10.1 Overview 10.1.1 Slope of a line If θ is the angle made by a line with positive direction ofx-axis in anticlockwise direction, then the value of tan θ is called the slope of the line and is denoted by m. The slope of a line passing through points P(x1, y1) and Q (x2, y2) is given by y2 −y1m=tan θ= x2 −x1 10.1.2 Angle between two lines The angle θ between the two lines having slopes m1 and m2 is given by (mm)−tan θ= ± 12 121+mm mm1 − 2If we take the acute angle between two lines, then tan θ = 1+ 12mm If the lines are parallel, then m = m.1 2If the lines are perpendicular, then m= – 1.1m2 10.1.3 Collinearity of three points If three points P (h, k), Q (x1, y1) and R (x2,y2) y −ky − y1 21are such that slope of PQ = slope of QR, i.e., = x − hx − x1 21 or (h– x) (y – y) = (k– y) (x – x) then they are said to be collinear. 1 21121 10.1.4 Various forms of the equation of a line (i) If a line is at a distance aand parallel to x-axis, then the equation of the line is y= ± a. (ii) If a line is parallel to y-axis at a distance b from y-axis then its equation is x= ± b (iii) Point-slope form : The equation of a line having slope m and passing through the point (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 (x, y) is given by22y2 −y1 y – y1 = (x – x1)x2 −x1 (v) Slope intercept form : The equation of the line making an intercept con y-axis and having slope mis given by y = mx + c Note that the value of c will be positive or negative as the intercept is made on the positive or negative side of the y-axis, respectively. (vi) Intercept form : The equation of the line making intercepts a and b on x- and yxyaxis respectively is given by +=1. ab (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 ωwhich normal makes with the positive direction of x-axis. Then the equation of such a line is given by x cos ω+ y sin ω= p 10.1.5 General equation of a line Any equation of the form Ax + By + C = 0, where A and B are simultaneously not zero, is called the general equation of a line. Different 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 as −A −C −A −C y= x+ or y = mx + c, where m= and c= BB BB −C If B = 0,then x= which is a vertical line whose slope is not defined andx-interceptA −C is .A STRAIGHT LINES 167 xy(ii) Intercept form : If C ≠ 0, then Ax + By + C = 0 can be written as +−C −C AB xy −C −C = 1 or +=1, where a= and b=. ab AB If C = 0, then Ax + By + C = 0 can be written as Ax + By = 0 which is a line passing through the origin and therefore has zero intercepts on the axes. (iii) Normal Form : The normal form of the equation Ax + By + C = 0 is x cos ω+ y sin ω= p where, AB C cos ω=± ,sin ω=± and p = ±22 22 22 . A+B A+B A+B Note: Proper choice of signs is to be made so that p should be always positive. 10.1.6 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 by Distance between two parallel lines The distance d between two parallel lines y = mx + c and y = mx + c is given by12cc1 − 2d = 10.1.7 Locus and Equation of Locus The curve described by a point which moves under certain given condition is called its locus. To find the locus of a point P whose coordinates are (h, k), express the condition involving h and k. Eliminate variables if any and finally replace h by x and k by y to get the locus of P. 10.1.8 Intersection of two given lines Two lines ax + by + c = 0 and ax + by +1 1122c2 = 0 are a1 b1(i) intersecting if ≠ a2 b2 a1 b1 c1(ii) parallel and distinct if =≠ a2 b2 c2 a1 b1 c1(iii) coincident if == a2 b2 c2 Remarks (i) The points (x1, y1) and (x2, y2) are on the same side of the line or on the opposite side of the line ax + by + c = 0, if ax1 + by1 + c and ax2 + by2 + c are of the same sign or of opposite signs respectively. (ii) The condition that the lines ax + by + c = 0 and ax + by + c = 0 are 11122 perpendicular 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 value of k is determined from extra condition given in the problem. 