STRAIGHT LINES

1 Slope of a lineIf is the angle made by a line with positive direction of x-axis in anticlockwise direction,then the value of tan is called the slope of the line and is denoted by slope of a line passing through points P(x1, y1) and Q (x2, y2) is given by2121tanyymxx = = Angle between two LINES The angle between the two LINES having slopes m1 andm2 is given by1212()tan1mmm m = +If we take the acute angle between two LINES , then tan = 12121mmm m +If the LINES are parallel, then m1 = the LINES are perpendicular, then m1m2 = Collinearity of three points If three points P (h, k), Q (x1, y1) and R (x2,y2)are such that slope of PQ = slope of QR, , 121121ykyyxhxx = or(h x1) (y2 y1) = (k y1) (x2 x1) then they are said to be Various forms of the equation of a line (i)If a line is at a distance a and parallel to x-axis, then the equation of the line isy = a.

2 (ii)If a line is parallel to y-axis at a distance b from y-axis then its equation isx = bChapter10 STRAIGHT LINES18/04/18166 EXEMPLAR PROBLEMS MATHEMATICS(iii) Point-slope form : The equation of a line having slope m and passing through thepoint (x0, y0) is given by y y0 = m (x x0)(iv)Two-point-form : The equation of a line passing through two points (x1, y1) and(x2, y2) is given byy y1 = 2121yyxx (x x1)(v)Slope intercept form : The equation of the line making an intercept c on y-axis andhaving slope m is given byy = mx + cNote that the value of c will be positive or negative as the intercept is made onthe positive or negative side of the y-axis, respectively.

3 (vi)Intercept form : The equation of the line making intercepts a and b on x- and y-axis respectively is given by 1xyab+ =.(vii)Normal form : Suppose a non-vertical line is known to us with following data:(a)Length of the perpendicular (normal) p from origin to the line .(b)Angle which normal makes with the positive direction of the equation of such a line is given by x cos + y sin = General equation of a lineAny equation of the form Ax + By + C = 0, where A and B are simultaneously not zero,is called the general equation of a forms of Ax + By + C = 0 The general form of the line can be reduced to various forms as given below:(i)Slope intercept form : If B 0, then Ax + By + C = 0 can be written asACBByx =+ or y = mx + c, where ACand =BBmc =If B = 0, then x = CA which is a vertical line whose slope is not defined and x-interceptis CA.

4 18/04/18 STRAIGHT LINES 167(ii)Intercept form : If C 0, then Ax + By + C = 0 can be written as CCABxy+ = 1 or 1xyab+ =, where CC=and =ABab .If C = 0, then Ax + By + C = 0 can be written as Ax + By = 0 which is a linepassing through the origin and therefore has zero intercepts on the axes.(iii)Normal Form : The normal form of the equation Ax + By + C = 0 isx cos + y sin = p where,2222 ABcos, sinA + BA + B = = and p = 22CA + B .Note: Proper choice of signs is to be made so that p should be always Distance of a point from a line The perpendicular distance (or simply distance)d of a point P (x1, y1) from the line Ax + By + C = 0 is given by1122 ABCA + Bxyd++=Distance between two parallel linesThe distance d between two parallel LINES y = mx + c1 and y = mx + c2 is given by1221+ccdm =.

5 Locus and Equation of Locus The curve described by a point which movesunder certain given condition is called its locus. To find the locus of a point P whosecoordinates are (h, k), express the condition involving h and k. Eliminate variables ifany and finally replace h by x and k by y to get the locus of Intersection of two given LINES Two LINES a1x + b1y + c1 = 0 and a2x + b2y +c2 = 0 are(i)intersecting if 1122abab 18/04/18168 EXEMPLAR PROBLEMS MATHEMATICS(ii)parallel and distinct if 111222abcabc= (iii)coincident if 111222abcabc==Remarks(i)The points (x1, y1) and (x2, y2) are on the same side of the line or on the oppositeside of the line ax + by + c = 0, if ax1 + by1 + c and ax2 + by2 + c are of the samesign or of opposite signs respectively.

