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 ≡ ax^{2} + 2hxy + by^{2} = 0 represents a pair of straight lines if and only if h^{2} ≥ ab.

→ If ax^{2} + 2hxy + by^{2} = 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 ax^{2} + 2hxy + by^{2} = (y – m_{1}x) (y – m_{2}x) then m_{1} + m_{2} = \(\frac{-2h}{b}\) and m_{1} m_{2} = \(\frac{a}{b}\).

→ If θ is the angle between the lines represented by ax^{2} + 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 h
^{2}= ab then ax^{2}+ 2hxy + by^{2}= 0 represents coincident or parallel lines. - ax
^{2}+ 2hxy + by^{2}= 0 represents a pair of perpendicular lines ⇔ a + b = 0 i.e., coefficient of x^{2}+ coefficient of y^{2}= 0.

→ (i) The equation of pair < >f lines passing! Iirough origin and perpendicular to ax^{2} + 2hxy + by^{2} = 0 is bx^{2} – 2hxy + ay^{2} = 0,

(ii) The equation of pair of lines passing through (x_{1}, y_{1}) and perpendicular to ax^{2} + 2hxy + by^{2} = 0 is b(x – x_{1})^{2} – 2h (x – x_{1}) (y – y_{1}) – a(y – y_{1})^{2} = 0.

(iii) The equation of pair of lines passing through (x_{1}, y_{1}) and parallel to ax^{2} + 2hxy + by^{2} = 0 is a(x – x_{1})^{2}+ 2h (x – x_{1}) (y – y_{1}) + b(y – y_{1})^{2} = 0.

→ The equation of bisectors of angles between the lines a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 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 ax^{2} + 2hxy + by^{2} = 0 is h(x^{2} – y^{2}) – (a – b)xy

→ The area of the triangle formed by ax^{2} + 2hxy + by^{2} = 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 ax^{2} + 2hxy + by^{2} = 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 = ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 represent the equation of pair of lines then

- Δ = abc + 2fgh – af
^{2}– bg^{2}– ch^{2}= 0 - h
^{2}≥ ab, g^{2}≥ ac, f^{2}≥ 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 ≡ ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 represent a pair of parallel lines then

- h
^{2}= ab - bg
^{2}= af^{2} - 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 ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 and the line lx + my + n = 0 is obtained by homogenisation ax^{2} + 2hxy + by^{2} + 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