Question

4. Find the ratio in which the line segment joining the points
(4,-3) and (8,5) is divided by (7,3).​

Answer 1

Question:

• Find the ratio in which the line segments joining the points (4,-3) and (8,5) is divided by (7,3).

Solution:

To find: Ratio of the line segments

Step-by-step explanation:

❍ Let A(4,-3) and B(8,5) be points and the point at which it is divided be C(7,3).

➜ AC² = (7 - 4)² + (3 + 3)²

➜ AC² = (3)² + (6)²

➜ AC² = 9 + 36

➜ AC² = 45

➜ BC² = (7 - 8)² + (3 - 5)²

➜ BC² = (-1)² + (-2)²

➜ BC² = 1 + 4

➜ BC² = 5

• Dividing the values of AC/BC, weget:

➜ AC/BC = 45/5

➜ AC/BC = 9/1

• The ratio in which the line segment is divided is 9:1.

______________________

Answer 2

$\mathfrak{{\pmb{{\underline{To~find}}:}}}$

• The ratio in which the line segment joining the points $\sf{\pmb{(4,-3)}}$ and $\sf{\pmb{(8,5)}}$ is divided by $\sf{\pmb{(7,3)}}$

$\mathfrak{{\pmb{{\underline{Solution}}:}}}$

Let the points be :–

• A(4, –3)
• B(8, 5)
• C(7, 3)

Here, in line segment AB ,

• $\sf x_{1} = { \bf{4}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: x_{2} = { \bf{8}}$
• $\sf y_{1} = {\bf{ - 3}}\: \: \: \: \: \: \: \: \: \: y_{2} = {\bf{5}}$

Using section formula, we get :–

$\sf x = \dfrac{m_{1} \:x_{2} + m_{2} \: x_{1}}{m_{1} + m_{2}} \: \: \\ \\ \sf \implies 7 = \dfrac{m_{1} \times 8 + m_{2} \times 4}{m_{1} + m_{2}} \: \: \: \: \: \: \: \: \\ \\ \sf \implies \frac{8m_{1} + 4m_{2}}{m_{1} + m_{2}} = 7 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \implies 8m_{1} + 4m_{2} = 7(m_{1} + m_{2}) \\ \\ \sf \implies 8m_{1} + 4m_{2} = 7m_{1} + 7m_{2} \: \\ \\ \sf \implies 8m_{1} - 7m_{1} = 7m_{2} - 4m_{2} \: \\ \\ \sf \implies 1m_{1} = 3m_{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \implies \frac{1}{3} = \frac{m_{2}}{m_{1}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \implies \frac{m_{1}}{m_{2}} = \frac{3}{1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \therefore \blue{ \underline{ \boxed{ \sf{ \pmb{m_{1} = 3 \: \: \: \: \: m_{2} = 1}}}}} \: \: \: \: \: \: \: \:$

$\therefore\mathfrak{{\pmb{{\underline{Required~answer}}:}}}$

The ratio is :-

• $\sf{\pmb{m_{1}:m_{2}=3:1}}$

$\sf{}$

• Refer the attachment for the figure