मामूच्या स्वभावातील विविध पैलूंचे विश्लेषण करा.
Answers
Answer:
Given :
Two points (5,6) and (3,-4)
To Find :
A point on the y axis which is equidistant from the point A(5,6) and (3,-4).
To Find :
Firstly we will find The Distance of AP :
\longmapsto\tt{{x}_{1}=5}⟼x
1
=5
\longmapsto\tt{{y}_{1}=6}⟼y
1
=6
\longmapsto\tt{{x}_{2}=0}⟼x
2
=0
\longmapsto\tt{{y}_{2}=y}⟼y
2
=y
Using Formula :
\longmapsto\tt\boxed{Distance\:Formula=\sqrt{({{x}_{2}-{x}_{1)}}^{2}+({{y}_{2}-{y}_{1)}}^{2}}}⟼
DistanceFormula=
(x
2
−x
1)
2
+(y
2
−y
1)
2
Putting Values :
\longmapsto\tt{\sqrt{({0-5)}^{2}+({y-6)}^{2}}}⟼
(0−5)
2
+(y−6)
2
\longmapsto\tt\bf{\sqrt{25+{y}^{2}+36-12y}}⟼
25+y
2
+36−12y
Similarly ,
For PB :
\longmapsto\tt{{x}_{1}=0}⟼x
1
=0
\longmapsto\tt{{y}_{1}=y}⟼y
1
=y
\longmapsto\tt{{x}_{2}=3}⟼x
2
=3
\longmapsto\tt{{y}_{2}=-4}⟼y
2
=−4
Using Formula :
\longmapsto\tt\boxed{Distance\:Formula=\sqrt{({{x}_{2}-{x}_{1)}}^{2}+({{y}_{2}-{y}_{1)}}^{2}}}⟼
DistanceFormula=
(x
2
−x
1)
2
+(y
2
−y
1)
2
Putting Values :
\longmapsto\tt{\sqrt{({3-0)}^{2}+({-4-y)}^{2}}}⟼
(3−0)
2
+(−4−y)
2
\longmapsto\tt\bf{\sqrt{9+16+{y}^{2}-8y}}⟼
9+16+y
2
−8y
Now ,
\longmapsto\tt{AP=PB}⟼AP=PB
\longmapsto\tt{\sqrt{25+{y}^{2}+36-12y}=\sqrt{9+16+{y}^{2}-8y}}⟼
25+y
2
+36−12y
=
9+16+y
2
−8y
\longmapsto\tt{25+{{\cancel{y}^{2}}}+36-12y-9-16-{{\cancel{y}^{2}}}+8y=0}⟼25+
y
2
+36−12y−9−16−
y
2
+8y=0
\longmapsto\tt{25-9-16+36-12y+8y=0}⟼25−9−16+36−12y+8y=0
\longmapsto\tt{16-16+36-4y=0}⟼16−16+36−4y=0
\longmapsto\tt{36=4y}⟼36=4y
\longmapsto\tt{\cancel\dfrac{36}{4}=y}⟼
4
36
=y
\longmapsto\tt\bf{9=y}⟼9=y
So ,The Points are (0,9) .