{"id":35122,"date":"2022-08-25T10:44:27","date_gmt":"2022-08-25T05:14:27","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35122"},"modified":"2022-08-25T10:44:27","modified_gmt":"2022-08-25T05:14:27","slug":"inter-1st-year-maths-1b-the-straight-line-formulas","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-1st-year-maths-1b-the-straight-line-formulas\/","title":{"rendered":"Inter 1st Year Maths 1B The Straight Line Formulas"},"content":{"rendered":"<p>Use these <a href=\"https:\/\/apboardsolutions.guru\/inter-1st-year-maths-1b-formulas\/\">Inter 1st Year Maths 1B Formulas<\/a> PDF Chapter 3 The Straight Line to solve questions creatively.<\/p>\n<h2>Intermediate 1st Year Maths 1B The Straight Line Formulas<\/h2>\n<p>\u2192 Equation of a horizontal line is y = k<\/p>\n<p>\u2192 Equation of a vertical line is x = h<\/p>\n<p>\u2192 If &#8216;\u03b8 &#8216; is the inclination of a line then slope = m = tan \u03b8<\/p>\n<p>\u2192 If m<sub>1<\/sub> m<sub>2<\/sub> are slopes of two lines and &#8216;\u03b8 &#8216; is angle between them then tan \u03b8 = \\(\\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\\)<\/p>\n<p>\u2192 If two lines are parallel then their slopes are equal i.e., m<sub>1<\/sub> = m<sub>2<\/sub><\/p>\n<p>\u2192 If two lines are perpendicular, then m<sub>1<\/sub>m<sub>2<\/sub> = -1<\/p>\n<p>\u2192 Equation of a line passing through (x<sub>1<\/sub>, y<sub>1<\/sub>) with slope &#8216;m&#8217; is y &#8211; y<sub>1<\/sub> = m(x &#8211; x<sub>1<\/sub>)<\/p>\n<p>\u2192 Equation of a line passing through origin with slope m is y = mx<\/p>\n<p>\u2192 Equation of the line passing through (x<sub>1<\/sub>, y<sub>1<\/sub>) and (x<sub>2<\/sub>, y<sub>2<\/sub>) is \\(\\) (or) (x &#8211; x<sub>1<\/sub>)(y<sub>1<\/sub> &#8211; y<sub>2<\/sub>) = (y &#8211; y<sub>1<\/sub>)(x<sub>1<\/sub> &#8211; x<sub>2<\/sub>)<\/p>\n<p>\u2192 Equation of the line with slope&#8217;m&#8217; and having<\/p>\n<ul>\n<li>y-intercept &#8216;c&#8217; is y = mx + c (slope intercept form)<\/li>\n<li>x-intercept &#8216;a&#8217; is y = m(x &#8211; a)<\/li>\n<\/ul>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas\" width=\"196\" height=\"17\" \/><\/p>\n<p>\u2192 Equation of a line in intercept form is \\(\\frac{x}{a}+\\frac{y}{b}\\) = 1<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19033\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-1.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 1\" width=\"249\" height=\"210\" \/><\/p>\n<ul>\n<li>Intersecting point on X &#8211; axis = (a, 0)<\/li>\n<li>Intersecting point on Y &#8211; axis = (0, b)<\/li>\n<\/ul>\n<p>\u2192 Area of \u0394<sup>le<\/sup> formed by the line with co-ordinate axis is = \\(\\frac{1}{2}\\)|ab|<\/p>\n<p>\u2192 Area of \u0394<sup>le<\/sup> formed by the line ax + by + c = 0 with co-ordinate axis is \\(\\frac{c^{2}}{2|a b|}\\)<\/p>\n<p>\u2192 If a point (x<sub>1<\/sub>, y<sub>1<\/sub>) divides the line segment between the co-ordinate axes in the ratio l: m then equation of the line is \\(\\frac{m x}{x_{1}}+\\frac{l y}{y_{1}}\\) = l + m<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19032\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-2.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 2\" width=\"207\" height=\"179\" \/><\/p>\n<p>\u2192 Equation of a line in normal form or perpendicular form is x cos \u03b1 + y sin \u03b1 = p where &#8216;\u03b1&#8217; is the angle made by the perpendicular with +ve X &#8211; axis. &#8216;p&#8217; is length of normal form origin to the line.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19031\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-3.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 3\" width=\"191\" height=\"186\" \/><\/p>\n<p>\u2192 Perpendicular distance from (x<sub>1<\/sub>, y<sub>1<\/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<p>\u2192 Perpendicular distance from origin to the line ax + by + c = 0 is \\(\\frac{|c|}{\\sqrt{a^{2}+b^{2}}}\\)<\/p>\n<p>\u2192 The ratio in which the line L ax + by + c = 0 (ab \u2260 0) divides the line segment AB joining the points A(x<sub>1<\/sub>, y<sub>1<\/sub>) and B(x<sub>2<\/sub>, y<sub>2<\/sub>) is &#8211;\\(\\left(\\frac{a x_{1}+b y_{1}+c}{a x_{2}+b y_{2}+c}\\right)\\) or &#8211;\\(\\frac{L_{11}}{L_{22}}\\)<br \/>\nwhere L<sub>11<\/sub> = a<sub>1<\/sub>x + b<sub>1<\/sub>y + c; L<sub>22<\/sub> = a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<\/p>\n<p>\u2192 If L<sub>11<\/sub>, L<sub>22<\/sub> are having same sign or opposite sign&#8217;s then those points lies same side or opposite sides of the line L = 0 respectively.