Class 10: Coordinate Geometry

$\displaystyle \textbf{Question 1: }\text{On which axis do the following points lie?}$
$\displaystyle \text{(i) }P(5,0)\qquad\text{(ii) }Q(0,-2)\qquad\text{(iii) }R(-4,0)\qquad\text{(iv) }S(0,5)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }P(5,0)\text{ lies on the }x\text{-axis, since its }y\text{-coordinate is }0.$
$\displaystyle \text{(ii) }Q(0,-2)\text{ lies on the }y\text{-axis, since its }x\text{-coordinate is }0.$
$\displaystyle \text{(iii) }R(-4,0)\text{ lies on the }x\text{-axis, since its }y\text{-coordinate is }0.$
$\displaystyle \text{(iv) }S(0,5)\text{ lies on the }y\text{-axis, since its }x\text{-coordinate is }0.$
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$\displaystyle \textbf{Question 2: }\text{Let }ABCD\text{ be a square of side }2a.\text{ Find the coordinates of the}$
$\displaystyle \text{vertices of this square when:}$
$\displaystyle \text{(i) }A\text{ coincides with the origin and }AB\text{ and }AD\text{ are along }OX\text{ and }OY$
$\displaystyle \text{respectively.}$

$\displaystyle \text{(ii) The centre of the square is at the origin and the coordinate axes are parallel to}$
$\displaystyle \text{the sides }AB\text{ and }AD\text{ respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Since }A\text{ is at the origin, }A=(0,0).$ $\displaystyle \text{Since }AB=2a\text{ and }AB\text{ lies along }OX,\quad B=(2a,0).$
$\displaystyle \text{Since }AD=2a\text{ and }AD\text{ lies along }OY,\quad D=(0,2a).$
$\displaystyle \therefore C=(2a,2a).$
$\displaystyle \therefore A=(0,0),\quad B=(2a,0),\quad C=(2a,2a),\quad D=(0,2a).$

$\displaystyle \text{(ii) Since the centre of the square is the origin and its side is }2a,$
$\displaystyle \text{each side is at a distance }a\text{ from the corresponding coordinate axis.}$
$\displaystyle \therefore A=(-a,-a),\quad B=(a,-a),\quad C=(a,a),\quad D=(-a,a).$
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$\displaystyle \textbf{Question 3: }\text{The base }PQ\text{ of two equilateral triangles }PQR\text{ and }PQR'\text{ with side }$
$\displaystyle 2a\text{ lies along the }y\text{-axis such that the mid-point of }PQ\text{ is at the origin. Find the}$
$\displaystyle \text{coordinates of the vertices }R\text{ and }R'\text{ of the triangles.}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Since the mid-point of }PQ\text{ is the origin and }PQ=2a,$
$\displaystyle P=(0,a)\quad\text{and}\quad Q=(0,-a).$
$\displaystyle \text{The vertices }R\text{ and }R'\text{ lie on the perpendicular bisector of }PQ,$
$\displaystyle \text{which is the }x\text{-axis.}$
$\displaystyle \text{Let }OR=x.$
$\displaystyle \text{In right-angled }\triangle POR,\quad PR=2a\text{ and }OP=a.$
$\displaystyle PR^2=OP^2+OR^2$
$\displaystyle (2a)^2=a^2+x^2$
$\displaystyle x^2=4a^2-a^2=3a^2$
$\displaystyle x=\sqrt{3}a$
$\displaystyle \therefore R=(\sqrt{3}a,0)\quad\text{and}\quad R'=(-\sqrt{3}a,0).$
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