maths-straight lines iv

Family of lines

 The general equation of the family of lines through the point of intersection of two given lines is L + λL’ = 0, where L = 0 and L’ = 0 are the two given lines, and λ is a parameter.  Illustration: A variable line through the point of intersection of the lines x/a + y/b = 1 and x/b + y/a = 1 meets the coordinate axes in A and B. Show that the locus of the midpoint of AB is the curve 2xy(a + b) = ab(x + y).  Solution:         Let (h, k) be the midpoint of the variable line AB.         The equation of the variable line AB is                (bx + ay – ab) + λ(ax + by – ab) = 0 

         Coordinates of A are (ab(1+λ)b+λa).         Coordinate of B are (0, ab(1+λ)b+λa).         Mid point of AB is (ab(1+λ)2(b+λa), ab(1+λ)2(a+λb))                 ⇒ h = ab(1+λ)2(b+λa); k = ab(1+λ)2(a+λb) ⇒ 1/2h = b+λa/ab(1+λ); 1/2k = a+λb/ab(1+λ)                 ⇒ 1/2h + 1/2k = a+b/ab ⇒ (h + k)ab = 2hk (a + b).         Hence the locus of the midpoint of AB is (x + y) ab = 2xy (a + b).  To find the equation to the straight lines which pass through a given point (x1, y1) and make equal angles with the given straight line y = m1x + c. If m is the slope of the required line and α is the angle which this line makes with the given line, then tana = + m1–m/1+m1m. 

  (i)     The above expression for tanα, gives two values of m, say mA and mB.  (ii)    The required equations of the lines through the point (x1, y1) and making equal angles α with the given line are y – y1 = mA (x – x1), y – y1 = mB(x – x1).  Illustration: Find the equations to the sides of an isosceles right-angled triangle, the equation of whose hypotenuse is 3x + 4y = 4 and the opposite vertex is the point (2, 2).  Solution:         The problem can be restated as:         Find the equations to the straight lines passing through the given point (2, 2) and making equal angles of 45o with the given straight line 3x + 4y – 4 = 0         Slope of the line 3x + 4y – 4 = 0 is m1 = –3/4 

                 ⇒ tan 45o = + m1–m/1+m1m, i.e., 1 = + m+3/4/1–3/4m

         So that mA = 1/7,           and   mB = –7.         Hence the required equations of the two lines are                 y – 2 = mA (x – 2) and y – 2 = mB(x – 2).                 ⇒ 7y – x – 12 = 0 and 7x + y = 16.  Illustration: The straight lines 3x + 4y = 5 and 4x – 3y = 15 intersect at the point A. On these lines points B and C are chosen so that AB = AC. Find the possible equations of the line BC passing through (1, 2).  Solution:         The two given straight lines are at right angles.         Since AB = AC, the triangle is an isosceles right angled triangle.         The required equation is of the form y – 2 = m(x – 1)              … (1)                 with tan 45o = + m+3/4/1–3m/4 = + m–4/3/1+4m/3                 ⇒ 1 = + m+3/4/1–3m/4 = + m–4/3/1+4m/3 ⇒ m = – 7, 1/7.         Substitute the value of m in (1). We get the required equations.  Illustration: Find the equation of the straight line passing through (–2, 7) and having intercept of length 3 units between the straight lines 4x + 3y = 12 and 4x + 3y = 3.  Solution:         Distance AC between the two given parallel lines               = |c1–c2/√a2+b2| = 12–3/√16+9 = 9/5. 

         Let AB be the intercept of length 3 units.         ⇒ BC = 12/5. If θ is the angle between BC and AB, then tanθ = 9/12 = 3/4.         Slope of the parallel lines = –4/3 = m2. If m1 is the slope of the required line, then tanθ = m1–m/1+m1m ⇒ 3/4 = + m1–4/3/1+4/3m1