<\/p>\n<p>\u2192 If the foot of the perpendicular from (x<sub>1<\/sub>, y<sub>1<\/sub>) to the line ax + by + c = 0 is (\u03b1, \u03b2) then \\(\\frac{\\alpha-x_{1}}{a}=\\frac{\\beta-y_{1}}{b}=\\frac{-\\left(a x_{1}+b y_{1}+c\\right)}{a^{2}+b^{2}}\\)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19030\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-4.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 4\" width=\"211\" height=\"136\" \/><\/p>\n<p>\u2192 If the image of(x<sub>1<\/sub>, y<sub>1<\/sub>) w.r.t. the line ax +by + c = 0 is (\u03b1, \u03b2) then<br \/>\n\\(\\frac{\\alpha-x_{1}}{a}=\\frac{\\beta-y_{1}}{b}=\\frac{-2\\left(a x_{1}+b y_{1}+c\\right)}{a^{2}+b^{2}}\\)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19029\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-5.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 5\" width=\"211\" height=\"157\" \/><\/p>\n<p>\u2192 The point of intersection of two lines a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0, a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0 is<br \/>\n\\(\\left(\\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}} \\cdot \\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\\right)\\)<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas\" width=\"196\" height=\"17\" \/><\/p>\n<p>\u2192 If \u03b8 |0 \u2264 \u03b8 \u2264 \u03c0| is angle between the lines a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0, a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0 then cos \u03b8 = \\(\\frac{a_{1} a_{2}+b_{1} b_{2}}{\\sqrt{a_{1}^{2}+b_{1}^{2}} \\sqrt{a_{2}^{2}+b_{2}^{2}}}\\)<br \/>\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}}}\\)<br \/>\nand tan \u03b8 = \\(\\frac{a_{1} b_{2}-a_{2} b_{1}}{a_{1} a_{2}+b_{1} b_{2}}\\)<\/p>\n<ul>\n<li>The lines are perpendicular \u21d4 a<sub>1<\/sub>a<sub>2<\/sub> + b<sub>1<\/sub>b<sub>2<\/sub> = 0<\/li>\n<li>Then lines are parallel \u21d4 \\(\\frac{a_{1}}{a_{2}}=\\frac{b_{1}}{b_{2}}\\)<\/li>\n<\/ul>\n<p>\u2192 The equation of a line passing through the point (x<sub>1<\/sub>, y<sub>1<\/sub>) and parallel to the line ax + by + c = 0 is a(x &#8211; x<sub>1<\/sub>) + b(y &#8211; y<sub>1<\/sub>) = 0<\/p>\n<p>\u2192 The equation of a line passing through the point (x<sub>1<\/sub>, y<sub>1<\/sub>) and perpendicular to ax + by + c = 0 is b(x &#8211; x<sub>1<\/sub>) &#8211; a(y &#8211; y<sub>1<\/sub>) = 0<\/p>\n<p>\u2192 The distance between the two parallel straight lines a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0 and a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>2<\/sub> = 0 is \\(\\frac{\\left|c_{1}-c_{2}\\right|}{\\sqrt{a_{1}^{2}+b_{1}^{2}}}\\)<\/p>\n<p>\u2192 Area of the parallelogram formed by the lines a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0,a <sub>2<\/sub>x + b<sub>2<\/sub>y + d<sub>1<\/sub>= 0, a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0 and a<sub>2<\/sub>x + b<sub>2<\/sub>y + d<sub>2<\/sub> = 0 is \\(\\left|\\frac{\\left(d_{1}-c_{1}\\right)\\left(d_{2}-c_{2}\\right)}{a_{1} b_{2}-a_{2} b_{1}}\\right|\\)<\/p>\n<p>\u2192 Equation of the line parallel to X &#8211; axis and passing through (x<sub>1<\/sub>, y<sub>1<\/sub>) is y = y<sub>1<\/sub><\/p>\n<p>\u2192 Equation of the line parallel to Y- axis and passing through (x<sub>1<\/sub>, y<sub>1<\/sub>) is x = x<sub>1<\/sub><\/p>\n<p><strong>Concurrent lines-properties related to a Triangle<br \/>\n<\/strong>Theorem:<br \/>\nThe medians of a triangle are concurrent.<br \/>\nProof:<br \/>\nLet A(x<sub>1<\/sub>, y<sub>1<\/sub>), B(x<sub>2<\/sub>, y<sub>2<\/sub>), C(x<sub>3<\/sub>, y<sub>3<\/sub>) be the vertices of the triangle<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19028\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-6.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 6\" width=\"277\" height=\"188\" \/><br \/>\nLet D,E,F be the mid points of BC, CA, AB respectively<br \/>\n\u2234 D = \\(\\left(\\frac{x_{2}+x_{3}}{2}, \\frac{y_{2}+y_{3}}{2}\\right)\\)<br \/>\nE = \\(\\left(\\frac{x_{3}+x_{1}}{2}, \\frac{y_{3}+y_{1}}{2}\\right)\\)<br \/>\nF = \\(\\left(\\frac{x_{1}+x_{2}}{2}, \\frac{y_{1}+y_{2}}{2}\\right)\\)<\/p>\n<p>Slope of \\(\\overline{A D}\\) is \\(\\frac{\\frac{y_{2}+y_{3}}{2}-y_{1}}{\\frac{x_{2}+x_{3}}{2}-x_{1}}=\\frac{y_{2}+y_{3}-2 y_{1}}{x_{2}+x_{3}-2 x_{1}}\\)<\/p>\n<p>Equation of \\(\\overline{A D}\\) is<br \/>\ny &#8211; y<sub>1<\/sub> = \\(\\frac{y_{2}+y_{3}-2 y_{1}}{x_{2}+x_{3}-2 x_{1}}\\)(x &#8211; x<sub>1<\/sub>)<br \/>\n\u21d2 (y &#8211; y<sub>1<\/sub>)(x<sub>2<\/sub> + x<sub>3<\/sub> &#8211; 2x<sub>1<\/sub>) = (x &#8211; x<sub>1<\/sub>)(y<sub>2<\/sub> + y<sub>3<\/sub> &#8211; 2y<sub>1<\/sub>)<br \/>\n\u21d2 L<sub>1<\/sub> = (x &#8211; x<sub>1<\/sub>)(y<sub>2<\/sub> + y<sub>3<\/sub> &#8211; 2y<sub>1<\/sub>) &#8211; (y &#8211; y<sub>1<\/sub>)(x<sub>2<\/sub> + x<sub>3<\/sub> &#8211; 2x<sub>1<\/sub>) = 0<\/p>\n<p>Similarly, the equations to \\(\\overline{B E}\\) and \\(\\overline{C F}\\) respectively are L<sub>2<\/sub> s (x &#8211; x<sub>2<\/sub>)(y<sub>3<\/sub> + y<sub>1<\/sub> &#8211; 2y<sub>2<\/sub>) &#8211; (y &#8211; y<sub>2<\/sub>) (x<sub>3<\/sub> + x<sub>1<\/sub> &#8211; 2x<sub>2<\/sub>) = 0.