Question

find the ratio in which p (4,m)divides the line segment joining the points A (2,3) and b (6, - 3) hence find m

Answer 1

A( 2 , 3 )  _p( 4 , m )  _B( 6 , - 3 )

              m₁               m₂

Given points : A( 2 , 3 ) and B( 6 , - 3 ).

Point which divides A and B = p( 4 ,m )

Let the ratio in which the point p( 4 , m ) divides A and B be m₁ : m₂

Let, x₁ = 2 , x₂ = 6 , y₁ = 3 , y₂ = - 3

X = 4 , Y = m

By section formula,

⇒ X = $\dfrac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}}$

⇒ 4 = $\dfrac{(m_{1} \times 6 ) + ( m_{2} \times 2 )}{m_{1}+m_{2}}$

⇒ 4( m₁ + m₂ ) = 6m₁ + 2m₂

⇒ 4m₁ + 4m₂ = 6m₁ + 2m₂

⇒ 4m₂ - 2m₂ = 6m₁ - 4m₁

⇒ 2m₂ = 2m₁

⇒ m₂ = m₁

⇒ $\dfrac{1}{1} =\dfrac{m_{1}}{m_{2}}$

∴ m₁ = 1 and m₂ = 1

∴ The ratio in which point p divides A and B is 1 : 1 .

Again by section formula,

⇒ Y = $\dfrac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}}$

⇒ m = $\dfrac{(1 \times - 3 ) + ( 1 \times 3 )}{1+1}$

⇒ m = $\dfrac{-3+3}{3}$

⇒ m = 0

Therefore the value of m is 0 .

Answer 2

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the point are A (2 ,3 ) B (6 , -3 )

the distance AB= ( 6 - 2) " 2 + ( - 3 - 3)"2 ) " 0 .5 = ( 4 " 2 + 6 " 2 )" 0.5= 52 "0.5 or 2 *13" 0.5.

The equation of line AB: (x-2)/(6–2) = (y-3)/(-3–3)

(x-2)/4 = (y-3)/(-6), or

12–6x = 4y-12, or

4y=-6x+24, or

2y = -3x+12 …(1)

If the point P (4,m) lies on the line AB, then

2m = -12+12 = 0, or m = 0

Hence the coordinates of P are (4,0)

A (2,3), P (4,0)

Distance AP = [(4–2)^2+(3–0)^2]^0.5 = [4+9]^0.5 = 13^0.5


AP/AB =13^0.5 /2*13^0.5 = 1:2

Hence P is the midpoint of AB and divides the line in the ratio of 1:1.

Also m = (-3k+3)/(k+1) = (-3+3)/1+1 = 0. So m= 0.