+b)}}\)

→ The equation to the pair of lines joinmg the ongin to the points of intersection of the curve ax² + 2hxy + by² + 2gx + 2fy + c = 0 and the line lx + my + n = 0 is obtained by homogenisation ax² + 2hxy + by² + 2gx\(\left(\frac{l x+m y}{-n}\right)\) + 2fy\(\left(\frac{l x+m y}{-n}\right)\) + c\(\left(\frac{l x+m y}{-n}\right)^2\) = 0