<br \/>\nL<sub>3<\/sub> = (x &#8211; x<sub>3<\/sub>)(y<sub>1<\/sub> + y<sub>2<\/sub> &#8211; 2y<sub>3<\/sub>) &#8211; (y &#8211; y<sub>3<\/sub>) (x<sub>1<\/sub> + x<sub>2<\/sub> &#8211; 2x<sub>3<\/sub>) = 0.<br \/>\nNow 1. L<sub>1<\/sub> + 1.L<sub>2<\/sub> + 1. L<sub>3<\/sub> = 0<br \/>\nThe medians L<sub>1<\/sub> = 0, L<sub>2<\/sub> = 0, L<sub>3<\/sub> = 0 are concurrent.<\/p>\n<p>Theorem:<br \/>\nThe altitudes of a triangle are concurrent.<br \/>\nProof:<br \/>\nLet A(x<sub>1<\/sub>, y<sub>1<\/sub>), B(x<sub>2<\/sub>, y<sub>2<\/sub>), C(x<sub>3<\/sub>, y<sub>3<\/sub>) be the vertices of the triangle ABC. Let AD, BE,CF be the altitudes.<br \/>\nSlope of \\(\\overline{B C}\\) is \\(\\frac{y_{3}-y_{2}}{x_{3}-x_{2}}\\) and AD \u22a5 BC<\/p>\n<p>Slope of the altitude through A is &#8211; \\(\\frac{x_{3}-x_{2}}{y_{3}-y_{2}}\\)<\/p>\n<p>Equation of the altitude through A is y &#8211; y<sub>1<\/sub> = \\(\\frac{x_{3}-x_{2}}{y_{3}-y_{2}}\\) (x &#8211; x<sub>1<\/sub>)<br \/>\n(y &#8211; y<sub>1<\/sub>) (y<sub>3<\/sub> &#8211; y<sub>2<\/sub>) = -(x &#8211; x<sub>1<\/sub>) (x<sub>3<\/sub> &#8211; x<sub>2<\/sub>)<br \/>\nL<sub>1<\/sub> = (x &#8211; x<sub>1<\/sub>)(x &#8211; x<sub>3<\/sub>) + (y &#8211; y<sub>1<\/sub>)(y &#8211; y<sub>3<\/sub>) = 0.<\/p>\n<p>Similarly equations of the altitudes through B,C are<br \/>\nL<sub>2<\/sub> = (x &#8211; x<sub>2<\/sub>) (x<sub>3<\/sub> &#8211; x<sub>1<\/sub>) + (y &#8211; y<sub>2<\/sub>) (y<sub>2<\/sub> &#8211; y<sub>3<\/sub>) = 0,<br \/>\nL<sub>3<\/sub> = (x &#8211; x<sub>3<\/sub>) (x<sub>1<\/sub> &#8211; x<sub>2<\/sub>) + (y &#8211; y<sub>3<\/sub>) (y<sub>1<\/sub> &#8211; y<sub>2<\/sub>) = 0.<br \/>\nNow 1.L<sub>1<\/sub> + 1.L<sub>2<\/sub> + 1.L<sub>3<\/sub> = 0<br \/>\nThe altitudes L<sub>1<\/sub> = 0, L<sub>2<\/sub> = 0, L<sub>3<\/sub> = 0 are concurrent.<\/p>\n<p>Theorem:<br \/>\nThe internal bisectors of the angles of a triangle are concurrent.<\/p>\n<p>Theorem:<br \/>\nThe perpendicular bisectors of the sides of a triangle are concurrent<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas\" width=\"196\" height=\"17\" \/><\/p>\n<p><strong>Point of Intersection of Two Straight Lines:<\/strong><br \/>\nTheorem:<br \/>\nThe point of intersection of the two non parallel lines<br \/>\na<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0, a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/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}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\\right)\\)<br \/>\nProof:<br \/>\nThe lines are not parallel ^ Slopes are not equal<br \/>\n\u21d2 \\(\\frac{-a_{1}}{b_{1}} \\neq \\frac{-a_{2}}{b_{2}}\\) \u21d2 a<sub>1<\/sub>b<sub>2<\/sub> \u2260 a<sub>2<\/sub>b<sub>1<\/sub> \u21d2 a<sub>1<\/sub>b<sub>2<\/sub> &#8211; a<sub>2<\/sub>b<sub>1<\/sub> \u2260 0<br \/>\nLet P (\u03b1, \u03b2) be the point of intersection. Then a<sub>1<\/sub> \u03b1 + b<sub>1<\/sub>\u03b2 + C<sub>1<\/sub> = 0 and a<sub>2<\/sub> \u03b1 + b<sub>2<\/sub>\u03b2 + c<sub>2<\/sub> = 0.<br \/>\nSolving by the method of cross multiplication<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19027\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-7.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 7\" width=\"315\" height=\"218\" \/><br \/>\nPoint of intersection is P = \\(\\left(\\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}}, \\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\\right)\\)<\/p>\n<p>Theorem:<br \/>\nThe ratio in which the line L = ax + by + c = 0 divides the line segment joining A(x<sub>1<\/sub>, y<sub>1<\/sub>), B(x<sub>2<\/sub>, y<sub>2<\/sub>) is -L<sub>11<\/sub> : L<sub>22<\/sub>, where L<sub>11<\/sub> = L(x<sub>1<\/sub>, y<sub>1<\/sub>) = ax<sub>1<\/sub> + by<sub>1<\/sub> + c and L<sub>22<\/sub> = L(x<sub>2<\/sub>, y<sub>2<\/sub>) = ax<sub>2<\/sub> + by<sub>2<\/sub> + c.<br \/>\nProof:<br \/>\nLet k : 1 be the ratio in which the line divides the line segment.