Math, asked by SAIRAofficial, 6 months ago

no spams......solve it plz​

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Answers

Answered by Anonymous
1

Answer:

Step-by-step explanation:

GIVEN=

DISTANCE BTW THE POINT=√10

GIVEN POINT=(4,3)

AND POINT A HAS ITS ORDINATE TWICE OF ITS ABSISCA

TO FIND=

CORDINATES OF POINT A

SOLUTION=

LET THE ABSSISCA OF POINT A BE X

A=(X,2X)

X CORDINTES OF POITS =

X1=4

X2=X

Y CORDINATES=

Y1=3

Y2=2X

DISTANCE=√(X2-X1)²+(Y2-Y1)²

√10=√(X-4)²+(2X-3)²

SQUARING BOTH SIDE

10=5X²-20X+25

5X²-20X+15=0

5(X²-4X+3)=0

BY SPLITING MIDDLE TERM WE GET

5(X-3)(X-1)=0

X-3=0      &          X-1=0

X=3,1

Y=6,2

THEREFORE THERE ARE TWO SUCH POINTS POSSIBLE

(3,6) &(1,2)

HOPE IT HELPS U BUDDY✌✌

Answered by Anonymous
2

Step-by-step explanation:

Answer:

Step-by-step explanation:

GIVEN=

DISTANCE BTW THE POINT=√10

GIVEN POINT=(4,3)

AND POINT A HAS ITS ORDINATE TWICE OF ITS ABSISCA

TO FIND=

CORDINATES OF POINT A

SOLUTION=

LET THE ABSSISCA OF POINT A BE X

A=(X,2X)

X CORDINTES OF POITS =

X1=4

X2=X

Y CORDINATES=

Y1=3

Y2=2X

DISTANCE=√(X2-X1)²+(Y2-Y1)²

√10=√(X-4)²+(2X-3)²

SQUARING BOTH SIDE

10=5X²-20X+25

5X²-20X+15=0

5(X²-4X+3)=0

BY SPLITING MIDDLE TERM WE GET

5(X-3)(X-1)=0

X-3=0      &          X-1=0

X=3,1

Y=6,2

THEREFORE THERE ARE TWO SUCH POINTS POSSIBLE

(3,6) &(1,2)

HOPE IT HELPS U BUDDY✌✌

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