Find the ratio
in which the line
segment jolning the points
(1, -3) and (45) is divided by x-axis
Answers
Answer:
?????????...............
Step-by-step explanation:
To find:
Ratio in which y-axis divides the line segment joining points (5, -6) and (-1, -4).
And coordinates of point of intersection
Formula required:
Section formula
\purple{\bigstar}\;\;\boxed{\sf{(x,y)=\bigg(\dfrac{mx_2+nx_1}{m+n}\;,\;\dfrac{my_2+ny_1}{m+n}\bigg)}}★
(x,y)=(
m+n
mx
2
+nx
1
,
m+n
my
2
+ny
1
)
[ where (x,y) are the coordinates of point which divide line segment joining points (x₁, y₁) and (x₂, y₂) in the ratio m : n ]
Solution:
Let, us assume that \sf{\dfrac{m}{n}=k}
n
m
=k
then, m : n = k : 1 will be the ratio in which given line segment is divided
and, Let the coordinates of point of division be ( 0, y)
[ x coordinate would be zero because point of division lies in y-axis ]
Now,
Using Section Formula
\implies\sf{(0,y)=\bigg(\dfrac{(k)(-1)+(1)(5)}{k+1}\;,\;\dfrac{(k)(-4)+(1)(-6)}{k+1}\bigg)}⟹(0,y)=(
k+1
(k)(−1)+(1)(5)
,
k+1
(k)(−4)+(1)(−6)
)
so,
\implies\sf{0=\dfrac{(k)(-1)+(1)(5)}{k+1}\;,\;y=\dfrac{(k)(-4)+(1)(-6)}{k+1}}⟹0=
k+1
(k)(−1)+(1)(5)
,y=
k+1
(k)(−4)+(1)(−6)
Taking
\implies\sf{0=\dfrac{(k)(-1)+(1)(5)}{k+1}}⟹0=
k+1
(k)(−1)+(1)(5)
\implies\sf{0=-k+5}⟹0=−k+5
\implies\boxed{\boxed{\purple{\sf{k=5}}}}⟹
k=5
Now, taking
\implies\sf{y=\dfrac{(k)(-4)+(1)(-6)}{k+1}}⟹y=
k+1
(k)(−4)+(1)(−6)
putting value of k
\implies\sf{y=\dfrac{(5)(-4)+(1)(-6)}{5+1}}⟹y=
5+1
(5)(−4)+(1)(−6)
\implies\sf{y=\dfrac{-20-6}{6}}⟹y=
6
−20−6
\implies\boxed{\boxed{\purple{\sf{y=\dfrac{-13}{3}}}}}⟹
y=
3
−13
Hence,
Ratio in which line segment is divided is, k : 1 = 5 : 1.
and, Coordinates of point of intersection are \bf{\bigg(0,\dfrac{-13}{3}\bigg)}(0,
3
−13
) .