Question

- ( Intercepts form two points form )

Answer 1

Answer:

Solution given:

A(a,b) lie in 6x-y =1

substituting value of A in equation ,we get

6a-b=1...................(1)

B(b,a) lie in 2x - 5y = 5

substituting value of B in equation ,we get

2b-5a=5

b=5(a+1)/2............(2)

substituting value of b in equation 1 ,we get

6a-5(a+1)/2=1

12a-5a-5=2

7a=2+5

a=$\frac{7}{7}$=1

substituting value of a in equation 1 ,we get

6×1-b=1

6-1=b

b=5

so points are A(1,5) and B(5,1)

we have

equation of line having two points

$(y - y1) = \frac{y2 - y1}{x2 - x1} (x - x1)$

$(y - 5) = \frac{1-5}{5-1} (x - 1)$

$(y - 5) = -1 (x - 1)$

y-5=1-x

x+y=6 is a required equation of line AB.

Answer 2

ANSWER:

If A(a,b) lies on the line 6x - y = 1,then equation of line becomes

• 6a - b = 1 ...(1)

If B(b,a) lies on the line 2x - 5y = 5, then equation of line becomes

• 2b - 5a = 5 ...(2)

Multiplying eq(1) by 2 and subtracting from eq(2), we get

• (12a - 2b) - (2b - 5a) = 2 - (-5)
• 12a - 2b - 2b + 5a = 2 + 5
• 7a = 7

So, a = 1

Putting the value of a = 1, in eq(1), we get

• 6 × 1 - b = 1
• 6 - b = 1

So, b = 5

So, the points are A(1,5) and B(5,1).

Now, the equation of line AB is

y - y1 = (y2 - y1)/(x2 - x1) × (x - x1)

• y - 5 = - 1(x - 1)
• x + y = 6

Hence, x + y = 6 is the equation of line AB from given points.