if the point (0,c), ( -c/m,0) and (x,y) are on the same line prove that y=mx+c
Answers
Step-by-step explanation:
The lines makes an angle α with the line y+x=0
Slope of given line that is m
1
=−1
Let the slope of the other line be m
2
The angle between straight lines thatt is tanθ=
∣
∣
∣
∣
∣
1+m
1
m
2
m
1
−m
2
∣
∣
∣
∣
∣
tanα=
∣
∣
∣
∣
∣
1−1.m
2
−1−m
2
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
m
2
−1
m
2
+1
∣
∣
∣
∣
∣
m
2
−1
m
2
+1
=tanα,
m
2
−1
m
2
+1
=−tanα
⇒m
2
=
tanα−1
tanα+1
,
tanα+1
tanα−1
Equation of straght line with given slope and a point is y=mx+c
Lines passes through origin 0=0.m+c
⇒c=0
So the equation of lines are y=
tanα−1
tanα+1
x and y=
tanα+1
tanα−1
x
y=
sinα−cosα
sinα+cosα
x,y=
sinα+cosα
sinα−cosα
x
Combined equation of straight lines
(y−
sinα−cosα
sinα+cosα
x)(y−
sinα+cosα
sinα−cosα
x)=0
y
2
−xy(
sinα+cosα
sinα−cosα
)−xy(
sinα−cosα
sinα+cosα
)+(
sinα−cosα
sinα+cosα
)(
sinα+cosα
sinα−cosα
)x
2
=0
y
2
−xy(
(sinα+cosα)(sinα−cosα)
(sinα−cosα)
2
+(sinα+cosα)
2
)+x
2
=0
y
2
−xy(
sin
2
α−cos
2
α
2
)+x
2
=0
y
2
−xy(
−cos2α
2
)+x
2
=0
y
2
+2xysec2α+x
2
=0
Hence proved