10.2 Solved Examples Short Answer Type Example 1 Find the equation of a line which passes through the point (2, 3) and makes an angle of 30° with the positive direction of x-axis. 1 Solution Here the slope of the line is m = tanθ = tan 30° = 3 and the given point is (2, 3). Therefore, using point slope formula of the equation of a line, we have 1 y – 3 = (x – 2) or x – 3 y + (3 3 – 2) = 0.3Example 2 Find the equation of the line where length of the perpendicular segment from the origin to the line is 4 and the inclination of the perpendicular segment with the positive direction ofx-axis is 30°. Solution The normal form of the equation of the line is x cosω + y sin ω = p. Here p = 4 and ω = 30°. Therefore, the equation of the line is x cos 30° + y sin 30° = 4 31 x+ y = 4 or 3 x + y = 822 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 constants. Proof Given a straight line, either it cuts the y-axis, or is parallel to or coincident with it. We know that the equation of a line which cuts the y-axis (i.e., it has y-intercept) can be 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 of these equations are of the form given in the problem and hence the proof. Example 4 Find the equation of the straight line passing through (1, 2) and perpendicular to the line x + y + 7 = 0. Solution Let m be the slope of the line whose equation is to be found out which is perpendicular to the line x + y + 7 = 0. The slope of the given line y = (– 1) x – 7 is – 1. Therefore, using the condition of perpendicularity of lines, we have m × (– 1) = – 1 or m = 1 (Why?) Hence, the required equation of the line is y – 1 = (1) (x – 2) or y – 1 = x – 2 ⇒ x – y – 1 = 0. Example 5 Find the distance between the lines 3x + 4y = 9 and 6x + 8y = 15. Solution The equations of lines 3x + 4y = 9 and 6x + 8y = 15 may be rewritten as 153x + 4y – 9 = 0 and 3x + 4y − = 0 2 Since, the slope of these lines are same and hence they are parallel to each other. Therefore, the distance between them is given by Example 6 Show that the locus of the mid-point of the distance between the axes of 114the variable line x cosα + y sinα = p is += where p is a constant.22 2xyp Solution Changing the given equation of the line into intercept form, we have xy ⎛ p ⎞⎛ p ⎞+=1 which gives the coordinates ,0 and 0, , where the⎜ ⎟⎜⎟pp ⎝ cos α⎠ ⎝ sin α⎠ cos α sin αline intersects x-axis and y-axis, respectively. Let (h, k) denote the mid-point of the line segment joining the points ⎛ p ⎞⎛ p ⎞ ⎜ ,0⎟ and ⎜ 0, ,⎟⎝cos α⎠ ⎝ sin α⎠ ppThen h = and k = (Why?)2cos α2sin α This gives cos α=2 ph and sin α= 2 pk Squaring and adding we get p2 p2 114+=1 or += .2 2 2224h 4k hkp 11 4Therefore, the required locus is += .22 2xy p Example 7 If the line joining two points A(2, 0) and B(3, 1) is rotated about A in anticlock wise direction through an angle of 15°. Find the equation of the line in new position. 