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Coordinate Geometry
CH07 Question Bank
CBSE_CH07_QUESTION_BANK
Grade 10

Question:

If the points $A(k,3)$, $B(6,-2)$ and $C(-3,4)$ are such that $A$ is equidistant from $B$ and $C$, find the value of $k$. Also find the length $AB$ for this value of $k$.

Step-by-Step Solution

Key Concept: Set AB = AC using the distance formula and solve the resulting linear equation for k, then substitute back to find AB.
$AB^2=(k-6)^2+(3-(-2))^2=(k-6)^2+25$. [1.0 Mark]

$AC^2=(k-(-3))^2+(3-4)^2=(k+3)^2+1$. [1.0 Mark]

Setting $AB^2=AC^2$: $(k-6)^2+25=(k+3)^2+1\Rightarrow k^2-12k+36+25=k^2+6k+9+1$. [1.5 Marks]

$-12k+61=6k+10\Rightarrow-18k=-51\Rightarrow k=\dfrac{51}{18}=\dfrac{17}{6}$. [1.0 Mark]

$AB^2=\left(\dfrac{17}{6}-6\right)^2+25=\left(-\dfrac{19}{6}\right)^2+25=\dfrac{361}{36}+\dfrac{900}{36}=\dfrac{1261}{36}$, so $AB=\sqrt{\dfrac{1261}{36}}=\dfrac{\sqrt{1261}}{6}\approx5.92$ units. [0.5 Mark]

Correct Answer:
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