in what ratio does the point (-4, 6) divide the line segment joining the points A (-6, 10) and (3, -8)
Answers
\pink{ \huge{❀ \: lol \: ❀}}❀lol❀
Step-by-step explanation:
In what ratio does the point \sf (-4,6)(−4,6) divide the line segment joining the points \sf (-6,10)(−6,10) and \sf (3,6)(3,6) ?
\begin{gathered} \\ \end{gathered}
\huge{\underline{\textsf{\textbf{\red{Solution...}}}}}
Solution...
Here, we are given three points as follows :
\green{\textsf{\textbf{A:(-6,10)}}}A:(-6,10)
\green{\textsf{\textbf{B:(3,6)}}}B:(3,6)
\green{\textsf{\textbf{C:(-4,6)}}}C:(-4,6)
Now, we shall find in what ratio the point C divides the line joining the points A and C.
We should find the ratio between AC and CB :
Let the required ratio be : \green{\sf{k:1}}k:1 .
So, now :
\sf m_1: km
1
:k
\sf m_2: 1m
2
:1
\sf x_1: -6x
1
:−6
\sf x_2: 3x
2
:3
\sf y_1:10y
1
:10
\sf y_2:6y
2
:6
\sf x:-4x:−4
\sf y:6y:6
Using the section formula :
\green{:\implies{\pmb{\sf{x = \dfrac{ m_{1} x_{2} + m_{2} x_{1}}{ m_{1} + m_{2}}}}}}:⟹
x=
m
1
+m
2
m
1
x
2
+m
2
x
1
x=
m
1
+m
2
m
1
x
2
+m
2
x
1
\sf: \implies - 4 = \dfrac{k(3) + 1( - 6)}{k + 1}:⟹−4=
k+1
k(3)+1(−6)
\sf: \implies - 4 = \dfrac{3k - 6}{k + 1}:⟹−4=
k+1
3k−6
\sf: \implies - 4(k + 1) = 3k - 6:⟹−4(k+1)=3k−6
\sf: \implies - 4k - 3k = - 6 + 4:⟹−4k−3k=−6+4
\sf: \implies - 7k = -2:⟹−7k=−2
\green{:\implies{\pmb{\sf{k = \dfrac{2}{7}}}}}:⟹
k=
7
2
k=
7
2
Now, substituting the value of k in k : 1 :
\sf : \implies \dfrac{2}{7} : 1:⟹
7
2
:1
We'll multiply the ratio with 7 to get the non-fractional values :
\sf: \implies7 (\dfrac{2}{7}) : 1(7):⟹7(
7
2
):1(7)
\green{:\implies{\pmb{\sf{2 : 7}}}}:⟹
2:7
2:7
\begin{gathered} \\ \end{gathered}
___________________
Therefore, the point (-4,6) divides the line segment joining the points (-6,10) and (3,6) in the ratio of 2 : 7.