10Solution The slope of the line AB is −=1or tan 45 ° (Why?) (see Fig.). After 32−rotation of the line through 15°, the slope of the line AC in new position is tan 60° = 3 Fig. 10.1 STRAIGHT LINES 171 Therefore, the equation of the new line AC is 3x +2y – 0 = 3(x − 2) or y – 3 = 0 Long Answer Type 3 Example 8 If the slope of a line passing through the point A(3, 2) is , then find4 points on the line which are 5 units away from the point A. 3Solution Equation of the line passing through (3, 2) having slope is given by43 y – 2 = (x – 3)4or 4y – 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 have 4k – 3h + 1 = 0 (3) (Why?) 3h −1 or k = (4)4 Putting the value of k in (2) and on simplifying, we get 25h2 – 150h – 175 = 0 (How?) or h2 – 6h – 7 = 0 or (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 of the required points are either (–1, –1) or (7, 5). Example 9 Find the equation to the straight line passing through the point of intersection of the lines 5x – 6y – 1 = 0 and 3x + 2y + 5 = 0 and perpendicular to the line 3x – 5y + 11 = 0. Solution First we find the point of intersection of lines 5x – 6y – 1 = 0 and 3x + 2y + 35 = 0 which is (– 1, – 1). Also the slope of the line 3x – 5y + 11 = 0 is . Therefore,5−5the slope of the line perpendicular to this line is (Why?). Hence, the equation of the3 required line is given by −5 y + 1 = (x + 1)3 or 5x + 3y + 8 = 0 Alternatively The equation of any line through the intersection of lines 5x – 6y – 1 = 0 and 3x + 2y + 5 = 0 is 5x – 6y – 1 + k(3x + 2y + 5) = 0 (1) −+ (5 3) k or Slope of this line is –62+k 3Also, slope of the line 3x – 5y + 11 = 0 is 5 Now, both are perpendicular (5 3) k−+ 3 so ×=–1–6 +2k 5 or k =45 Therefore, equation of required line in given by 5x – 6y – 1 + 45 (3x + 2y + 5) = 0 or 5x + 3y + 8 = 0 Example 10 A ray of light coming from the point (1, 2) is reflected at a point A on the x-axis and then passes through the point (5, 3). Find the coordinates of the point A. Solution Let the incident ray strike x-axis at the point A whose coordinates be (x, 0). From the figure, the slope of the reflected ray is given by 3 tan θ = (1)5– x Fig. 10.2 STRAIGHT LINES 173 Again, the slope of the incident ray is given by −2 Therefore, the required coordinates of the point A are ,0⎜⎟ tan (π – θ) = x −1 (Why?) or – tan θ = −2 x −1 (2) Solving (1) and (2), we get 3 2 13 5 − x = x −1 or x = 5 ⎛13 ⎞ ⎠ .⎝ 5 Example 11 If one diagonal of a square is along the line 8x – 15y = 0 and one of its vertex is at (1, 2), then find the equation of sides of the square passing through this vertex. Solution Let ABCD be the given square and the coordinates of the vertex D be (1, 2). We are required to find the equations of its sides DC and AD. Fig. 10.3 8 Given that BD is along the line 8x – 15y = 0, so its slope is (Why?). The angles15made by BD with sides AD and DC is 45° (Why?). Let the slope of DC be m. Then 8 m − 15tan 45° = (Why?)8m1 + 15 or 15 + 8m =15m – 8 23 or 7m = 23, which gives m = 7 Therefore, the equation of the side DC is given by 23 y – 2 = (x – 1) or 23x – 7y – 9 = 0.7Similarly, the equation of another side AD is given by −7 y – 2 = (x – 1) or 7x + 23y – 53 = 0.23 Objective Type Questions Each of the Examples 12 to 20 has four possible options out of which only one option is correct. Choose the correct option (M.C.Q.). Example 12 The inclination of the line x – y + 3 = 0 with the positive direction of x-axis is (A) 45° (B) 135° (C) – 45° (D) –135° Solution (A) is the correct answer. The equation of the line x – y + 3 = 0 can be rewritten as y = x + 3 ⇒ m = tan θ = 1 and hence θ = 45°. Example 13 The two lines ax + by = c and a′x + b′y = c′ are perpendicular if (A) aa′ + bb′ = 0 (B) ab′ = ba′ (C) ab + a′b′ = 0 (D) ab′ + ba′ = 0 −aSolution (A) is correct answer. Slope of the line ax + by = c is ,b a−′and the slope of the line a′x + b′y = c′ is . The lines are perpendicular if b′ 3 tan θ = (1)5– x ⎛−a⎞⎛ − ′ a ⎞⎜ ⎟⎜ ⎟=−1 or aa′+bb′=0 (Why?)