<\/p>\n<p>The point which divides in the ratio k : 1 is P = \\(\\left(\\frac{k x_{2}+x_{1}}{k+1}, \\frac{k y_{2}+y_{1}}{k+1}\\right)\\)<\/p>\n<p>Since P lies on the line ax + by + c = 0 \u21d2 a\\(\\left(\\frac{k x_{2}+x_{1}}{k+1}\\right)\\) + b\\(\\left(\\frac{k y_{2}+y_{1}}{k+1}\\right)\\) + c = 0<br \/>\n\u21d2 a(kx<sub>2<\/sub> + x<sub>1<\/sub>) + b(ky<sub>2<\/sub> + y<sub>1<\/sub>) + c(k + 1) = 0<br \/>\n\u21d2 k(ax<sub>2<\/sub> + by<sub>2<\/sub> + c) = -(ax<sub>1<\/sub> + by<sub>1<\/sub> + c)<br \/>\n\u21d2 k = &#8211;\\(\\frac{\\left(a x_{1}+b y_{1}+c\\right)}{a x_{2}+b y_{2}+c}\\)<br \/>\nRequired Ratio = -(ax<sub>1<\/sub> + by<sub>1<\/sub> + c) : (ax<sub>2<\/sub> + by<sub>2<\/sub> + c)<br \/>\n= -L<sub>11<\/sub>: L<sub>22<\/sub>.<\/p>\n<p>Note:<br \/>\nThe points A,B lie in the same side or opposite side of the line L = 0 according as L<sub>11<\/sub>, L<sub>22<\/sub> have the same sign or opposite signs.<\/p>\n<ul>\n<li>The points A,B are opposite sides of the line L = 0 iff L<sub>11<\/sub> and L<sub>22<\/sub> have opposite signs.<\/li>\n<li>The points A,B are same side of the line L = 0 iff L<sub>11<\/sub> and L<sub>22<\/sub> have same sign.<\/li>\n<\/ul>\n<p><strong>Concurrent Lines:<\/strong><br \/>\nThree or more lines are said to be concurrent if they have a point in common. The common point is called the point of concurrence.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19026\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-8.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 8\" width=\"219\" height=\"179\" \/><\/p>\n<p><strong>Condition for Concurrency of Three Straight Lines:<\/strong><br \/>\nTheorem:<br \/>\nThe condition that the lines a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0, a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0, a<sub>3<\/sub>x + b<sub>3<\/sub>y + c<sub>3<\/sub> = 0 to be concurrent is a<sub>3<\/sub>(b<sub>1<\/sub>c<sub>2<\/sub> &#8211; b<sub>2<\/sub>c<sub>1<\/sub>) + b<sub>3<\/sub>(c<sub>1<\/sub>a<sub>2<\/sub> &#8211; c<sub>2<\/sub>a<sub>1<\/sub>) + c<sub>3<\/sub>(a<sub>1<\/sub>b<sub>2<\/sub> &#8211; a<sub>2<\/sub>b<sub>1<\/sub>) = 0.<br \/>\nProof:<br \/>\nSuppose the given lines are concurrent.<br \/>\nLet P(\u03b1, \u03b2) be the point of concurrence.<br \/>\nThen a<sub>1<\/sub> \u03b1 + b<sub>1<\/sub> \u03b2 + c<sub>1<\/sub> = 0 &#8230;&#8230;&#8230;..(1)<br \/>\na<sub>2<\/sub> \u03b1 + b<sub>2<\/sub> \u03b2 + c<sub>2<\/sub>= 0 &#8230;&#8230;&#8230;..(2)<br \/>\na<sub>3<\/sub> \u03b1 + b<sub>3<\/sub> \u03b2 + c<sub>3<\/sub>= 0 &#8230;&#8230;&#8230;&#8230;.(3)<\/p>\n<p>By solving (1) and (2) we get<br \/>\n\u03b1 = \\(\\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}}\\)<br \/>\n\u03b2 = \\(\\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\\)<\/p>\n<p>Therefore P = \\(\\left(\\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}}, \\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\\right)\\)<\/p>\n<p>Substituting P in eq.(3), we get<br \/>\na<sub>3<\/sub>\\(\\left(\\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}}\\right)\\) + b<sub>3<\/sub>\\(\\left(\\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\\right)\\) + c<sub>3<\/sub> = 0<br \/>\na<sub>3<\/sub>(b<sub>1<\/sub>c<sub>2<\/sub> &#8211; b<sub>2<\/sub>c<sub>1<\/sub>) + b<sub>3<\/sub>(c<sub>1<\/sub>a<sub>2<\/sub> &#8211; c<sub>2<\/sub>a<sub>1<\/sub>) + C3 (a<sub>1<\/sub>b<sub>2<\/sub> &#8211; a<sub>2<\/sub>b<sub>1<\/sub>) = 0 which is required Condition.<\/p>\n<p>Above condition can be written in a determinant form as \\(\\left|\\begin{array}{lll}<br \/>\na_{1} &amp; b_{1} &amp; c_{1} \\\\<br \/>\na_{2} &amp; b_{2} &amp; c_{2} \\\\<br \/>\na_{3} &amp; b_{3} &amp; c_{3}<br \/>\n\\end{array}\\right|\\) = 0<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas\" width=\"196\" height=\"17\" \/><\/p>\n<p>Problem:<br \/>\nFind the Condition that the lines ax + hy + g = 0, hx + by + f = 0, gx + fy + c = 0 to be Concurrent.<br \/>\nAnswer:<br \/>\nLet P be the point of ConCurrenCe<br \/>\na\u03b1 + h \u03b2 + g = 0 &#8230;&#8230;&#8230;&#8230;..(1)<br \/>\nh\u03b1 + b \u03b2 + f = 0 &#8230;&#8230;&#8230;&#8230;.(2)<br \/>\ng\u03b1 + f \u03b2 + c = 0 &#8230;&#8230;&#8230;&#8230;(3)<\/p>\n<p>solving (1) and (2) we get<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19025\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-9.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 9\" width=\"421\" height=\"104\" \/><\/p>\n<p>Sub these values in eq.(3)<br \/>\ng\\(\\left(\\frac{h g-b g}{a b-h^{2}}\\right)\\) + f\\(\\left(\\frac{g h-a f}{a b-h^{2}}\\right)\\) + c = 0<br \/>\n\u21d2 g(hf &#8211; bg) + f(gh &#8211; af) + C(ab &#8211; h<sup>2<\/sup>) = 0<br \/>\n\u21d2 fgh &#8211; bg<sup>2<\/sup> + fgh &#8211; af<sup>2<\/sup> + abc &#8211; ch<sup>2<\/sup> = 0<br \/>\n\u21d2 abc + 2fgh &#8211; af<sup>2<\/sup> &#8211; bg<sup>2<\/sup> &#8211; ch<sup>2<\/sup> = 0<br \/>\nThe Condition is abc + 2fgh &#8211; af<sup>2<\/sup> &#8211; bg<sup>2<\/sup> &#8211; ch<sup>2<\/sup> = 0<\/p>\n<p><strong>Family of Straight Lines &#8211; Concurrent Straight Lines:<\/strong><br \/>\nTheorem:<br \/>\nSuppose L<sub>1<\/sub> = a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0, L<sub>2<\/sub> = a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0 are two intersection lines.