CUET UG 2025 — General Test Geometry

A point (x, y) is equidistant from (-1, 1) and (4, 3) when the distances are equal.

Using the distance formula:

$$\sqrt{(x - (-1))^2 + (y - 1)^2} = \sqrt{(x - 4)^2 + (y - 3)^2}$$(x−(−1))2+(y−1)2​=(x−4)2+(y−3)2​

$$\sqrt{(x + 1)^2 + (y - 1)^2} = \sqrt{(x - 4)^2 + (y - 3)^2}$$(x+1)2+(y−1)2​=(x−4)2+(y−3)2​

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Squaring both sides:

$$(x + 1)^2 + (y - 1)^2 = (x - 4)^2 + (y - 3)^2$$(x+1)2+(y−1)2=(x−4)2+(y−3)2

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Expanding the left side:

$$(x + 1)^2 = x^2 + 2x + 1$$(x+1)2=x2+2x+1

$$(y - 1)^2 = y^2 - 2y + 1$$(y−1)2=y2−2y+1

$$x^2 + 2x + 1 + y^2 - 2y + 1 = x^2 + 2x + y^2 - 2y + 2$$x2+2x+1+y2−2y+1=x2+2x+y2−2y+2

Expanding the right side:

$$(x - 4)^2 = x^2 - 8x + 16$$(x−4)2=x2−8x+16

$$(y - 3)^2 = y^2 - 6y + 9$$(y−3)2=y2−6y+9

$$x^2 - 8x + 16 + y^2 - 6y + 9 = x^2 - 8x + y^2 - 6y + 25$$x2−8x+16+y2−6y+9=x2−8x+y2−6y+25

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The equation becomes:

$$x^2 + 2x + y^2 - 2y + 2 = x^2 - 8x + y^2 - 6y + 25$$x2+2x+y2−2y+2=x2−8x+y2−6y+25

Canceling $x^2$x2 and $y^2$y2 from both sides:

$$2x - 2y + 2 = -8x - 6y + 25$$2x−2y+2=−8x−6y+25

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Collecting like terms:

$$2x + 8x - 2y + 6y = 25 - 2$$2x+8x−2y+6y=25−2

$$10x + 4y = 23$$10x+4y=23

Therefore, the equation is $10x + 4y = 23$10x+4y=23.