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The Straight Line

1. Find the equation of the straight line passing through the point (x1,  y1) and making X and Y intercepts which are in the ratio m : n.

Sol: Given that X and Y intercepts are in the ratio m : n

           The intercepts may be taken as ma and na  such that  ma : na = m : n.

           The line in intercept form takes the equation =  

                                                                    (or)   nx + my = mna  ---------- (1)

        If this line passes through (x1, y1), we have nx1 + my1 = mna

                  (1) 

 nx + my = nx1 + my1 is the required equation.

2. A(10, 4), B(−4, 9), C(−2, −1) are the vertices of a triangle. Find the  equations of

     (i)                           (ii) The median through A       (iii) The altitude through B

     (iv) The perpendicular bisector of the side  .

Sol: (i) A = (10, 4), B = (−4, 9)

                The equation of AB =    
                                                       

                         5(x − 10) = −14(y − 4)    5x + 14y − 50 − 56 = 0

                         5x + 14y − 106 = 0

(ii) The median through A bisects the side BC i.e., passes through the  mid point of BC.

         B = (−4, 9), C = (−2, −1) implies mid point of BC = (−3, 4) = D  (say) and A = (10, 4).

        

   Equation of AD is y = 4  (or)  y - 4 = 0.

This is the eq. of the median through A.

(iii) The altitude through B is perpendicular to the side AC.

           A = (10, 4), C = (−2, −1) 

         Slope AC = 

           The slope of the altitude is  and the altitude passes through B = (−4, 9)

           Equation of the altitude through B is y − y1 = m(x − x1)

                     y − 9 =   (x + 4)

                    5y − 45 = −12x − 48

                    12x + 5y + 3 = 0

(iv) A = (10, 4), B = (−4, 9)

          

       Slope of its perpendicular =   

                           Mid point of AB = 

  Perpendicular bisector of AB passes through    and with slope  .

  Equation is y−y1 = m(x−x1)

                     y−  =  (x−3)

                     5y−  = 14x − 42

                     10y − 65 = 28x − 84    28x − 10y − 19 = 0.

3. If the portion of a line intercepted between the coordinate axes is cut at a point in the ratio m : n, then find the equation of the line.

Sol: Let   (or) bx + ay = ab be the line in intercept form.

 'a' is the X − intercept,   A = (a, 0) is a point on the X − axis.

 'b' is the Y - intercept,    B = (0, b) is a point on the Y − axis.

This line AB is cut at (x1, y1) say in m : n ratio.

                           

Note: (i) If (x1, y1) is the mid point of AB, then m = n and hence, the  equation takes the form   

(ii) If the point of bisection is (2p, 2q), then the equation takes the  form  

4. If the linear equations ax + by + c = 0 (abc ≠ 0) and lx + my + n = 0  represent  the same line and 

 write the value of r in terms of m and b. 

Sol:  ax + by + c = 0 and lx + my + n = 0 represent the same line.

         

5. Transform the equation (2 + 5k)x − 3(1 + 2k)y + (2 − k) = 0 into L1 + λL2 = 0 and find the point of concurrence of the family of straight lines.

Sol: (2 + 5k)x − 3(1 + 2k)y + (2 − k) = 0 can be written as (2x − 3y + 2)+ k(5x − 6y − 1) = 0

           This is of the form L1 + λL2 = 0 where   L1 ≡ 2x − 3y + 2 = 0, L2 ≡ 5x − 6y − 1 = 0

           Solving these two equations, we get the point of concurrence.

                                     4x − 6y + 4 = 0

                                    5x − 6y − 1 = 0                                  

                                      x − 5   = 0

                                   x = 5, y = 4

                                                               Point = (5, 4).

6. Find the value of 'p', if the straight lines x + p = 0 , y + 2 = 0 and 3x + 2y +5 = 0 are concurrent.

Sol:    x + p = 0  x = −p

            y + 2 = 0 

 y = −2

            3x + 2y + 5 = 0

 3(−p) + 2(-2) + 5 = 0

                  −3p + 1 = 0 

                            p = 

7. Find the area of the triangle formed by the line 3x − 4y+ 12 = 0 with  the coordinate axes.

Sol:  3x − 4y + 12 = 0

          a = X − intercept =  = −4

           b = Y−intercept =  = 3

           Area =    square units.