⎝ b ⎠⎝ b′⎠Example 14 The equation of the line passing through (1, 2) and perpendicular to x + y + 7 = 0 is (A) y – x + 1 = 0 (B) y – x – 1 = 0 STRAIGHT LINES 175 (C) y – x + 2 = 0 (D) y – x – 2 = 0. Solution (B) is the correct answer. Let the slope of the line be m. Then, its equation passing through (1, 2) is given by y – 2 = m (x – 1) (1) Again, this line is perpendicular to the given line x + y + 7 = 0 whose slope is – 1 (Why?) Therefore, we have m ( – 1) = –1 or m =1 Hence, the required equation of the line is obtained by putting the value of m in (1), i.e., y – 2 = x – 1 or y – x – 1 = 0 Example 15 The distance of the point P (1, – 3) from the line 2y – 3x = 4 is 7 (A) 13 (B) 13 13 (C) 13 (D) None of these Solution (A) is the correct answer. The distance of the point P (1, – 3) from the line 2y – 3 x – 4 = 0 is the length of perpendicular from the point to the line which is given by 2(–3) –3–4 = 13 13 Example 16 The coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y – 11 = 0 are (A) (–6, 5) (B) (5, 6) (C) (–5, 6) (D) (6, 5) Solution (B) is the correct choice. Let (h, k) be the coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y – 11 = 0. Then, the slope of the k –3 perpendicular line is . Again the slope of the given line x + y – 11 = 0 is – 1 h −2 (why?) Using the condition of perpendicularity of lines, we have ⎛ k –3⎞ ⎜ −⎠⎟ (– 1) = – 1 (Why?)⎝ h 2 or k – h = 1 (1) Since (h, k) lies on the given line, we have, h+ k – 11 = 0 or h+ k = 11 (2) Solving (1) and (2), we get h= 5 and k= 6. Thus (5, 6) are the required coordinates of the foot of the perpendicular. Example 17 The intercept cut off by a line from y-axis is twice than that from x-axis, and the line passes through the point (1, 2). The equation of the line is (A) 2x+ y= 4 (B)2x+ y+ 4 = 0 (C) 2x– y = 4 (D)2x– y+ 4 = 0 Solution (A) is the correct choice. Let the line make intercept ‘a’ on x-axis. Then, it makes intercept ‘2a’ on y-axis. Therefore, the equation of the line is given by xy+=1 a2a It passes through (1, 2), so, we have 12+=1or a=2 a2a Therefore, the required equation of the line is given by xy+=1 or 2x+ y= 424 Example 18 A line passes through P (1, 2) such that its intercept between the axes is bisected at P. The equation of the line is (A) x+ 2y= 5 (B) x– y+ 1 = 0 (C) x+ y– 3 = 0 (D) 2x+ y– 4 = 0 Solution The correct choice is (D). We know that the equation of a line making intercepts aand bwith x-axis and y-axis, respectively, is given by xy+=1.aba+00+bHere we have 1 = and 2 = , (Why?)22 which give a= 2 and b= 4. Therefore, the required equation of the line is given by STRAIGHT LINES 177 xy+=1 or 2x + y – 4 = 024 Example 19 The reflection of the point (4, – 13) about the line 5x + y + 6 = 0 is (A) (– 1, – 14) (B) (3, 4) (C) (0, 0) (D) (1, 2) Solution The correct choice is (A). Let (h, k) be the point of reflection of the given point (4, – 13) about the line 5x +y + 6 = 0. The mid-point of the line segment joining points (h, k) and (4, – 13) is given by ⎛ h+ 4 k −13 ⎞ ⎜ , ⎟ (Why?)⎝ 22 ⎠This point lies on the given line, so we have ⎛ h+ 4⎞ k −13 5⎜⎟+ + 6 = 0⎝ 2 ⎠ 2 or 5 h + k + 19 = 0 (1) k +13 Again the slope of the line joining points (h, k) and (4, –13) is given by . This lineh −4 ⎛ k +3⎞(5) = –1 −is perpendicular to the given line and hence ⎜⎟ (Why?)⎝ h − 4⎠This gives 5k + 65 = h – 4 or h – 5k – 69 = 0 (2) On solving (1) and (2), we get h = –1 and k = –14. Thus the point (–1, – 14) is the reflection of the given point. Example 20 A point moves such that its distance from the point (4, 0) is half that of its distance from the line x = 16. The locus of the point is (A) 3x2 + 4y2 = 192 (B) 4x2 + 3y2 = 192 (C) x2 + y2 = 192 (D) None of these Solution The correct choice is (A). Let (h, k) be the coordinates of the moving point. Then, we have ⎛⎞ 2 21 h −16 (–4) h + k =⎜ ⎟2 (Why?)