Question

the ratio in which the x-axis divides the line segment joining the points (4,6)and(3,-8)is​

Answer 1

Given:

• A line segment joining points (4, 6) and (3, -8)

To find:

• Ratio in which x-axis divides the given line segment

Formula required:

• Section formula

$\purple{\bigstar}\boxed{\sf{(x\;,\;y)=\left(\dfrac{mx_2+nx_1}{m+n}\;,\;\dfrac{my_2+ny_1}{m+n}\right)}}$

[ Where point (x, y) divide the line segment joining points  (x₁, y₁) and (x₂, y₂) in the ratio m : n ]

Solution:

Let, point at which x-axis and given line segment interesect be (x, 0)

and, Let x-axis divide the line segment in ratio k : 1.

then,

On comparison with section formula we will get,

• x = x ; y = 0
• x₁ = 4 ; y₁ = 6
• x₂ = 6 ; y₂ = -8
• m = k ; n = 1

Using section formula

$\implies\sf{(x\;,\;0)=\left(\dfrac{(k)(6)+(1)(4)}{k+1}\;,\;\dfrac{(k)(-8)+(1)(6)}{k+1}\right)}$

$\implies\sf{(x\;,\;0)=\left(\dfrac{6k+4}{k+1}\;,\;\dfrac{-8k+6}{k+1}\right)}$

$\implies\sf{x=\dfrac{6k+4}{k+1}\;\;,\;\;0=\dfrac{-8k+6}{k+1}}$

taking

$\implies\sf{0=\dfrac{-8k+6}{k+1}}$

$\implies\sf{-8k+6=0}$

$\implies\sf{-8k=-6}$

$\implies\boxed{\boxed{\sf{k=\dfrac{6}{8}=\dfrac{3}{4}}}}\;\;\;\purple{\bigstar}$

therefore,

• Ratio in which x -axis will divide the given line segment is k : 1 = 3 : 4.