8. If 3a + 2b + 4c = 0 then, show that the equation ax + by + c = 0  represents a family of concurrent straight lines and find the point of concurrency.

Sol:  3a+ 2b + 4c = 0  c = 

          ax + by + c = 0  ax + by 

           4ax + 4by − 3a − 2b = 0

           a(4x − 3) + b(4y − 2) = 0

           (4x − 3) +  (4y − 2) = 0

   For various values of a and b, this equation represents a set of  concurrent lines.

                           The point of Concurrence = 

9. Astraight line through P(3, 4) makes an angle of 60° with the positive  direction of X−axis. Find the coordinates of the points on the line which are 5 units  away from P.

Sol:  This problem is related to the straight line in symmetric form which is

           given as 

           here (x1, y1) = (3, 4); θ = 60°;   = 5

  
 

10. Find the angle between the lines y = 4 − 2x and y = 3x + 7.

Sol: m1 = −2; m2 = 3

11. Find the equation of the straight line  parallel to the line 2x + 3y +7 = 0 and passing through the point (5, 4).

Sol:  The equation of the straight line passing through (x1, y1) and  parallel to the line

          ax + by + c = 0 is a(x − x1) + b(y − y1) = 0.

          Here, a = 2, b = 3, x1 = 5, y1 = 4.

           The equation required:

                                                          2(x − 5) + 3(y − 4) = 0

 2x + 3y − 22 = 0
 

12. Find the equation of the line perpendicular to the line 5x − 3y + 1 =0 and passing through the point (4, −3).

Sol:  The equation of the line passing through (x1, y1) and perpendicular  to the line

          ax + by + c = 0 is b(x − x1) − a(y − y1) = 0.

          Here, a = 5; b = −3; x1 = 4; y1 = −3.

          The equation required:

                                                   −3(x − 4) − 5(y + 3 ) = 0

                                                             −3x − 5y − 3 = 0

                                                              3x + 5y + 3 = 0

13. Find the value of 'k', if the straight lines y − 3kx + 4 = 0 and (2k − 1)x − (8k − 1)y −6 = 0  are perpendicular.

Sol:  Slope of y − 3kx + 4 = 0  is  3k = m1 (say)

          Slope of (2k − 1)x − (8k − 1)y − 6 = 0 is    = m2 (say)

          for perpendicular lines, m1m2 = −1

         

         6k2−3k = 1−8k (or) 6k2 + 5k−1 = 0

       

 6k(k +1)−1 (k+1) = 0

          k = −1  or .

14. Find the equations of the straight lines passing through (1, 3) and (i) parallel to

(ii) perpendicular to the line passing through the points (3, −5) and (−6, 1).

Sol:  Slope of the line passing through (3, −5) and (−6, 1) = 

Equation of the line passing through (1, 3) and parallel to the line with slope =

 isy − y1 = m(x − x1y − y1 = m(x − x1

 y − 3 =   (x − 1)

  3y − 9 = −2x + 2

  2x + 3y − 11 = 0

Equation of the line passing through (1, 3) and perpendicular to the line with slope =  is

y − y1 =   (x − x1)

 y − 3 =   (x−1)

  2y − 6 = 3x − 3

  3x − 2y + 3 = 0.

15. The line  

meets the X − axis at P. Find the equation of the line per pendicular to this line at P.

Sol:    represents the line in intercept form with X − intercept = a and Y − intercept = b in the 4th quadrant.

 P = (a, 0) and Q = (0, −b)

 

 16. Find the foot of the perpendicular drawn from (3, 0) upon the straight line 5x + 12y − 41 = 0.

Sol: If (h, k) is the foot of the perpendicular drawn from (x1, y1) upon the straight line ax + by + c = 0, then we have

17. Show that the distance of the point (6, −2) from the line 4x + 3y = 12 is half of the distance of the point (3, 4) from the line 4x − 3y = 12.

Sol:  The distance of the point (6, −2) from the line 4x + 3y − 12 = 0 is given by

The distance of the point (3, 4) from the line 4x−3y−12 = 0 is given by

 d2 = d1  the result proved.

18. Find the locus of the foot of the perpendicular from the origin to a variable straight line which always passes through a fixed point (a, b).