⎟2 ⎜⎝ 1 + 0 ⎠ 1 ⇒ (h – 4)2 + k2 = 4 (h – 16)2 4 (h2 – 8h + 16 + k2) = h2 – 32h + 256 or 3h2 + 4 k2 = 192 Hence, the required locus is given by 3x2 + 4y2 = 192 Short Answer Type Questions 1. Find the equation of the straight line which passes through the point (1, – 2) and cuts off equal intercepts from axes. 2. Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, – 1). 3. Find the angle between the lines y = (2 – 3) (x + 5) and y = (2 + 3) (x – 7). 4. Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14. 5. Find the points on the line x + y = 4 which lie at a unit distance from the line 4x + 3y = 10. x y xy6. Show that the tangent of an angle between the lines +=1 and −=1 is ab ab 2ab . a2 −b2 7. Find the equation of lines passing through (1, 2) and making angle 30° with y-axis. 8. Find the equation of the line passing through the point of intersection of 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x + 4y = 7. 9. For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x – 3y + 6 = 0 on the axes. 10. If the intercept of a line between the coordinate axes is divided by the point (–5, 4) in the ratio 1 : 2, then find the equation of the line. 11. Find the equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of 120° with the positive direction of x-axis. STRAIGHT LINES 179 [Hint: Use normal form, here ω=30°.] 12. Find the equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by 3x+ 4y= 4 and the opposite vertex of the hypotenuse is (2, 2). Long Answer Type 13. If the equation of the base of an equilateral triangle is x+ y= 2 and the vertex is (2, – 1), then find the length of the side of the triangle. [Hint: Find length of perpendicular (p) from (2, – 1) to the line and use p = lsin 60°, where lis the length of side of the triangle]. 14. A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2, 0), (0, 2) and (1, 1) on the line is zero. Find the coordinates of the point P. [Hint: Let the slope of the line be m. Then the equation of the line passing through the fixed point P (x1, y1) is y– y1 = m(x– x1). Taking the algebraic sum of perpendicular distances equal to zero, we get y– 1 = m(x– 1). Thus (x1, y1) is (1, 1).] 15. In what direction should a line be drawn through the point (1, 2) so that its point 6of intersection with the line x+ y= 4 is at a distance from the given point.316. A straight line moves so that the sum of the reciprocals of its intercepts made on axes is constant. Show that the line passes through a fixed point. xy 11 1[Hint: +=1 where +=constant = (say). This implies thatabab kkk+=1⇒line passes through the fixed point (k, k).]ab 17. Find the equation of the line which passes through the point (– 4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point. 18. Find the equations of the lines through the point of intersection of the lines 7 x– y+ 1 = 0 and 2x– 3y+ 5 = 0 and whose distance from the point (3, 2) is .5 19. If the sum of the distances of a moving point in a plane from the axes is 1, then find the locus of the point. [Hint: Given that x+y=1 , which gives four sides of a square.] 