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

Answer:

$\bigstar{\bold{Ratio=1:1}}$

$\bigstar{\bold{Value\:of\:m=0}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Point A = (2,3)
• Point B = (6, -3)
• Point P = (4, m)

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Ratio in which the line segment is divided
• The value of m

$\Large{\underline{\underline{\bf{Solution:}}}}$

➞Here we have to find the ratio in which the line segment is divided.

➞ Let us assume the ratio as k : 1

➞ By section formula we know that,

    $\tt{(x,y)=\bigg(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2}\bigg)}$

    where x₁ = 2, x₂ = 6, y₁ = 3, y₂ = -3, m₁ = k, m₂ = 1, x = 4, y = m

➞ Substitute the data,

    $\tt{(4,m)=\bigg(\dfrac{6k+2}{k+1},\dfrac{3k-3}{k+1} \bigg)}$

➞ Equating the x coordinate,

    $\tt{\dfrac{6k+2}{k+1}=4}$

➞ Cross multiplying,

    6k + 2 = 4 (k + 1)

    6k + 2 = 4k + 4

    6k - 4k = 4 - 2

    2k = 2

      k = 1

➞ Hence the line segment is divided in the ratio 1 : 1

    $\boxed{\bold{Ratio=1:1}}$

➞ Now equate the y coordinate,

    $\tt{\dfrac{3k-3}{k+1}=m}$

➞ Substitute the value of k,

    3 × 1 - 3 = m (1 + 1)

    3 - 3 = 2 m

    0 = 2 m

     m = 0

➞ Hence the value of m is 0

    $\boxed{\bold{Value\:of\:m=0}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

➞ Section formula is given by,

    $\tt{(x,y)=\bigg(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2}\bigg)}$

Answer 2

The value of m is 0

Step-by-step explanation:

• Let the point P divides the line AB in k :1.

• The coordinates of line AB are A (2, 3) and B (6, -3).

• The coordinates of P are (4,m).

• By section formula, if a point divides a line in m:n, then

$=> \sf P = \bigg \lgroup \dfrac{x_2 m + x_1 n}{m+n}, \dfrac{y_2 m + y_1 n}{m+n} \bigg \rgroup$

$=> \sf 4 = \dfrac{6k+ 2}{k + 1}$

$=> \sf 4 \: (k + 1) = 6k + 2$

$=> \sf 4k + 4= 6k + 2$

$=> \sf 4 - 2= 6k - 4k$

$=> \sf 2k = 2$

$=> \underline { \boxed{ \sf k = 1}}$

It means the point P divides the line AB in 1 : 1. So, P is the midpoint of AB.

$= > \sf m=\dfrac{3+(-3)}{2}$

$= > \sf m = \dfrac{0}{2}$

$= > \underline{ \boxed{ \sf m = 0}}$