Notes On Equations of Lines (Part II) - CBSE Class 11 Maths
Slope-Intercept Form Consider a line L with slope m that intersects the Y-axis at a point A, which is at distance C from the origin. That is the given line passes through the point zero, C.   Point-Slope Form: y – y1 = m(x – x1)   Thus, the point-slope form of line l  is: y – c = m(x – 0)   ⇒ y = mx + c        This equation is the Slope-Intercept Form of the line, where: m = Slope of the line c = y-intercept of the line        Consider a line L with slope m, whose x-intercept is given as D. The given line passes through the point B (d, 0).   Thus, the point-slope form of line l: y – y1 = m(x – x1) ⇒y – 0 = m(x –d)         ⇒ y = m(x – d)        This equation is the Slope-Intercept Form, where: m = Slope of the line d = x-intercept of the line      Intercept  Form   Consider a line l with x-intercept A and y-intercept B. Let’s denote the points where the line intersects the X-axis and the Y-axis as P and Q, respectively. Therefore, the coordinates of point ‘P’ are (a, 0) and that of point ‘Q’ are (0, b).   Two-Point Form: y -  y1 = {(y2 – y1)/ (x2 – x1)} x (x – x1) Thus, the two-point form of line l is: y -  0 = {(b – 0)/ (0 – a)} x (x – a) ⇒ ay = -bx + ab     ⇒ bx + ay = ab     Dividing both sides by ab:  (bx + ay)/ab = ab/ab     ⇒ x/a + y/b = 1    The above equation is the Intercept Form of equation of the straight line.     Normal Form Consider a non-vertical line L. Let OA be the normal or perpendicular drawn to line L from the origin. Let the perpendicular distance of line L from the origin, or the length of perpendicular OA, be P.   ∠AOX = ω AM ⊥ OX Distance OM represents the x-coordinate, and distance AM represents the y-coordinate of point A.   From right angled triangle OMA Cos ω = $\frac{\text{OM}}{\text{OA}}$ ⇒ Cos ω = $\frac{\text{OM}}{\text{P}}$ ⇒ OM = P Cos ω Sin ω = $\frac{\text{AM}}{\text{OA}}$ ⇒ Sin ω = $\frac{\text{AM}}{\text{P}}$ ⇒ AM = P Sin ω   For the given values of P and omega, the coordinates of point A will remain the same no matter the quadrant in which line L lies.   The coordinates of a point on line L, we can find its equation if we know its slope, using the point-slope form. Let M be the slope of the given line L. Given OA ⊥ l: Slope of l x slope of OA = -1 Slope of OA = tan ω ∴ m tan ω = -1                         ⇒ m = -1/ tan ω ⇒ m = -cot ω ⇒ m = -cos ω/ sin ω          Point-Slope Form: y – y1 = m(x – x1) Thus, the point-slope form of line l is: y – p sin ω = (-cos ω/ sin ω)(x – p cos ω)   ⇒ y sin ω – p sin2 ω = -x cos ω + p cos2 ω) ⇒ x cos ω + y sin ω = p(sin2 ω + cos2 ω)⇒ x cos ω + y sin ω = p     [$\because$ sin2 ω + cos2 ω = 1] Normal Form: x cos ω + y sin ω = p     where: p = normal distance of the line from origin ω = Inclination of the normal

#### Summary

Slope-Intercept Form Consider a line L with slope m that intersects the Y-axis at a point A, which is at distance C from the origin. That is the given line passes through the point zero, C.   Point-Slope Form: y – y1 = m(x – x1)   Thus, the point-slope form of line l  is: y – c = m(x – 0)   ⇒ y = mx + c        This equation is the Slope-Intercept Form of the line, where: m = Slope of the line c = y-intercept of the line        Consider a line L with slope m, whose x-intercept is given as D. The given line passes through the point B (d, 0).   Thus, the point-slope form of line l: y – y1 = m(x – x1) ⇒y – 0 = m(x –d)         ⇒ y = m(x – d)        This equation is the Slope-Intercept Form, where: m = Slope of the line d = x-intercept of the line      Intercept  Form   Consider a line l with x-intercept A and y-intercept B. Let’s denote the points where the line intersects the X-axis and the Y-axis as P and Q, respectively. Therefore, the coordinates of point ‘P’ are (a, 0) and that of point ‘Q’ are (0, b).   Two-Point Form: y -  y1 = {(y2 – y1)/ (x2 – x1)} x (x – x1) Thus, the two-point form of line l is: y -  0 = {(b – 0)/ (0 – a)} x (x – a) ⇒ ay = -bx + ab     ⇒ bx + ay = ab     Dividing both sides by ab:  (bx + ay)/ab = ab/ab     ⇒ x/a + y/b = 1    The above equation is the Intercept Form of equation of the straight line.     Normal Form Consider a non-vertical line L. Let OA be the normal or perpendicular drawn to line L from the origin. Let the perpendicular distance of line L from the origin, or the length of perpendicular OA, be P.   ∠AOX = ω AM ⊥ OX Distance OM represents the x-coordinate, and distance AM represents the y-coordinate of point A.   From right angled triangle OMA Cos ω = $\frac{\text{OM}}{\text{OA}}$ ⇒ Cos ω = $\frac{\text{OM}}{\text{P}}$ ⇒ OM = P Cos ω Sin ω = $\frac{\text{AM}}{\text{OA}}$ ⇒ Sin ω = $\frac{\text{AM}}{\text{P}}$ ⇒ AM = P Sin ω   For the given values of P and omega, the coordinates of point A will remain the same no matter the quadrant in which line L lies.   The coordinates of a point on line L, we can find its equation if we know its slope, using the point-slope form. Let M be the slope of the given line L. Given OA ⊥ l: Slope of l x slope of OA = -1 Slope of OA = tan ω ∴ m tan ω = -1                         ⇒ m = -1/ tan ω ⇒ m = -cot ω ⇒ m = -cos ω/ sin ω          Point-Slope Form: y – y1 = m(x – x1) Thus, the point-slope form of line l is: y – p sin ω = (-cos ω/ sin ω)(x – p cos ω)   ⇒ y sin ω – p sin2 ω = -x cos ω + p cos2 ω) ⇒ x cos ω + y sin ω = p(sin2 ω + cos2 ω)⇒ x cos ω + y sin ω = p     [$\because$ sin2 ω + cos2 ω = 1] Normal Form: x cos ω + y sin ω = p     where: p = normal distance of the line from origin ω = Inclination of the normal

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