CBSE BoardEnglish MediumSTD 11 ScienceMathsModel Paper 102 Marks
Question
Find a point on the $y-$axis which is equidistant from $A (-4, 3)$ and $B(5, 2).$
✓
Answer
Let the point on the $y-$axis be $P(0, y)$
Here, it is given that $P$ is equidistant from $A(-4, 3)$ and $B(5, 2).$
$i.e., PA = PB$
$\Rightarrow \sqrt{(-4-0)^2+(3-y)^2}=\sqrt{(5-0)^2+(2-y)^2}$
Squaring both sides, we obtain
$\Rightarrow(-4-0)^2+(3-y)^2=(5-0)^2+(2-y)^2$
$\Rightarrow 16+9-6 y+y^2=25+4-4 y+y^2$
$\Rightarrow 25-6 y=29-4 y$
$\Rightarrow 2 y =-4$
$\Rightarrow y =-2$
Thus, the required point on the y-axis is $(0, -2).$
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