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a point whose abscissa and ordinate are equal is √10 units from (-1,-3). find the point

Sagot :

Formula for the distance formula
d=[tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} [/tex]
d=[tex]\sqrt{10}[/tex]

[tex]\sqrt{10} = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} [/tex]
Since the abscissa and ordinate are equal we can either say [tex]x_2=y_2[/tex]
Substitute the (-1,-3)
[tex]\sqrt{10} = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} [/tex]
[tex]\sqrt{10} = \sqrt{(x_2+1)^2+(y_2+3)^2} [/tex]
square both sides
10=[tex](x_2+1)^2+(y_2+3)^2[/tex]
Distribute
10=[tex](x_2)^2+2x_2+1+(y_2)^2+6y_2+9[/tex]
Substitute
10=[tex](x_2)^2+2x_2+1+(x_2)^2+6x_2+9[/tex]
Add like terms
10=[tex]2(x_2)^2+8x_2+10[/tex]
Subtract both sides by 10, then divide both sides by 2
0=[tex](x_2)^2+4x_2[/tex]
Add 4 to both sides (completing the square)
4=[tex](x_2)^2+4x_2+4[/tex]
4=[tex](x_2+2)^2[/tex]
Square root both sides
posineg2=[tex]x_2+2[/tex]
Subtract both sides by 2
posineg2-2=[tex]x_2[/tex]
[tex]x_2[/tex]=0 or [tex]x_2[/tex]=-4
(0,0) or (-4,-4)

Hope this helps ^-^