Question

17. A straight line cuts the coordinate axes at A and B. If the midpoint of AB is (3, 2), then
find the equation of AB.

Answer 1

Final Answer :
$\frac{x}{6} + \frac{y}{4} = 1$
Steps :
1) Let the x - intercept and y-intercept be 'a'and 'b'respectively .
Then,
A = (a, 0)
B = (0,b)

2) Now,
Mid -point of AB is given by P(x,y)
$= > x = \frac{(0 + a)}{2} = \frac{a}{2} \\ = > y = \frac{(b + 0)}{2} = \frac{b}{2}$

3) According to the Question,
P(x,y) = (3,2)
$= > \: \frac{a}{2} = 3 \\ = > a = 6 \: \: and \\ \frac{b}{2} = 2 \\ = > b = 4$


4) We have, a = 6 , b = 4
Then, Equation of line by Intercept Form :
$\frac{x}{a} + \frac{y}{b} = 1\\ \frac{x}{6} + \frac{y}{4} = 1$

Answer 2

Solution:

given by:- A straight line cuts the coordinate axes at A and B. If the midpoint of AB is (3, 2),

we have:-

Let
》x -intercept be= a

》 y-intercept be= b

》the coordinate of A is (a , 0) and coordinate of B is (0 , b)

Midpoint of AB = (x₁ + x₂)/2 , (y₁ + y₂)/2

》   (3 , 2) = (a + 0)/2 , (0 + b)/2

》    3 = a/2       2 = b/2

    a = 6          b = 4

Intercept form:

》(x/a) + (y/b) = 1

》(x/6) + (y/4) = 1

》(2 x + 3 y)/12 = 1

》2 x + 3 y = 12

》(2 x + 3 y - 12 = 0)

 
the equation of AB. ( 2 x + 3 y - 12 = 0) ans

☆i hope its help☆