<br \/>\n(i) If (\u03bb<sub>1<\/sub>, \u03bb<sub>2<\/sub>) \u2260 (0,0) then \u03bb<sub>1<\/sub>L<sub>1<\/sub> + \u03bb<sub>2<\/sub>L<sub>2<\/sub> = 0represents a straight line passing through the point of intersection of L<sub>1<\/sub> = 0 and L<sub>2<\/sub> = 0.<br \/>\n(ii) The equation of any line passing through the point of intersection of<br \/>\nL<sub>1<\/sub> = 0, L<sub>2<\/sub> = 0 is of the form where (\u03bb<sub>1<\/sub>, \u03bb<sub>2<\/sub>) \u2260 (0,0).<br \/>\nNote: If L<sub>1<\/sub> = 0, L<sub>2<\/sub> = 0 are two intersecting lines, then the equation of any line other than L<sub>2<\/sub> = 0 passing through their point of intersection can be taken as L<sub>1<\/sub> + \u03bbL<sub>2<\/sub> = 0 where \u03bb is a parameter.<\/p>\n<p>Theorem:<br \/>\nIf there exists three constants p,q,r not all zero such that<br \/>\np(a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub>) + q(a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub>) + r(a<sub>3<\/sub>x + b<sub>3<\/sub>y + c<sub>3<\/sub>) = 0 for all x and y then the three lines a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0, a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0 and a<sub>3<\/sub>x + b<sub>3<\/sub>y + c<sub>3<\/sub> = 0 in which no two of them are parallel, are concurrent.<\/p>\n<p>Theorem:<br \/>\nLet L<sub>1<\/sub> = a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0, L<sub>2<\/sub> = a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0 represent two parallel lines.Then the straight line represented by \u03bb<sub>1<\/sub>L<sub>1<\/sub> + \u03bb<sub>2<\/sub>L<sub>2<\/sub> = 0 is parallel to each of the straight line L<sub>1<\/sub> = 0 and L<sub>2<\/sub> = 0.<\/p>\n<p><strong>A Sufficient Condition for Concurrency of Three Straight Lines:<\/strong><br \/>\nTheorem:<br \/>\nIf L<sub>1<\/sub> = a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0, L<sub>2<\/sub> = a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0, L<sub>3<\/sub> = a<sub>3<\/sub>x + b<sub>3<\/sub>y + c<sub>3<\/sub> = 0 are three straight lines, no two of which are parallel, and<br \/>\nif non-zero real numbers \u03bb<sub>1<\/sub>, \u03bb<sub>2<\/sub>, \u03bb<sub>3<\/sub> exist such that \u03bb<sub>1<\/sub> L<sub>1<\/sub> + \u03bb<sub>2<\/sub> L<sub>2<\/sub> + \u03bb<sub>3<\/sub> L<sub>3<\/sub> = 0 then the straight lines L<sub>1<\/sub> = 0, L<sub>2<\/sub> = 0 and L<sub>3<\/sub> = 0are concurrent.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas\" width=\"196\" height=\"17\" \/><\/p>\n<p><strong>Length of The Perpendicular From A Point to A Straight Line and Distance Between Two Parallel Lines<\/strong><br \/>\nTheorem:<br \/>\nThe perpendicular distance from a point P(x<sub>1<\/sub>, y<sub>1<\/sub>) to the line ax + by + c = 0 is \\(\\frac{\\left|a x_{1}+b y_{1}+c\\right|}{\\sqrt{a^{2}+b^{2}}}\\)<br \/>\nProof:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19024\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-10.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 10\" width=\"156\" height=\"112\" \/><br \/>\nLet the axes be translated to the point P(x<sub>1<\/sub>, y<sub>1<\/sub>).<br \/>\nLet (X,Y) be the new coordinates of (x, y). Then x = X + x<sub>1<\/sub>, y = Y + y<sub>1<\/sub><br \/>\nThe transformed equation of the given line is<br \/>\na(X + x<sub>1<\/sub>) + b(Y + y<sub>1<\/sub>) + c = 0<br \/>\n\u21d2 aX + bY + (ax<sub>1<\/sub> + by<sub>1<\/sub> + c) = 0<br \/>\nThe perpendicular distance from the new origin P to the line is (from normal form) The perpendicular distance from a point<br \/>\nP(x<sub>1<\/sub>, y<sub>1<\/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<p><strong>Distance Between Parallel Lines Theorem:<\/strong><br \/>\nThe distance between the two parallel lines ax + by + c<sub>1<\/sub> = 0 and ax + by + c<sub>2<\/sub> = 0 is \\(\\frac{\\left|c_{1}-c_{2}\\right|}{\\sqrt{a^{2}+b^{2}}}\\)<br \/>\nProof:<br \/>\nGiven lines are ax + by + c<sub>1<\/sub> = 0 &#8230;&#8230;&#8230;&#8230;(1)<br \/>\nax + by + c<sub>2<\/sub> = 0 &#8230;&#8230;&#8230;&#8230;(2)<br \/>\nLet P(x<sub>1<\/sub>, y<sub>1<\/sub>) be a point on the line (2).<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19023\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-11.