Question

Consider the family of lines 5x+3y-2+lambda 1(3x-y-4)=0 and x-y+1+lambda 2(2x-y-2)=0.Equation of a straight lines that belong to both families is.?

Answer 1

hence, equation of a straight lines that belongs to both families is 9x - 17y + 16 = 0

first find point of intersection of of (5x + 3y - 2) = 0 and (x - y + 1) = 0

so, (5x + 3y - 2) + 3(x - y + 1) = 0

⇒5x + 3y - 2 + 3x - 3y + 3 = 0

⇒8x + 1 = 0

⇒x = -1/8

and y = x + 1 = 7/8

so, point of intersection is (-1/8, 7/8)

again find point of intersection of (3x - y - 4) = 0 and (2x - y - 2) = 0

so, (3x - y - 4) - (2x - y - 2) = 0

⇒3x - y - 4 - 2x + y + 2 = 0

⇒x - 2 = 0, x = 2

now, y = 2x - 2 = 2

hence, point of intersection is (2, 2)

now find equation of line passing through point (-1/8, 7/8) and (2,2) .

i.e., (y - 2) = (2 - 7/8)/(2 + 1/8) (x - 2)

⇒(y - 2) = 9/17 (x - 2)

⇒17y - 34 = 9x - 18

⇒9x - 17y + 16 = 0

hence, equation of a straight lines that belongs to both families is 9x - 17y + 16 = 0

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Answer 2

Equation of a straight lines that belong to both families is 5x-2y=7

Step-by-step explanation:

Consider the family of lines,

$5x+3y-2+\lambda (3x-y-4)=0$

Two equations of this family.

$5x+3y-2=0$

$3x-y-4=0$

Family of equation of lines are passing from same points.

• Point of intersection of 5x+3y-2=0 and 3x-y-4=0

5x + 3y = 2

9x - 3y = 12

     14x = 14

        x = 1

Put x = 1 into 3x - y = 4

y = -1

Point of intersection: (1,-1)

Another family of equation of lines,

$x-y+1+\lambda 2(2x-y-2)=0$

System of equation of family of lines

$x-y=-1$

$2x-y=2$

• Point of intersection of x-y=-1 and 2x-y=2

  x - y = -1

-2x + y = -2

      -x = -3

      x = 3

Put x = 3 into x-y=-1

3 - y = -1

    y = 4

Point of intersection: (3,4)

Equation of a straight lines that belong to both families must be passes from these two points.

Passing point: (1,-1) and (3,4)

Equation of line of two point form:-

$y-4=\dfrac{-1-4}{1-3}(x-3)$

$y-4=\dfrac{5}{2}(x-3)$

$2y-8=5x-15$

$5x-2y=7$

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Family of lines

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