Learning these TS Inter 1st Year Maths 1B Formulas Chapter 3 The Straight Lines will help students to solve mathematical problems quickly.

## TS Inter 1st Year Maths 1B The Straight Lines Formulas

→ The equation of a horizontal line which is parallel to X – axis and at a distance of k’ from X – axis and lying above X – axis is given by y = k.

→ Similarly, y = -k is the equation of the horizontal line which is at a distance of k from X – axis and lying below X -axis.

→ The equation of X – axis is y = 0.

→ The equation of a vertical line which is parallel to Y – axis and at a distance of k from Y – axis and lying left of Y – axis is x = k.

→ Similarly, x = -k is the equation of the vertical line which is at a distance of k units from Y – axis and lying right of Y – axis is x = -k.

→ Equation of Y- axis is x = 0.

→ If a non vertical straight line L makes an angle θ with X – axis measured anti-clockwise from the positive direction of the X – axis then tan θ is called the slope or gradient of the line L denoted by ‘m’.

→ Slope of horizontal line is 0 since tan 0 – 0 and slope of vertical line is not defined.

→ If m_{1}, m_{2}, are slopes of two lines and θ is called the angle between them then tan θ = \(\left(\frac{m_1-m_2}{1+m_1 m_2}\right)\)

→ If two lines are parallel then slopes are equal, m_{1} = m_{2}, and if two lines are perpendicular then m_{1}. m_{1} = -1.

→ Equation of a line passing through (x_{1}; y_{1}) with slope m’ is y – y_{1} = m (x – x_{1}).

→ Equation of a line passing through origin with slope in is y = mx.

→ Equation of a line passing through the points A (x_{1}, y_{1}) and B (x_{2}, y_{2}) is \(\frac{y-y_1}{y_1-y_2}=\frac{x-x_1}{x_1-x_2}\)

→ Equation of a line with Y – intercept ‘c’ and slope m is y = mx + c.

→ Equation of a line in intercept form is \(\frac{x}{a}+\frac{y}{b}\) = 1.

→ Reduction of a straight line ax + by + c = 0 in intercept form is \(\frac{x}{-\left(\frac{c}{a}\right)}+\frac{y}{-\left(\frac{c}{b}\right)}\) = 1

→ Area of the triangle formed by the line ax + by + c = 0 with coordinate axes is \(\frac{c^2}{2|a b|}\).

→ Equation of a line in normal form or perpendicular form is x cos α + y sin α = p where p is the length of the perpendicular from origin to line and a. is the angle made by the perpendicular with + ve X – axis.

→ Reduction of the equation ax + by + c = 0 of a line to the normal form is \(\pm\left(\frac{a}{\sqrt{a^2+b^2}}\right) x+\left(\pm \frac{b}{\sqrt{a^2+b^2}}\right)=\frac{\pm c}{\sqrt{a^2+b^2}}\)

→ Perpendicular distance from (x_{1}; y_{1}) to the line ax + by + c = 0 is \(\frac{\left|a x_1+b y_1+c\right|}{\sqrt{a^2+b^2}}\)

→ Perpendicular distance from origin to the line ax + by + c = 0 is points A (x_{1}, y_{1}) and B (x_{2}, y_{2}) is \(\frac{|c|}{\sqrt{a^2+b^2}}\)

→ The ratio in which the line L = ax + by + c = 0 (ab ≠ 0) divides the line segment AB joining points A(x_{1}, y_{1}) and B(x_{2}, y_{2}) is \(-\left(\frac{a x_1+b y_1+c}{a x_2+b y_2+c}\right)=-\frac{L_{11}}{L_{22}}\)

If L_{11} and L_{22} are having same sign or opposite sign then the points on same side or opposite sides of the line L = 0.

→ If (h, k) is the foot of the perpendicular from (x_{1}, y_{1}) to the line ax + by + c = 0. then \(\frac{h-x_1}{a}=\frac{k-y_1}{b}=-\left(\frac{a x_1+b y_1+c}{a^2+b^2}\right)\)

→ If (h, k) is the image of the point (x_{1}, y_{1}) with respect to the line ax + by + c = 0, then \(\frac{h-x_1}{a}=\frac{k-y_1}{b}=-2\left(\frac{a x_1+b y_1+c}{a^2+b^2}\right)\)

→ The point of intersection of lines a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 is \(\left(\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}, \frac{c_1 a_2-a_1 c_2}{a_1 b_2-a_2 b_1}\right)\)

→ If angle between lines a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 is 0 where (0 ≤ θ ≤ π), then

cos θ = \(\frac{a_1 a_2+b_1 b_2}{\sqrt{a_1^2+b_1^2} \sqrt{a_2^2+b_2^2}}\)

sin θ = \(\frac{a_1 b_2-a_2 b_1}{\sqrt{a_1^2+b_1^2} \sqrt{a_2^2+b_2^2}}\)

and tan θ = \(\frac{a_1 b_2-a_2 b_1}{a_1 a_2+b_1 b_2}\)

- Lines are perpendicular ⇔ a
_{1}a_{2}+ b_{1}b_{2}= 0 - Lines are parallel ⇔ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\)

→ The equation of a line passing through (x_{1}, y_{1}) and parallel to the line ax + by + c = 0 is a (x – x_{1}) – b (y – y_{1}) = 0.

→ The equation of a line passing through (x_{1}, y_{1}) and perpendicular to ax + by + c = 0 is b(x – x_{1}) – a(y – y_{1}) = 0.

→ If a_{1}x + b_{1}y + c_{1} = 0. a_{2}x + b_{2}y + c_{2} = 0, and a_{3}x + b_{3}y + c_{3} = 0 represent three lines, no two of which are parallel, then a necessary and sufficient condition for these lines to be concurrent is Σa_{1}(b_{2}c_{3} – b_{3}c_{2}) = 0 (0r) \(\left|\begin{array}{lll}

a_1 & b_1 & c_1 \\

a_2 & b_2 & c_2 \\

a_3 & b_3 & c_3

\end{array}\right|\) = 0

→ The distance between parallel lines a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 is \(\frac{\left|c_1-c_2\right|}{\sqrt{a^2+b^2}}\)