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 11\" width=\"231\" height=\"132\" \/><br \/>\nThen<br \/>\nax<sub>1<\/sub> + by<sub>1<\/sub> + c<sub>2<\/sub> = 0<br \/>\nax<sub>1<\/sub> + by<sub>1<\/sub> = -c<sub>2<\/sub>.<br \/>\nDistance between the parallel lines = Perpendicular distance from P to line (i)<br \/>\n= \\(\\frac{\\left|a x_{1}+b y_{1}+c_{1}\\right|}{\\sqrt{a^{2}+b^{2}}}=\\frac{\\left|c_{1}-c_{2}\\right|}{\\sqrt{a^{2}+b^{2}}}\\)<\/p>\n<p><strong>Foot of The Perpendicular:<\/strong><br \/>\nTheorem:<br \/>\nIf (h, k) is the foot of the perpendicular from (x<sub>1<\/sub>, y<sub>1<\/sub>) to the line ax + by + c = 0 (a0, b0) then \\(\\frac{h-x_{1}}{a}=\\frac{k-y_{1}}{b}=\\frac{-\\left(a x_{1}+b y_{1}+c\\right)}{a^{2}+b^{2}}\\).<br \/>\nProof:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19022\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-12.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 12\" width=\"216\" height=\"101\" \/><br \/>\nLet A = (x<sub>1<\/sub>, y<sub>1<\/sub>) P = (h, k)<br \/>\nP lies on ax + by + c = 0<br \/>\nah + bk + c = 0<br \/>\nah + bk = &#8211; c<br \/>\nSlope of \\(\\overline{A P}\\) is \\(\\frac{k-y_{1}}{h-x_{1}}\\)<br \/>\nSlope of given line is \\(-\\frac{a}{b}\\)<\/p>\n<p>\\(\\overline{A P}\\) is perpendicular to the given line<br \/>\n\u21d2 \\(\\left(\\frac{k-y_{1}}{h-x_{1}}\\right)\\left(-\\frac{a}{b}\\right)\\) = -1<br \/>\n\u21d2 \\(\\frac{k-y_{1}}{b}=\\frac{h-x_{1}}{a}\\)<\/p>\n<p>By the law of multipliers in ratio and proportion<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19021\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-13.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 13\" width=\"320\" height=\"236\" \/><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas\" width=\"196\" height=\"17\" \/><\/p>\n<p><strong>Image of A Point<\/strong><br \/>\nTheorem:<br \/>\nIf (h, k) is the image of (x<sub>1<\/sub>, y<sub>1<\/sub>) w.r.t the line ax + by + c = 0 (a \u2260 0, b \u2260 0), then \\(\\frac{h-x_{1}}{a}=\\frac{k-y_{1}}{b}=\\frac{-2\\left(x_{1}+b y_{1}+c\\right)}{a^{2}+b^{2}}\\).<br \/>\nProof:<br \/>\nLet A(x<sub>1<\/sub>, y<sub>1<\/sub>), B(h, k)<br \/>\nMid Point of is P = \\(\\left(\\frac{x_{1}+h}{2}, \\frac{y_{1}+k}{2}\\right)\\)<br \/>\nSince B is the image of A,therefore mid point P lies on ax + by + c = 0.<br \/>\na\\(\\left(\\frac{x_{1}+h}{2}\\right)\\) + b\\(\\left(\\frac{y_{1}+k}{2}\\right)\\) + c = 0<br \/>\n\u21d2 ax<sub>1<\/sub> + by<sub>1<\/sub> + ah + bk + 2c = 0<br \/>\n\u21d2 ah + bk = -ax<sub>1<\/sub> + by<sub>1<\/sub> &#8211; 2c.<\/p>\n<p>Slope of \\(\\overline{A B}\\) is \\(\\frac{k-y_{1}}{h-x_{1}}\\)<br \/>\nSlope of given line is \\(-\\frac{a}{b}\\)<br \/>\n\\(\\overline{A B}\\) is perpendicular to the given line a<br \/>\n\u21d2 \\(\\left(\\frac{k-y_{1}}{h-x_{1}}\\right)\\left(-\\frac{a}{b}\\right)\\) = -1<br \/>\n\u21d2 \\(\\frac{k-y_{1}}{b}=\\frac{h-x_{1}}{a}\\)<\/p>\n<p>By the law of multipliers in ratio and proportion<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19020\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-14.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 14\" width=\"336\" height=\"296\" \/><br \/>\nNote :<\/p>\n<ul>\n<li>The image of (x<sub>1<\/sub>, y<sub>1<\/sub>) w.r.t the line x = y is (y<sub>1<\/sub>, x<sub>1<\/sub>)<\/li>\n<li>The image of (x<sub>1<\/sub>, y<sub>1<\/sub>) w.r.t the line x + y = 0 is (-y<sub>1<\/sub>, -x<sub>1<\/sub>)<\/li>\n<\/ul>\n<p>Theorem:<br \/>\nIf the four straight lines ax + by + p = 0, ax + by + q = 0, cx + dy + r = 0 and cx + dy + s = 0 form a parallelogram. Then the area of the parallelogram so formed is<br \/>\n\\(\\left|\\frac{(p-q)(r-s)}{b c-a d}\\right|\\)<br \/>\nProof:<br \/>\nLet L<sub>1<\/sub> = ax + by + p = 0<br \/>\nL<sub>2<\/sub> = ax + by + q = 0<br \/>\nL<sub>3<\/sub> = cx + dy + r = 0<br \/>\nL<sub>4<\/sub> = cx + dy + s = 0<br \/>\nClearly<br \/>\nL<sub>1<\/sub> \u2225 L<sub>2<\/sub> and L<sub>3<\/sub> \u2225 L<sub>4<\/sub>. So L<sub>1<\/sub> and L<sub>3<\/sub> are nonparallel. Let be the angle between L<sub>1<\/sub> and L<sub>3<\/sub>.