{"id":37037,"date":"2022-12-01T10:28:25","date_gmt":"2022-12-01T04:58:25","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=37037"},"modified":"2022-12-01T10:28:25","modified_gmt":"2022-12-01T04:58:25","slug":"ts-inter-1st-year-maths-1b-the-straight-lines-formulas","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-maths-1b-the-straight-lines-formulas\/","title":{"rendered":"TS Inter 1st Year Maths 1B The Straight Lines Formulas"},"content":{"rendered":"

Learning these TS Inter 1st Year Maths 1B Formulas<\/a> Chapter 3 The Straight Lines will help students to solve mathematical problems quickly.<\/p>\n

TS Inter 1st Year Maths 1B The Straight Lines Formulas<\/h2>\n

\u2192 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.<\/p>\n

\u2192 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.<\/p>\n

\u2192 The equation of X – axis is y = 0.<\/p>\n

\u2192 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.<\/p>\n

\u2192 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.<\/p>\n

\u2192 Equation of Y- axis is x = 0.<\/p>\n

\u2192 If a non vertical straight line L makes an angle \u03b8 with X – axis measured anti-clockwise from the positive direction of the X – axis then tan \u03b8 is called the slope or gradient of the line L denoted by ‘m’.<\/p>\n

\"TS<\/p>\n

\u2192 Slope of horizontal line is 0 since tan 0 – 0 and slope of vertical line is not defined.<\/p>\n

\u2192 If m1<\/sub>, m2<\/sub>, are slopes of two lines and \u03b8 is called the angle between them then tan \u03b8 = \\(\\left(\\frac{m_1-m_2}{1+m_1 m_2}\\right)\\)<\/p>\n

\u2192 If two lines are parallel then slopes are equal, m1<\/sub> = m2<\/sub>, and if two lines are perpendicular then m1<\/sub>. m1<\/sub> = -1.<\/p>\n

\u2192 Equation of a line passing through (x1<\/sub>; y1<\/sub>) with slope m’ is y – y1<\/sub> = m (x – x1<\/sub>).<\/p>\n

\u2192 Equation of a line passing through origin with slope in is y = mx.<\/p>\n

\u2192 Equation of a line passing through the points A (x1<\/sub>, y1<\/sub>) and B (x2<\/sub>, y2<\/sub>) is \\(\\frac{y-y_1}{y_1-y_2}=\\frac{x-x_1}{x_1-x_2}\\)<\/p>\n

\u2192 Equation of a line with Y – intercept ‘c’ and slope m is y = mx + c.<\/p>\n

\u2192 Equation of a line in intercept form is \\(\\frac{x}{a}+\\frac{y}{b}\\) = 1.<\/p>\n

\u2192 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<\/p>\n

\u2192 Area of the triangle formed by the line ax + by + c = 0 with coordinate axes is \\(\\frac{c^2}{2|a b|}\\).<\/p>\n

\u2192 Equation of a line in normal form or perpendicular form is x cos \u03b1 + y sin \u03b1 = 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.<\/p>\n

\u2192 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}}\\)<\/p>\n

\u2192 Perpendicular distance from (x1<\/sub>; y1<\/sub>) to the line ax + by + c = 0 is \\(\\frac{\\left|a x_1+b y_1+c\\right|}{\\sqrt{a^2+b^2}}\\)<\/p>\n

\u2192 Perpendicular distance from origin to the line ax + by + c = 0 is points A (x1<\/sub>, y1<\/sub>) and B (x2<\/sub>, y2<\/sub>) is \\(\\frac{|c|}{\\sqrt{a^2+b^2}}\\)<\/p>\n

\u2192 The ratio in which the line L = ax + by + c = 0 (ab \u2260 0) divides the line segment AB joining points A(x1<\/sub>, y1<\/sub>) and B(x2<\/sub>, y2<\/sub>) is \\(-\\left(\\frac{a x_1+b y_1+c}{a x_2+b y_2+c}\\right)=-\\frac{L_{11}}{L_{22}}\\)
\nIf L11<\/sub> and L22<\/sub> are having same sign or opposite sign then the points on same side or opposite sides of the line L = 0.<\/p>\n

\u2192 If (h, k) is the foot of the perpendicular from (x1<\/sub>, y1<\/sub>) 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)\\)<\/p>\n

\u2192 If (h, k) is the image of the point (x1<\/sub>, y1<\/sub>) 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)\\)<\/p>\n

\u2192 The point of intersection of lines a1<\/sub>x + b1<\/sub>y + c1<\/sub> = 0 and a2<\/sub>x + b2<\/sub>y + c2<\/sub> = 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)\\)<\/p>\n

\"TS<\/p>\n

\u2192 If angle between lines a1<\/sub>x + b1<\/sub>y + c1<\/sub> = 0 and a2<\/sub>x + b2<\/sub>y + c2<\/sub> = 0 is 0 where (0 \u2264 \u03b8 \u2264 \u03c0), then
\ncos \u03b8 = \\(\\frac{a_1 a_2+b_1 b_2}{\\sqrt{a_1^2+b_1^2} \\sqrt{a_2^2+b_2^2}}\\)
\nsin \u03b8 = \\(\\frac{a_1 b_2-a_2 b_1}{\\sqrt{a_1^2+b_1^2} \\sqrt{a_2^2+b_2^2}}\\)
\nand tan \u03b8 = \\(\\frac{a_1 b_2-a_2 b_1}{a_1 a_2+b_1 b_2}\\)<\/p>\n