 i.e. m1 + 4/3 = 3/4 (1 – 4/3 m1) and m1 + 4/3 = 3/4 (1 – 4/3 m1). The slopes are (i) m1 = –7/24  (ii) m1 = ∞ (the line is parallel to the y-axis). The required equations of the lines are 7x + 24y + 182 = 0 and x + 2 = 0.  Alternative solution: Equation of the line, through P(–2, 7) and making angle θ with the x-axis, is x+2/cosθ = y+7/sin θ = r. If this line intersects the given lines at A and B, with AB = 3, the points A and B are A(–2 + r1 cos θ, – 7 + r1 sin θ) and B (–2 + (r1 + 3) cos θ, – 7 + (r1 + 3) sin θ). Since A and B lie on the lines 4x + 3y = 3 and 4x + 3y = 12, we have 4r1 cos θ + 3r1 sin θ = 32 and 4r1 cos θ+ 3r1 sin θ + 12 cos θ + 9 sin θ = 41, so that 12 cos θ + 9 sin θ = 0 or 4 cos θ + 3 sin θ = 3. 

Solving this equation we find that θ = π/2 and tan θ = –7/24. Hence the required lines are x + 2 = 0 

and y + 7 = –7/24 (x + 2) i.e. 7x + 24y + 182 = 0.

 Concurrency of Straight Lines

 The condition for 3 lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0, a3x + b3y + c3 = 0 to be concurrent is 

        (i)      = 0.  (ii)    There exist 3 constants l, m, n (not all zero at the same time) such that IL1 + mL2 + nL3 = 0, where L1 = 0, L2 = 0 and L3 = 0 are the three given straight lines.  (iii)    The three lines are concurrent if any one of the lines passes through the point of intersection of the other two lines. Illustration:         Check if lines                 a1 x + b1 y + c1 = 0                                             …… (1)                 a2 x + b2 y + c2 = 0                                             …… (2)                 (2a1 – 3a2)x + (2b1 – 3b2)y + (2c1 – 3c2) = 0           …… (3)         are concurrent?Solution:        We can try to find α, ß and γ by observation as follow:                 L3 – 2L1 + 3L2 = 0 Enquiry:    many lines can pass through the intersection of two lines. Can we find a general equation of these lines? If L1 = 0 and L2 = 0 are two lines then equation of family of lines passing through their intersection is given by L1 + λ L2 = 0                                                       …… (A) 

Where ‘λ’ is any parameter. (Equation A is satisfied by the point of intersection of L1= 0 and L2 = 0) Note: To determine a particular line one more condition is required so as to determine or eliminate λ . Illustration:         If x (2q + p) + y(3q + p) = 0         (x + y – 1) + q/p (2x + 3y – 1) = 0, p ≠ 0Solution:This equation represents the family of lines passing through the intersection of lines x + y – 1 = 0 and 2x + 3y – 1 = which is fixed point i.e. (2, –1).         If p = 0 then equation becomes                 q(2x + 3y – 1) = 0         this also represents a line which passes through fixed points (2, –1). Hence the given equation represents family of lines passing through a fixed point (2, –1) for variable p, q. Illustration: Find the equation of a line, through the intersection of 2x + 3y – 7 = 0 and x + 3y – 5 = 0 and having distance from origin as large as possible. 

 Point of intersection of two lines is A(2, 1) 

Now, with OA as radius and O itself as centre draw a circle.         There will be infinitely many lines through A and each except one of them produces a chord of circle and hence their distance from origin i.e. centre of circle is less than OA i.e. radius of circle.         But the exceptional one which infact is a tangent to circle at A will be at a distance OA from O.         Thus, tangent to circle at A will be the line through A and is farthest from origin. Now, OA ⊥ tangent at A. ∴ (slope of OA) × (slope of tangent at A) = –1 Or, 1–0/2–0 (slope of tangent at A) = –2 ∴ equation of required line is         (y – 1) = –2(x – 2) Or     2x + y – 5 = 0 Illustration: Find the point of concurrency of the altitudes drawn from the vertices (at1t2, a(t1 + r2)), (at2t3, a2t2 + t3)) and (at3t1, a(t3 + t1)) respectively of a triangle ABC.  Solution:         Slope of AD = –t3.         Equation of AD is y – a(t1 + t2) = –t3(x + at1t2).                … (1)         Equation of CF is y – a(t3 + t1) = –t2(x – at3t1).                 … (2) 

