Question

# Find a point on the x−axis, which is equidistant from the points (7,6) and (3,4).

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Solution

## Step 1: Simplification of given data Let the given points be A(7,6) and B(3,4) Let C be a point on the x−axis Coordinates of C=(x,0) We know that distance between two points (x1,y1) and (x2,y2) is D=√(x2−x1)2+(y2−y1)2 Distance between A(7,6) & C(x,0) AC=√(x−7)2+(0−6)2 =√(x−7)2+36 Distance between B(3,4) & C(x,0) BC=√(x−3)2+(0−4)2 =√(x−3)2+16 Step 2: Solve for required point C Given that point C is equidistant from the points A & B Hence, Distance AC=Distance BC ⇒√(x−7)2+36=√(x−3)2+16 Squaring both sides ⇒(√(x−7)2+36)2=(√(x−3)2+16)2 ⇒(x−7)2+36=(x−3)2+16 ⇒(x−7)2−(x−3)2=16−36 ⇒x2+49−14x−(x2+9−6x)=−20 ⇒x2+49−14x−x2−9+6x=−20 ⇒−8x+40=−20 ⇒8x=40+20 ⇒x=608=152 Therefore, Required point = C(x,0)=(152,0)

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