                                                       

Sol:  Let F(x1, y1) be a variable foot of the perpendicular drawn from origin that passes through P(a, b).

19. Find the equations of the straight lines passing through the point (−3, 2) and making an angle of 45° with the straight line 3x−y+4 = 0

Sol:  Let 'm1' be the slope of the required straight line that passes through the point (−3, 2).

 The equation of the line is y−2 = m1(x + 3) (1)

Let 'm2' be the slope of the given line 3x−y + 4 = 0

 m2 = 3

The required line makes an angle of 45° with the given line.

 ± (1 + 3m1) = (m1 − 3)2

 1 + 6m1 + 9m12 = m1 2 − 6m1 + 9

 8m1 2 + 12m1 − 8 = 0

 2m1 2 + 3m1 − 2 = 0

 2m1 (m1 + 2) − 1(m1 + 2) = 0

 (2m1 − 1)(m1 + 2) = 0

m1  = 1/2 or −2.

Putting these values in (1), we get the equations as follows:

m1  =  1/2--- y−2 =  1/2(x + 3)

2y − 4 = x + 3

x − 2y + 7 = 0         If m1 = −2

y − 2 = −2(x + 3)

2x + y + 4 = 0

20. Find the equations of the straight lines passing through the point of intersection of the lines 3x + 2y + 4 = 0, 2x + 5y = 1 and whose distance from (2, −1) is 2.

Sol:  Let the line required be of the form L1 + λL2 = 0 where

L1 ≡ 3x + 2y + 4 = 0; L2 ≡ 2x + 5y − 1 = 0

 (3x + 2y + 4) + λ(2x + 5y − 1) = 0 (1)

 (3 + 2λ)x + (2 + 5λ)y + (4 − λ) = 0

This has a distance '2' from (2, −1)

(4 − λ)2 = 29λ2 + 32λ + 13

28λ2 + 40λ − 3 = 0

(2λ + 3)(14λ − 1) = 0

λ = 1/14 (or) λ = -3/2

λ = -3/2 , (1)   (3x + 2y + 4) +  (2x + 5y − 1) = 0

4x + 3y + 5 = 0

λ = -3/2, (1)  (3x + 2y + 4) −( -3/2)  (2x + 5y − 1) = 0

y − 1 = 0

21. Find the Orthocentre of the triangle formed by the lines 4x − 7y + 10 = 0, x + y = 5 and 7x + 4y = 15.

Sol: 4x − 7y + 10 = 0, 7x + 4y − 15 = 0 form a pair of mutually perpendicular lines.

Hence, the angle between them is a right angle. And the two lines intersect at the point (1, 2).

Therefore, the triangle is right angled at (1, 2) and hence, the Orthocentre is (1, 2).

22. Find the values of k, if the angle between the straight lines kx + y + 9 = 0 and

3x - y + 4 = 0 is

Sol:  a1 = k;   a2 = 3;

          b1 = 1; b2 = -1    

 5(k2 + 1) = (3k - 1)2

 5k2 + 5 = 9k2 - 6k + 1

4k2 - 6k - 4 = 0

 2k2 - 3k - 2 = 0

 2k2 - 4k + k - 2 = 0

 2k(k - 2) + 1(k - 2) = 0

 (2k + 1)(k - 2) = 0

 k = -1/2 or 2.

23. Find the equation of the straight line parallel to the line 3x + 4y = 7 and passing through the point of intersection of the lines x - 2y - 3 = 0 and x + 3y - 6 = 0.

Sol:  The required equation can take the form

L1 + λL2 = 0 which represents a line passing through the intersection of the lines L1 = 0 and L2  = 0 for various values of the parameter λ.

Here, we have

(x - 2y - 3) + λ(x + 3y - 6) = 0

i.e., (1 + λ)x + (3λ - 2)y - (3 + 6λ) = 0

This is parallel to 3x + 4y = 7 (given)

 Slopes are equal

 4 + 4λ = 9λ - 6

 10 = 5λ (or) λ = 2

 Line required: (x - 2y - 3) + 2(x + 3y - 6) = 0

 3x + 4y - 15 = 0

Posted Date : 16-11-2021

గమనిక : ప్రతిభ.ఈనాడు.నెట్‌లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి.

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