<br \/>\nLet d<sub>1<\/sub> = distance between L<sub>1<\/sub> and L<sub>2<\/sub>= \\(\\frac{|p-q|}{\\sqrt{a^{2}+b^{2}}}\\)<br \/>\nLet d<sub>2<\/sub> = distance between L<sub>3<\/sub> and L<sub>4<\/sub> = \\(\\frac{|r-s|}{\\sqrt{c^{2}+d^{2}}}\\)<br \/>\nNow cos \u03b8 = \\(\\frac{|a c+b d|}{\\sqrt{\\left(a^{2}+b^{2}\\right)\\left(c^{2}+d^{2}\\right)}}\\) and sin \u03b8 = \\(\\sqrt{\\frac{\\left(a^{2}+b^{2}\\right)\\left(c^{2}+d^{2}\\right)-(a c+b d)^{2}}{\\left(a^{2}+b^{2}\\right)\\left(c^{2}+d^{2}\\right)}}\\)<br \/>\n= \\(\\frac{|b c-a d|}{\\sqrt{\\left(a^{2}+b^{2}\\right)\\left(c^{2}+d^{2}\\right)}}\\)<br \/>\nNow area of the parallelogram is \\(\\frac{d_{1} d_{2}}{\\sin \\theta}=\\left|\\frac{(p-q))(r-s)}{b c-a d}\\right|\\)<\/p>\n<p><strong>Angle Between Two Lines<\/strong><br \/>\nTheorem:<br \/>\nIf \u03b8 is an angle between the lines a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0, a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0 then<br \/>\ncos \u03b8 = \u00b1\\(\\frac{a_{1} a_{2}+b_{1} b_{2}}{\\sqrt{a_{1}^{2}+b_{2}^{3}} \\sqrt{a_{2}^{2}+b_{2}^{2}}}\\)<br \/>\nProof:<br \/>\nThe lines passing through the origin and parallel to the given lines are<br \/>\na<sub>1<\/sub>x + b<sub>1<\/sub>y = 0, &#8230;&#8230;&#8230; (1)<br \/>\na<sub>2<\/sub>x + b<sub>2<\/sub>y = 0. &#8230;&#8230;&#8230;&#8230;&#8230;.(2)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19019\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-15.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 15\" width=\"342\" height=\"211\" \/><\/p>\n<p>Let \u03b8<sub>1<\/sub>, \u03b8<sub>2<\/sub> be the inclinations of (1) and (2) respectively (\u03b81 &gt; \u03b82)<br \/>\nNow \u03b8 is an angle between (1) and (2)<br \/>\n\u03b8 = \u03b8<sub>1<\/sub> &#8211; \u03b8<sub>2<\/sub><br \/>\nP(-b<sub>1<\/sub>, a<sub>1<\/sub>) satisfies eq(1), the point lies on (1)<br \/>\nSimilarly, Q(-b<sub>2<\/sub>, a<sub>2<\/sub>) lies on (2)<\/p>\n<p>Let L and M be the projection of P, Q respectively on the x &#8211; axis.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19018\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-16.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 16\" width=\"440\" height=\"403\" \/><\/p>\n<p>Note :<\/p>\n<ul>\n<li>If is the acute angle between the lines then cos \u03b8 = \\(\\frac{\\left|a_{1} a_{1}+b_{1} b_{2}\\right|}{\\sqrt{a_{1}^{2}+b_{1}^{2}} \\sqrt{a_{2}^{2}+b_{2}^{2}}}\\)<\/li>\n<li>If is an angle between two lines, then is another angle between the lines.<\/li>\n<li>If is an angle between two lines are not a right angle then the angle between the lines means the acute angle between the lines.<\/li>\n<li>If 0 is an angle between the lines a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0, a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0 then tan \u03b8 = \\(\\frac{a_{1} b_{2}-a_{2} b_{1}}{a_{1} a_{2}+b_{1} b_{2}}\\)<\/li>\n<li>If is the acute angle between the lines a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0, a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0 then<br \/>\ntan \u03b8 = \\(\\left|\\frac{a_{1} b_{2}-a_{2} b_{1}}{a_{1} a_{2}+b_{1} b_{2}}\\right|=\\left|\\frac{a_{1} \/ b_{1}-a_{2} \/ b_{2}}{\\left(a_{1} a_{2}\\right) \/\\left(b_{1} b_{2}\\right)+1}\\right|\\)<br \/>\n= \\(\\left|\\frac{\\left(-a_{1} \/ b_{1}\\right)-\\left(-a_{2} \/ b_{2}\\right)}{1+\\left(-a_{1} \/ b_{1}\\right)\\left(-a_{2} \/ b_{2}\\right)}\\right|\\) where mj, m2 are the slopes of the lines.<\/li>\n<\/ul>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas\" width=\"196\" height=\"17\" \/><\/p>\n<p>Theorem:<br \/>\nThe equation of the line parallel to ax + by + c = 0 and passing through (x<sub>1<\/sub>, y<sub>1<\/sub>) is a(x &#8211; x<sub>1<\/sub>) + b(y &#8211; y<sub>1<\/sub>) = 0.<br \/>\nProof:<br \/>\nSlope of the given line is -a\/b.<br \/>\n\u21d2 Slope of the required line is -a\/b.(lines are parallel)<br \/>\nEquation of the required line is<br \/>\ny &#8211; y<sub>1<\/sub> = &#8211;\\(\\frac{a}{b}\\)(x &#8211; x<sub>1<\/sub>)<br \/>\nb(y &#8211; y<sub>1<\/sub>) = -a(x &#8211; x<sub>1<\/sub>)<br \/>\na(x &#8211; x<sub>1<\/sub>) + b(y &#8211; y<sub>1<\/sub>) = 0.<\/p>\n<p>Note :<\/p>\n<ul>\n<li>The equation of a line parallel to ax + by + c = 0 may be taken as ax + by + k = 0.<\/li>\n<li>The equation of a line parallel to ax + by + c = 0 and passing through the origin is ax + by = 0.<\/li>\n<\/ul>\n<p>Theorem:<br \/>\nThe equation of the line perpendicular to ax + by + c = 0 and passing through (x<sub>1<\/sub>, y<sub>1<\/sub>) is b(x &#8211; x<sub>1<\/sub>) &#8211; a(y &#8211; y<sub>1<\/sub>) = 0.<br \/>\nProof:<br \/>\nSlope of the given line is -a\/b. \u21d2 Slope of the required line is b\/a. (since product of slopes = -1)<br \/>\nEquation of the required line is y &#8211; y<sub>1<\/sub> = \\(\\frac{b}{a}\\) (x &#8211; x<sub>1<\/sub>)<br \/>\na(y &#8211; y<sub>1<\/sub>) = b(x &#8211; x<sub>1<\/sub>)<br \/>\nb(x &#8211; x<sub>1<\/sub>) &#8211; a(y &#8211; y<sub>1<\/sub>) = 0.<\/p>\n<p>Note :<\/p>\n<ul>\n<li>The equation of a line perpendicular to ax + by + c = 0 may be taken as bx &#8211; ay + k = 0<\/li>\n<li>The equation of a line perpendicular to ax + by + c = 0 and passing through the origin is bx &#8211; ay = 0.