Co-ordinate Geometry Class 9 ICSE Important Questions with Answers

Answer: (b) Second.

Here $x=-4<0$x=-4<0 and $y=7>0$y=7>0, i.e. sign pattern $(-,+)$(-,+), which is the second quadrant.

Answer: (b) $0$0.

A point on the $y$y-axis has the form $(0,y)$(0,y), so its abscissa (the $x$x-coordinate) is $0$0.

Answer: (b) $10$10.

Distance $=\sqrt{(6-0)^2+(8-0)^2}=\sqrt{36+64}=\sqrt{100}=10$=√((6-0)²+(8-0)²)=√(36+64)=√100=10 units.

Answer: (c) $5$5 or $-1$-1.

$(k-2)^2+(3+1)^2=25\Rightarrow(k-2)^2=9\Rightarrow k-2=\pm3\Rightarrow k=5$(k-2)²+(3+1)²=25⇒(k-2)²=9⇒ k-2=±3⇒ k=5 or $k=-1$k=-1.

Answer: (a) Both A and R are true and R is the correct explanation of A. We choose values of $x$x freely (independent variable) and calculate $y$y from them, so $y$y depends on $x$x and is the dependent variable.

Each unit is one grid step from the origin $O$O. $A(3,2)$A(3,2) has $(+,+)$(+,+), so it lies in the first quadrant. $B(-2,-1)$B(-2,-1) has $(-,-)$(-,-), so it lies in the third quadrant.

$PQ=\sqrt{(3-(-1))^2+(1-4)^2}=\sqrt{4^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5\text{ units}.$PQ=√((3-(-1))²+(1-4)²)=√(4²+(-3)²)=√(16+9)=√25=5 units.

$AB=\sqrt{(4-1)^2+0^2}=3$AB=√((4-1)²+0²)=3, $BC=\sqrt{0^2+(6-2)^2}=4$BC=√(0²+(6-2)²)=4, $AC=\sqrt{(4-1)^2+(6-2)^2}=\sqrt{9+16}=5$AC=√((4-1)²+(6-2)²)=√(9+16)=5.

Since $AB^2+BC^2=9+16=25=AC^2$AB²+BC²=9+16=25=AC², by the converse of Pythagoras' theorem the triangle is right-angled at $B$B.

Make a table of values for each equation and plot the points.

The two lines meet at $(3, 2)$(3, 2), so the solution is $x=3$x=3, $y=2$y=2. Check: $3+2=5$3+2=5 and $3-2=1$3-2=1 ✓.

Let the point be $P(x, 0)$P(x, 0). Then $PA^2=PB^2$PA²=PB²:
$(x-2)^2+(0+5)^2=(x+2)^2+(0-3)^2$(x-2)²+(0+5)²=(x+2)²+(0-3)²
$x^2-4x+4+25=x^2+4x+4+9\Rightarrow-8x=-16\Rightarrow x=2.$x²-4x+4+25=x²+4x+4+9⇒-8x=-16⇒ x=2.
The point is $(2, 0)$(2, 0). Check: $PA=\sqrt{0+25}=5$PA=√(0+25)=5 and $PB=\sqrt{16+9}=5$PB=√(16+9)=5 ✓.

(i) $OABC$OABC is a rectangle (a $4\times3$4×3 rectangle).

(ii) $OA=4$OA=4 units and $AB=3$AB=3 units (along the grid). $OB=\sqrt{4^2+3^2}=5$OB=√(4²+3²)=5 units and $AC=\sqrt{(0-4)^2+(3-0)^2}=5$AC=√((0-4)²+(3-0)²)=5 units.

(iii) The diagonals are equal ($OB=AC=5$OB=AC=5); together with opposite sides being equal, this confirms that $OABC$OABC is a rectangle.

(a) Tables of values:

The lines meet at $(2, 3)$(2, 3): $x=2$x=2, $y=3$y=3.

(b) $2x+y=7$2x+y=7 cuts the $y$y-axis at $(0, 7)$(0, 7); $x-y=-1$x-y=-1 cuts it at $(0, 1)$(0, 1). The triangle has base $7-1=6$7-1=6 units along the $y$y-axis and height $2$2 units (the $x$x-coordinate of the meeting point).
$\text{Area}=\tfrac12\times6\times2=6\text{ square units}.$Area=1/2×6×2=6 square units.

(i) All coordinates are positive, i.e. sign pattern $(+,+)$(+,+), so all three places lie in the first quadrant.

(ii) $SL=\sqrt{0^2+(7-3)^2}=4$SL=√(0²+(7-3)²)=4 km; $SP=\sqrt{(8-2)^2+0^2}=6$SP=√((8-2)²+0²)=6 km.

(iii) $LP=\sqrt{(8-2)^2+(3-7)^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}\approx7.2$LP=√((8-2)²+(3-7)²)=√(36+16)=√52=2√13≈7.2 km. Since $SL^2+SP^2=16+36=52=LP^2$SL²+SP²=16+36=52=LP², the triangle is right-angled at $S$S, so the two roads meet at right angles at the school.