A point p divides the line segment joining A(3,-5) ,B(-4,8) withAP/PB=K/1 lies the line x+y=0
Answers
Answer:
Given a point p divides the line segment joining the points a(3,-5) and b (-4,8) such that \frac{ap}{pb}=\frac{k}{1}
pb
ap
=
1
k
if p lies on the line x+y=0, we have to find the value of k
By section formula,
when a point p(x,y) divides the line segment joining the points a(3,-5) and b(-4,8) in ratio m:n=k:1 then coordinates of p are
p(x,y)=(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})=(\frac{-4k+3}{k+1},\frac{8k-5}{k+1})p(x,y)=(
m+n
mx
2
+nx
1
,
m+n
my
2
+ny
1
)=(
k+1
−4k+3
,
k+1
8k−5
)
As the above point passing through the line x+y=0
hence, satisfy the above equation.
⇒ \frac{-4k+3}{k+1}+\frac{8k-5}{k+1}=0
k+1
−4k+3
+
k+1
8k−5
=0
⇒ -4k+3+8k-5=0−4k+3+8k−5=0
⇒ 4k-2=04k−2=0
⇒ k=\frac{2}{4}=\frac{1}{2}k=
4
2
=
2
1
\text{The value of k is }\frac{1}{2}The value of k is
2
1
Answer:
K=1/2
Step-by-step explanation:
