Question

The incentre of the triangle formed the lines x = 0 , y = 0 and 3x + 4y = 12 is (a , b), then a + b = ​

Answer 1

Solution :-

• Draw a line and draw a x - axis , y axis
• Draw a perpendicular line
• It forms a right angle triangle

Given line 3x + 4y = 12

If x = 0

y = 3

The co-ordinates of y = (0,3) x axis is zero because It is away from y axis

So, y = (0,3)

b = (0,0) (origin)

If y = 0

3x + 4y = 12

x = 4

x = (4,0) y axis is zero because it is away from y axis

x = (4,0)

Now find the diatance between A,B,C

Let distance between AB = c

BC = a

AC = b

Distance between AB = $\sf\sqrt{(x_1-x_2)^2+ (y_1-y_2)^2}$

c = $\sf\sqrt{(0-0)^2 + (3-0)^2}$

c = $\sf\sqrt{9}$

c = 3

Diatance between BC = a

a = $\sf\sqrt{(x_1-x_2)^2+ (y_1-y_2)^2}$

a = $\sf\sqrt{(0-4)^2 + (0-0)^2}$

a = $\sf\sqrt{16}$

a = 4

Distance between AC = b

b = $\sf\sqrt{(x_1-x_2)^2+ (y_1-y_2)^2}$

b = $\sf\sqrt{(0-4)^2+(0-3)^2}$

b = $\sf\sqrt{16+9}$

b = $\sf\sqrt{25}$

b = 5

• a = 4
• b = 5
• c = 3
• x1 = 0
• y1 =3
• x2 = 0
• y2 = 0
• x3 = 4
• y3 = 0

Finding incentre

Incentre =

$\sf\dfrac{ax1+bx2+cx3}{a+b+c}$,$\sf\dfrac{ay1+by2+cy3}{a+b+c}$

I = $\sf\dfrac{0+0+12}{4+5+3}$,$\sf\dfrac{12+0+0}{4+5+3}$

I = $\sf\dfrac{12}{12}$,$\sf\dfrac{12}{12}$

I = (1 ,1 )

But ATQ I = (a,b)

Therefore a = 1 b = 1

a + b = 1 +1

a + b = 2

Answer 2

Answer:

a+b=1+1=2

Step-by-step explanation: