EQUATION OF THE LINE PASSING THROUGH THE INTERSECTION OF THE LINES

Question 1 :

Find the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x − 4y = 1 and parallel to the line 13x + 5y +12 = 0

Solution :

First, let us find the point of intersection of the given lines 7x + 3y = 10, 5x − 4y = 1.

7x + 3y = 10  ----(1)

5x − 4y = 1  ----(2)

4(1) + 3(2)

28x + 12y  =  40

15x - 12y  =  3

------------------

43x  =  43

x  =  1

By applying the value of x in (1), we get 

7(1) + 3y  =  10

7 + 3y  =  10

3y  =  10 - 7

3y  =  3

y  =  1

So, the point of intersection of the given lines is (1, 1)

The required line is passing through the point (1, 1) and parallel to the line 13x + 5y + 12  =  0

Slope of the required line  =  -13/5

(y - y₁)  =  m(x - x₁)

(y - 1)  =  (-13/5)(x - 1)

5(y - 1)  =  -13(x - 1)

5y - 5  =  -13x + 13

13x + 5y - 5 - 13  =  0

13x + 5y - 18  =  0

Question 2 :

Find the equation of a straight line through the intersection of lines 5x −6y = 2, 3x + 2y = 10 and perpendicular to the line 4x −7y +13 = 0

Solution :

5x −6y = 2 -----(1) 

3x + 2y = 10 -----(2) 

        (1)  ==>  5x −6y = 2 

     3(2) ==>   9x + 6y = 30

                ------------------

                    14x  =  32

x  =  32/14  =  16/7

By applying the value of x in (1), we get

5(16/7) - 6y  =  2

(80/7) - 6y  =  2

6y  =  (80/7) - 2

6y  =  (80 - 14)/7

y =  11/7

The point of intersection of the given lines is (16/7, 11/7)

The required line is perpendicular to the line 4x −7y +13 = 0

slope of the required line  =  -4/(-7)  =  4/7

(y - y₁)  =  (-1/m)(x - x₁)

(y - (11/7))  =  (-7/4)(x - (16/7))

(7y - 11)/7  =  (-7/4)(7x - 16)/7)

4(7y - 11)  =  -7(7x - 16)

28y - 44  =  -49x + 112

49x + 28y - 44 - 112  =  0

49x + 28y - 156  =  0

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