20. P, P are points on either of the two lines y– = 2 at a distance of 5 units x12 3 from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from P1, P2 on the bisector of the angle between the given lines. [Hint: Lines are y= 3 x+ 2 and y= – 3 x+ 2 according as x≥0 or x< 0. y-axis is the bisector of the angles between the lines. P1, P2 are the points on these lines at a distance of 5 units from the point of intersection of these lines which have a point on y-axis as common foot of perpendiculars from these points. The y-coordinate of the foot of the perpendicular is given by 2 + 5 cos30°.] xy121. If pis the length of perpendicular from the origin on the line +=and a2,ab p2, b2 are in A.P, then show that a4 + b4 = 0. Objective Type Questions Choose the correct answer from the given four options in Exercises 22 to 41 22. A line cutting off intercept – 3 from the y-axis and the tengent at angle to the x3axis is , its equation is5 (A) 5y– 3x+ 15 = 0 (B) 3y– 5x+ 15 = 0 (C) 5y– 3x– 15 = 0 (D) None of these 23. Slope of a line which cuts off intercepts of equal lengths on the axes is (A) –1 (B) –0 (C) 2 (D) 3 24. The equation of the straight line passing through the point (3, 2) and perpendicular to the line y= xis (A) x– y = 5 (B) x+ y= 5 (C) x+ y = 1 (D) x– y= 1 25. The equation of the line passing through the point (1, 2) and perpendicular to the line x+ y+ 1 = 0 is (A) y– x+ 1 = 0 (B) y– x–1 = 0 (C) y– x+ 2 = 0 (D) y– x– 2 = 0 26. The tangent of angle between the lines whose intercepts on the axes are a, – b and b, – a, respectively, is STRAIGHT LINES 181 22 22a −b b −a(A) (B)ab 2 b2 −a2 (C) (D) None of these 2ab xy27. If the line +=1 passes through the points (2, –3) and (4, –5), then (a, b) isab (A) (1, 1) (B) (– 1, 1) (C) (1, – 1) (D) (– 1, –1) 28. The distance of the point of intersection of the lines 2x – 3y + 5 = 0 and 3x + 4y = 0 from the line 5x– 2y = 0 is 130 13 130 (A) (B) (C) (D) None of these 1729 729 7 29. The equations of the lines which pass through the point (3, –2) and are inclined at 60° to the line 3 x + y = 1 is (A) y + 2 = 0, 3 x – y – 2 – 3 3 = 0 (B) x – 2 = 0, 3 x – y + 2 + 3 3 = 0 (C) 3 x – y – 2 – 3 3 = 0 (D) None of these 330. The equations of the lines passing through the point (1, 0) and at a distance 2 from the origin, are (A) 3 x + y – 3 = 0, 3 x – y – 3 = 0 (B) 3 x + y + 3 = 0, 3 x – y + 3 = 0 (C) x + 3 y – 3 = 0, x – 3 y – 3 = 0 (D) None of these. 31. The distance between the lines y = mx + c1 and y = mx + c2 is cc− | cc− |12 12 −cc(A) 2 (B) 2 (C) 21 (D)0 m +11+m 21+m 32. The coordinates of the foot of perpendiculars from the point (2, 3) on the line y = 3x + 4 is given by ⎛ 37 −1⎞ ⎛−137 ⎞⎛ 10 ⎞⎛ 21⎞ ,, − , −,10 (A) ⎜⎟ (B) ⎜⎟ (C) ⎜⎟ (D) ⎜⎟⎝ 10 10 ⎠⎝ 10 10 ⎠⎝ 37 ⎠⎝ 33⎠ 33. If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is (3, 2), then the equation of the line will be (A) 2x + 3y = 12 (B) 3x + 2y = 12 (C) 4x – 3y = 6 (D)5x – 2y = 10 34. Equation of the line passing through (1, 2) and parallel to the line y = 3x – 1 is (A) y + 2 = x + 1 (B) y + 2 = 3 (x + 1) (C) y – 2 = 3 (x – 1) (D) y – 2 = x – 1 35. Equations of diagonals of the square formed by the lines x = 0, y = 0, x = 1 and y = 1 are (A) y = x, y + x = 1 (B) y = x, x + y = 2 (C) 2y = x, y + x = (D) y = 2x, y + 2x = 113 36. For specifying a straight line, how many geometrical parameters should be known? (A) 1 (B)2 (C)4 (D)3 37. The point (4, 1) undergoes the following two successive transformations : (i) Reflection about the line y = x (ii) Translation through a distance 2 units along the positive x-axis Then the final coordinates of the point are ⎛ 77 ⎞ ,(A) (4, 3) (B) (3, 4) (C) (1, 4) (D) ⎜⎟⎝ 22 ⎠ 38. A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is (A) (1, –1) (B) (1, 1) (C) (0, 0) (D) (0, 1) 39. A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is 12 4 (A) (B) (C) 1 (D) 33 3 STRAIGHT LINES 183 40. The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x +4y – 5 = 0 is (A) 1 : 2 (B) 3 : 7 (C) 2 : 3 (D) 2 : 5 41. One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is (A) (–1, –1) (B) (2, 2) (C) (–2, –2) (D) (2, –2) [Hint: Let ABC be the equilateral triangle with vertex A(h, k) and let D (α, β) 2α+h 2β+k ==be the point on BC. Then 0 . Also α + β – 2 = 0 and33 ⎛k −0⎞(1) −1×− =].