<\/li>\n<\/ul>\n<p><strong>Concurrent Lines- Properties Related to A Triangle<\/strong><br \/>\nTheorem:<br \/>\nThe medians of a triangle are concurrent.<br \/>\nProof:<br \/>\nLet A(x<sub>1<\/sub>, y<sub>1<\/sub>), B(x<sub>2<\/sub>, y<sub>2<\/sub>), C(x<sub>3<\/sub>, y<sub>3<\/sub>) be the vertices of the triangle<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19017\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-17.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 17\" width=\"295\" height=\"182\" \/><br \/>\nLet D, E, F be the mid points of \\(\\overline{B C}, \\overline{C A}, \\overline{A B}\\) respectively<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-19016\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/Inter-1st-Year-Maths-1B-The-Straight-Line-Formulas-18.png\" alt=\"Inter 1st Year Maths 1B The Straight Line Formulas 18\" width=\"412\" height=\"226\" \/><br \/>\nEquation of \\(\\overline{A D}\\)<br \/>\ny &#8211; y<sub>1<\/sub> = (x &#8211; x<sub>1<\/sub>)<br \/>\n(y &#8211; y<sub>1<\/sub>) (x<sub>2<\/sub> + x<sub>3<\/sub> &#8211; 2x<sub>1<\/sub>) = (x &#8211; x<sub>1<\/sub>)(y<sub>2<\/sub> + y<sub>3<\/sub> &#8211; 2y<sub>1<\/sub>)<br \/>\n\u21d2 L<sub>1<\/sub> \u2261 (x &#8211; x<sub>1<\/sub>)(y<sub>2<\/sub> + y<sub>3<\/sub> &#8211; 2y<sub>1<\/sub>) &#8211; (y &#8211; y<sub>1<\/sub>) (x<sub>2<\/sub> + x<sub>3<\/sub> &#8211; 2x<sub>1<\/sub>) = 0.<\/p>\n<p>Similarly, the equations to \\(\\overline{B E}\\) and \\(\\overline{C F}\\) respectively are L<sub>2<\/sub> \u2261 (x &#8211; x<sub>2<\/sub>)(y<sub>3<\/sub> + y<sub>1<\/sub> &#8211; 2y<sub>2<\/sub>) &#8211; (y &#8211; y<sub>2<\/sub>) (x<sub>3<\/sub> + x<sub>1<\/sub> &#8211; 2x<sub>2<\/sub>) = 0.<br \/>\nL<sub>3<\/sub> \u2261 (x &#8211; x<sub>3<\/sub>)(y<sub>1<\/sub> + y<sub>2<\/sub> &#8211; 2y<sub>3<\/sub>) &#8211; (y &#8211; y<sub>3<\/sub>) (x<sub>1<\/sub> + x<sub>2<\/sub> &#8211; 2x<sub>3<\/sub>) = 0.<br \/>\nNow 1. L<sub>1<\/sub> + 1.L<sub>2<\/sub> + 1. L<sub>3<\/sub> = 0<br \/>\nThe medians L<sub>1<\/sub> = 0, L<sub>2<\/sub> =0, L<sub>3<\/sub> = 0 are concurrent.<\/p>\n<p>Theorem:<br \/>\nThe altitudes of a triangle are concurrent.<br \/>\nProof:<br \/>\nLet A(x<sub>1<\/sub>, y<sub>1<\/sub>), B(x<sub>2<\/sub>, y<sub>2<\/sub>), C(x<sub>3<\/sub>, y<sub>3<\/sub>) be the vertices of the triangle ABC.<br \/>\nLet AD, BE,CF be the altitudes.<br \/>\nSlope of \\(\\overline{B C}\\) is \\(\\frac{y_{3}-y_{2}}{x_{3}-x_{2}}\\) and AD \u22a5 BC<br \/>\nSlope of the altitude through A is \\(-\\frac{x_{3}-x_{2}}{y_{3}-y_{2}}\\)<\/p>\n<p>Equation of the altitude through A is y &#8211; y<sub>1<\/sub> = \\(\\frac{x_{3}-x_{2}}{y_{3}-y_{2}}\\)(x &#8211; x<sub>1<\/sub>)<br \/>\n(y &#8211; y<sub>1<\/sub>) (y<sub>3<\/sub> &#8211; y<sub>2<\/sub>) = -(x &#8211; x<sub>1<\/sub>) (x<sub>3<\/sub> &#8211; x<sub>2<\/sub>)<br \/>\nL<sub>1<\/sub> = (x &#8211; x<sub>1<\/sub>)(x<sub>2<\/sub> &#8211; x<sub>3<\/sub>) + (y &#8211; y<sub>1<\/sub>)(y<sub>2<\/sub> &#8211; y<sub>3<\/sub>) = 0.<\/p>\n<p>Similarly equations of the altitudes through B,C are<br \/>\nL<sub>2<\/sub> = (x &#8211; x<sub>2<\/sub>) (x<sub>3<\/sub> &#8211; x<sub>1<\/sub>) + (y &#8211; y<sub>2<\/sub>) (y<sub>2<\/sub> &#8211; y<sub>3<\/sub>) = 0<br \/>\nL<sub>3<\/sub> = (x &#8211; x<sub>3<\/sub>) (x<sub>1<\/sub> &#8211; x<sub>2<\/sub>) + (y &#8211; y<sub>3<\/sub>) (y<sub>1<\/sub> &#8211; y<sub>2<\/sub>) = 0.<br \/>\nNow 1.L<sub>1<\/sub> + 1.L<sub>2<\/sub> + 1.L<sub>3<\/sub> = 0<br \/>\nThe altitudes L<sub>1<\/sub> = 0, L<sub>2<\/sub> =0, L<sub>3<\/sub> = 0 are concurrent.<\/p>\n<p>Theorem:<br \/>\nThe internal bisectors of the angles of a triangle are concurrent.<\/p>\n<p>Theorem:<br \/>\nThe perpendicular bisectors of the sides of a triangle are concurrent<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Use these Inter 1st Year Maths 1B Formulas PDF Chapter 3 The Straight Line to solve questions creatively. Intermediate 1st Year Maths 1B The Straight Line Formulas \u2192 Equation of a horizontal line is y = k \u2192 Equation of a vertical line is x = h \u2192 If &#8216;\u03b8 &#8216; is the inclination of &hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35122"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35122"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35122\/revisions"}],"predecessor-version":[{"id":35123,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35122\/revisions\/35123"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35122"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35122"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35122"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}