6 (ii)The condition that the LINES a1x + b1y + c1 = 0 and a2x + b2y + c = 0 areperpendicular is a1a2 + b1b2 = 0.(iii)The equation of any line through the point of intersection of two LINES a1x + b1y +c1 = 0 and a2x + b2y + c2 = 0 is a1x + b1y + c1 + k (ax2 + by2 + c2) = 0. The valueof k is determined from extra condition given in the Solved ExamplesShort Answer T ypeExample 1 Find the equation of a line which passes through the point (2, 3) andmakes an angle of 30 with the positive direction of Here the slope of the line is m = tan = tan 30 = 13 and the given point is(2, 3). Therefore, using point slope formula of the equation of a line , we havey 3 = 13 (x 2)orx 3y + (33 2) = 2 Find the equation of the line where length of the perpendicular segmentfrom the origin to the line is 4 and the inclination of the perpendicular segment with thepositive direction of x-axis is 30.

7 Solution The normal form of the equation of the line is x cos + y sin = p. Herep = 4 and = 30 . Therefore, the equation of the line isx cos 30 + y sin 30 = 432x + y 12= 4 or 3 x + y = 818/04/18 STRAIGHT LINES 169 Example 3 Prove that every STRAIGHT line has an equation of the form Ax + By + C = 0,where A, B and C are Given a STRAIGHT line , either it cuts the y-axis, or is parallel to or coincident withit. We know that the equation of a line which cuts the y-axis ( , it has y-intercept) canbe put in the form y = mx + b; further, if the line is parallel to or coincident with the y-axis, its equation is of the form x = x1, where x = 0 in the case of coincidence. Both ofthese equations are of the form given in the problem and hence the 4 Find the equation of the STRAIGHT line passing through (1, 2) and perpendicularto the line x + y + 7 = Let m be the slope of the line whose equation is to be found out which isperpendicular to the line x + y + 7 = 0.

8 The slope of the given line y = ( 1) x 7 is , using the condition of perpendicularity of LINES , we have m ( 1) = 1or m = 1 (Why?)Hence, the required equation of the line is y 1 = (1) (x 2) or y 1 = x 2 x y 1 = 5 Find the distance between the LINES 3x + 4y = 9 and 6x + 8y = The equations of LINES 3x + 4y = 9 and 6x + 8y = 15 may be rewritten as3x + 4y 9 = 0 and 3x + 4y 152 = 0 Since, the slope of these LINES are same and hence they are parallel to each , the distance between them is given by22159321034 =+Example 6 Show that the locus of the mid-point of the distance between the axes ofthe variable line x cos + y sin = p is 222114xyp+= where p is a Changing the given equation of the line into intercept form, we have1cossinxypp+= which gives the coordinates , 0 and 0,cossinpp.

9 Where theline intersects x-axis and y-axis, EXEMPLAR PROBLEMS MATHEMATICSLet (h, k) denote the mid-point of the line segment joining the points, 0 and0,,cossinpp Then2 cosph= and 2 sinpk= (Why?)This givescos2ph = and sin2pk =Squaring and adding we get2222144pphk+= or 222114hkp+=.Therefore, the required locus is 222114xyp+=.Example 7 If the line joining two points A(2, 0) and B(3, 1) is rotated about A inanticlock wise direction through an angle of 15 . Find the equation of the line in The slope of the line AB is 1 01 or tan 453 2 = (Why?) (see Fig.). Afterrotation of the line through 15 , the slope of the line AC in new position is tan 60 = 3 Fig.

10 LINES 171 Therefore, the equation of the new line AC isy 0 = 3(2)x ory 32 3x+ = 0 Long Answer T ypeExample 8 If the slope of a line passing through the point A(3, 2) is 34, then findpoints on the line which are 5 units away from the point Equation of the line passing through (3, 2) having slope 34 is given byy 2 =34 (x 3)or4y 3x + 1 =0(1)Let (h, k) be the points on the line such that(h 3)2 + (k 2)2 =25(2) (Why?)Also, we have4k 3h + 1 =0(3) (Why?)ork =314h (4)Putting the value of k in (2) and on simplifying, we get25h2 150h 175 =0(How?)orh2 6h 7 =0or(h + 1) (h 7) =0 h = 1, h = 7 Putting these values of k in (4), we get k = 1 and k = 5. Therefore, the coordinates ofthe required points are either ( 1, 1) or (7, 5).