⎜⎝h −0⎠⎟Fill in the blank in Exercises 42 to 47. 42. If a, b, c are in A.P., then the straight lines ax + by + c = 0 will always pass through ____. 43. The line which cuts off equal intercept from the axes and pass through the point (1, –2) is ____. 44. Equations of the lines through the point (3, 2) and making an angle of 45° with the line x – 2y = 3 are ____. 45. The points (3, 4) and (2, – 6) are situated on the ____ of the line 3x – 4y – 8 = 0. 46. A point moves so that square of its distance from the point (3, –2) is numerically equal to its distance from the line 5x – 12y = 3. The equation of its locus is ____. 47. Locus of the mid-points of the portion of the line x sin θ + y cos θ= p intercepted between the axes is ____. State whether the statements in Exercises 48 to 56 are true or false. Justify. 48. If the vertices of a triangle have integral coordinates, then the triangle can not be equilateral. 49. The points A (– 2, 1), B (0, 5), C (– 1, 2) are collinear. 50. Equation of the line passing through the point (a cos3θ, a sin3θ) and perpendicular to the line x sec θ + y cosec θ = a is x cos θ – y sin θ = a sin 2θ. 51. The straight line 5x + 4y = 0 passes through the point of intersection of the straight lines x + 2y – 10 = 0 and 2x + y + 5 = 0. 52. The vertex of an equilateral triangle is (2, 3) and the equation of the opposite side is x + y = 2. Then the other two sides are y – 3 = (2 ± 53. The equation of the line joining the point (3, 5) to the point of intersection of the lines 4x + y – 1 = 0 and 7x – 3y – 35 = 0 is equidistant from the points (0, 0) and (8, 34). xy 111 54. The line +=1 moves in such a way that += , where c is a constant.3) (x – 2). 2 2 2ab abc The locus of the foot of the perpendicular from the origin on the given line is x2 + y2 = c2. 55. The lines ax + 2y + 1 = 0, bx + 3y + 1 = 0 and cx + 4y + 1 = 0 are concurrent if a, b, c are in G.P. 56. Line joining the points (3, – 4) and (– 2, 6) is perpendicular to the line joining the points (–3, 6) and (9, –18). Match the questions given under Column C1 with their appropriate answers given under the Column C2 in Exercises 57 to 59. 57.Column C1 Column C2 (a) The coordinates of the points (i) (3, 1), (–7, 11) P and Q on the line x + 5y = 13 which are at a distance of 2 units from the line 12x – 5y + 26 = 0 are ⎛ 111 ⎞⎛ 47 ⎞ , ,,(b) The coordinates of the point on the line (ii) ⎜− ⎟⎜ ⎟⎝ 33 ⎠⎝ 33⎠ x + y = 4, which are at a unit distance from the line 4x + 3y – 10 = 0 are ⎛ 12⎞⎛ 16⎞1, ,–3, (c) The coordinates of the point on the line (iii) ⎜⎟⎜ ⎟⎝ 5 ⎠⎝ 5 ⎠ joining A (–2, 5) and B (3, 1) such that AP = PQ = QB are 58. The value of the λ, if the lines (2x + 3y + 4) + λ (6x – y + 12) = 0 are Column C1 Column C2 3 (a) parallel to y-axis is (i) λ=−4 STRAIGHT LINES 185 (b) perpendicular to 7x + y – 4 = 0 is (ii) 1 3 λ=− (c) passes through (1, 2) is (iii) 17 41 λ=− (d) parallel to x axis is (iv) λ = 3 59. (a) (b) (c) (d) The equation of the line through the intersection of the lines 2x – 3y = 0 and 4x – 5y = 2 and Column C1 Column C2 through the point (2, 1) is (i) 2x – y = 4 perpendicular to the line (ii) x + y – 5 = 0 x + 2y + 1 = 0 is parallel to the line (iii) x – y –1 = 0 3x – 4y + 5 = 0 is equally inclined to the axes is (iv) 3x – 4y – 1 = 0

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