The coordinates of the point on the straight line y=3x+1 that is equidistant from the origin and (−3, 4) is (-33/18, 39/6)
The distance formula is expressed as:
[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\[/tex]
For the coordinate point (x,y) to (-3, 4), the distance is expressed as:
[tex]D_1 = \sqrt{(x+3)^2+(y-4)^2}\\[/tex]
For the coordinate point (x,y) to (0, 0), the distance is expressed as:
[tex]D_2=\sqrt{(x-0)^2+(y-0)^2}\\D_2 = \sqrt{x^2+y^2}[/tex]
Since the distance are equidistant, then D1 = D2
[tex]\sqrt{(x+3)^2+(y-4)^2} = \sqrt{x^2+y^2}\\ (x+3)^2+(y-4)^2 = x^2+y^2\\x^2+6x+9+y^2-8y+16 = x^2+y^2\\6x+9-8y+16=0\\6x-8y+25=0[/tex]
Substitute y = 3x+1 and get x
6x - 8y + 25 = 0
6x-8(3x+1) - 25 = 0
6x -24x-8-25=0
-18x-33 = 0
-18x = 33
x =-33/18
Sincey= 3x + 1
y = 3(33/18) + 1
y = 33/6 + 1
y = 39/6
Hence the required coordinates is (-33/18, 39/6)
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