Orienting Yourself: The Use of Coordinates — CBSE Class 9 Maths Important Questions

Answer: (b) Quadrant II.

Here the abscissa is negative $(-4)$(-4) and the ordinate is positive $(+7)$(+7). The sign pattern $(-,+)$(-,+) corresponds to the second quadrant.

Answer: (b) $0$0.

A point on the $y$y-axis has no horizontal displacement from the origin, so its $x$x-coordinate (abscissa) is $0$0. Such a point is of the form $(0, y)$(0, y).

Answer: (c) $a>0,\ b<0$a>0, b<0.

In the fourth quadrant the abscissa is positive and the ordinate is negative, giving the sign pattern $(+,-)$(+,-).

Answer: (a) $(4,2)$(4,2). Midpoint $=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)=\left(\dfrac{2+6}{2},\dfrac{-4+8}{2}\right)=(4,2)$=((x₁+x₂)/2,(y₁+y₂)/2)=((2+6)/2,(-4+8)/2)=(4,2).

Answer: (a) Both A and R are true and R is the correct explanation of A. The point $(0,-6)$(0,-6) has abscissa $0$0, which is exactly the condition for a point to lie on the $y$y-axis.

A point $5$5 units to the right of the origin on the $x$x-axis has ordinate $0$0, so it is $(5, 0)$(5, 0).

A point $3$3 units below the origin on the $y$y-axis has abscissa $0$0 and a negative ordinate, so it is $(0, -3)$(0, -3).

$A(2,3)$A(2,3): both coordinates positive $(+,+)$(+,+) — Quadrant I.

$B(-1,-4)$B(-1,-4): both negative $(-,-)$(-,-) — Quadrant III.

$C(6,0)$C(6,0): ordinate $0$0 — on the $x$x-axis.

$D(-3,8)$D(-3,8): signs $(-,+)$(-,+) — Quadrant II.

The four points are one in each quadrant, symmetric about both axes.

Joining $A\to B\to C\to D\to A$A→ B→ C→ D→ A gives a rectangle (width $4$4 units along $x$x, height $6$6 units along $y$y). Since the sides are unequal it is a rectangle, not a square.

Distance formula: $AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$AB=√((x₂-x₁)²+(y₂-y₁)²).

$AB=\sqrt{(4-1)^2+(6-2)^2}=\sqrt{9+16}=5,\quad BC=\sqrt{(7-4)^2+(10-6)^2}=5,\quad AC=\sqrt{(7-1)^2+(10-2)^2}=\sqrt{100}=10.$AB=√((4-1)²+(6-2)²)=√(9+16)=5, BC=√((7-4)²+(10-6)²)=5, AC=√((7-1)²+(10-2)²)=√100=10.
Since $AB+BC=5+5=10=AC$AB+BC=5+5=10=AC, the three points lie on a straight line ($B$B is in fact the midpoint of $AC$AC).

All three points have the same abscissa $x=3$x=3, so they lie on a vertical line parallel to the $y$y-axis, $3$3 units to the right of it.

Since the points are collinear on a vertical line, the distance between two of them equals the difference of their ordinates.

$QR = |5 - (-1)| = |5 + 1| = 6$QR = |5 - (-1)| = |5 + 1| = 6 units.

Plotting: $A$A at the origin, $B$B on the positive $x$x-axis, $D$D on the positive $y$y-axis, and $C$C in Quadrant I.

$AB$AB is horizontal of length $4$4 cm; $BC$BC is vertical of length $3$3 cm; $CD=4$CD=4 cm and $DA=3$DA=3 cm. Opposite sides are equal and all angles are right angles, so $ABCD$ABCD is a rectangle.

Perimeter $= 2(l+b) = 2(4+3) = 14$= 2(l+b) = 2(4+3) = 14 cm.

Area $= l\times b = 4\times 3 = 12\ \text{cm}^2$= l× b = 4× 3 = 12 cm².

Plot the given points. $PQ$PQ is horizontal ($y=-2$y=-2) with length $|3-(-2)| = 5$|3-(-2)| = 5 units. $QR$QR is vertical ($x=3$x=3) with length $|3-(-2)| = 5$|3-(-2)| = 5 units. So the side of the square is $5$5 units.

For $PQRS$PQRS taken in order, $S$S must be directly above $P$P and directly left of $R$R. Thus $S$S has the abscissa of $P$P and the ordinate of $R$R:

$S = (-2,\ 3).$S = (-2, 3).

Check: $RS = |3-(-2)| = 5$RS = |3-(-2)| = 5 and $SP = |3-(-2)| = 5$SP = |3-(-2)| = 5 units, confirming all four sides equal.

Side $= 5$= 5 units, so Area $= 5^2 = 25$= 5² = 25 square units.

(i) The Park $P(-5, 4)$P(-5, 4) has signs $(-,+)$(-,+), so it lies in Quadrant II.

(ii) The School $S(2, 4)$S(2, 4) and the Library $L(2, -3)$L(2, -3) have the same abscissa $x=2$x=2, so they lie on a vertical line parallel to the $y$y-axis.

(iii) By the distance formula, $HS=\sqrt{(2-0)^2+(4-0)^2}=\sqrt{20}=2\sqrt5\approx4.47$HS=√((2-0)²+(4-0)²)=√20=2√5≈4.47 km. (iv) The Hospital $H(0, 0)$H(0, 0) is located at the origin (the city centre).