         Subtracting (1) from (2), we get                 x = –a Þ y = a(t1 + t2 + t1t2t3).         Hence the point of concurrency of the altitudes is         (–a, a(t1 + t2 + t3 + t1t2t3)). Position of two points with respect to a given line Let the line be ax + by + c = 0 and P(x1, y1), Q(x2, y2) be two points.  Case 1: If P(x1, y1) and Q(x2, y2) are on the opposite sides of the line ax + by + c = 0, then the point R on the line ax + by + c = 0 divides the line PQ internally in the ratio m1 : m2, where m1/m2 must be positive. Co-ordinates of R         are (m1x2+m2x1/m1+m2, m1y2+m2y1/m1+m2). Point R lies on the line ax + by + c = 0.         ⇒ m1/m2 = ax1+by1+c/ax2+by2+c > 0 

 So that ax1 + by1 + c and ax2 + by2 + c should have opposite signs.  Case 2: If ax1 + by1 + c and ax2 + by2 + c have the same signs then m1/m2 = –ve, so that the point R on the line ax + by + c = 0 will divide the line PQ externally in the

ratio m1 : m2 and the points P(x1, y1) and Q(x2, y2) are on the same side of the line ax + by + c = 0.  Illustration: Find the range of θ in the interval (0, π) such that the points (3, 5) and (sinθ, cosθ) lie on the same side of the line x + y – 1 = 0.  Solution:         3 + 5 – 1 =7 > 0 ⇒ sinθ + cosθ – 1 > 0         ⇒ sin(π/4 + θ) > 1/√2         ⇒ π/4 < π/4 + θ < 3π/4         ⇒ 0 < θ < π/2.  Illustration:         Find a, if (α, α2) lies inside the triangle having sides along the lines                 2x + 3y = 1, x + 2y – 3 = 0, 6y = 5x – 1.  Solution:         Let A, B, C be vertices of the triangle.         A ≡ (–7, 5), B ≡ (5/4, 7/8),         C ≡ (1/3, 1/9).         Sign of A w.r.t. BC is –ve. 

 If p lies in-side the ¦ABC, then sign of P will be the same as sign of a w.r.t. the line BC ⇒     5α – 6α2 – 1 < 0.                         …… (1)         Similarly    2α + 3α2 – 1 > 0.                                …… (2)         And,          α + 2α2 – 3 < 0.                          …… (3)         Solving, (1), (2) and (3) for α and then taking intersection,         We get      α ? (1/2, 1) ∪ (–3/2, –1).  Illustration: The equations of the perpendicular bisectors of the sides AB and AC of a triangle ABC are respectively x – y + 5 = 0 and x + 2y = 0. If the co-ordinates of A are (1, –2), find the equation of BC.  Solution:         From the figure,         E ≡ (x1+1/2, y1–2/2),         F ≡ (x2+1/2, y2–2/2).

 Alt text : equations of the perpendicular bisectors of sides of triangle

       Since E and F lie on OE and OF respectively,                 x1 – y1 + 13 = 0                                  … (1)         and   x2 + 2y2 – 3 = 0                                  … (2)         Also, slope of AB = –1 and slope of AC is 2, so that                 x1 + y1 + 1 = 0.                                  … (3)         And   2x2 – y2 – 4 = 0                                  … (4)

         Solving these equations, we get the co-ordinates of B and C as                 B ≡ (–7, 6) and C ≡ (11/5, 2/5) ⇒ Equation of BC is 14x + 23y – 40 = 0.  Illustration: Two fixed points A and B are taken on the co-ordinate axes such that OA = a and OB = b. Two variable points A’ and B’ are taken on the same axes such that OA’ + OB’ = OA + OB. Find the locus of the point of intersection of AB’ and A’B.  Solution:         Let A ≡ (a, 0), B (0, b), A’ ≡ (a’, 0), B’ ≡ (0, b’).         Equation of A’B is x/a' + y/b' = 1.                                          …. (1)         and equation of AB’ is x/a + y/b'  = 1.                                    …. (2)         Subtracting (1) from (2), we get, x (1/a – 1/a') + y(1/b' – 1/b) = 0.         ⇒ x(a'–a)/aa' + y(b–b')/bb' = 0.                             [Using a’ – a = b – b’]         ⇒ x/a(b–b'+a) + y/bb', 0 ⇒ b’ = a(a+b)y/ay–bx.                              ….. (3)         From (2) b’x + ay = (4) we get x + y = a + b         which is the required locus.