Question

Difficulty: MediumCoordinate Geometry of Straight Lines

Find the value of kk if the point P(k,3)P(k, 3) is equidistant from the points A(1,5)A(1, 5) and B(7,1)B(7, 1).

Answer: 4

Answer

The value of kk is 4.
Using the distance formula, the squared distance PA2=(k−1)2+(3−5)2=(k−1)2+4PA^2 = (k-1)^2 + (3-5)^2 = (k-1)^2 + 4, and PB2=(k−7)2+(3−1)2=(k−7)2+4PB^2 = (k-7)^2 + (3-1)^2 = (k-7)^2 + 4. Equating PA2=PB2PA^2 = PB^2 gives (k−1)2=(k−7)2(k-1)^2 = (k-7)^2. Expanding both sides yields k2−2k+1=k2−14k+49k^2 - 2k + 1 = k^2 - 14k + 49. Subtracting k2k^2 from both sides gives 12k=4812k = 48, which leads to k=4k = 4.

Step-by-Step Solution

1
Write the expressions for the squared distances PA2PA^2 and PB2PB^2 using the distance formula.
PA2=(k−1)2+4PA^2 = (k - 1)^2 + 4 and PB2=(k−7)2+4PB^2 = (k - 7)^2 + 4
The distance formula between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is d2=(x2−x1)2+(y2−y1)2d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2.
2
Equate PA2PA^2 and PB2PB^2 since point PP is equidistant from points AA and BB.
(k−1)2+4=(k−7)2+4  ⟹  (k−1)2=(k−7)2(k - 1)^2 + 4 = (k - 7)^2 + 4 \implies (k - 1)^2 = (k - 7)^2
Subtracting 4 from both sides simplifies the equality of squared distances.
3
Expand both sides and isolate kk to find its numerical value.
k2−2k+1=k2−14k+49  ⟹  12k=48  ⟹  k=4k^2 - 2k + 1 = k^2 - 14k + 49 \implies 12k = 48 \implies k = 4
Canceling k2k^2 terms yields a simple linear equation.

Key Concept

Equidistant points and the distance formula in coordinate geometry
Estimated Time:1m 30s
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