Co-ordinate Geometry — ICSE Class 9 Maths Important Questions

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) $2$2.

$m=\dfrac{11-3}{6-2}=\dfrac{8}{4}=2$m=11-3/6-2=8/4=2.

Answer: (a) $1$1.

Slope of $(3,4)$(3,4)–$(5,6)$(5,6) is $\dfrac{6-4}{5-3}=1$6-4/5-3=1. For collinearity, slope of $(k,2)$(k,2)–$(3,4)$(3,4) must also be $1$1: $\dfrac{4-2}{3-k}=1\Rightarrow 3-k=2\Rightarrow k=1$4-2/3-k=1 3-k=2 k=1.

Answer: (a) $m=\dfrac{5-5}{7-2}=\dfrac{0}{5}=0$m=5-5/7-2=0/5=0, and both points have $y=5$y=5 so the line is parallel to the $x$x-axis. R correctly explains A.

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.

$m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-4-4}{3-(-1)}=\dfrac{-8}{4}=-2.$m=y_2-y_1/x_2-x_1=-4-4/3-(-1)=-8/4=-2.

Since the gradient is negative, the line falls from left to right.

Gradient of $AB$AB: $m_{AB}=\dfrac{2-2}{4-1}=\dfrac{0}{3}=0.$m_AB=2-2/4-1=0/3=0.

A zero gradient means $AB$AB is parallel to the $x$x-axis, i.e. horizontal.

Gradient of $BC$BC: $m_{BC}=\dfrac{6-2}{4-4}=\dfrac{4}{0}$m_BC=6-2/4-4=4/0, which is undefined.

An undefined gradient means $BC$BC is parallel to the $y$y-axis, i.e. vertical.

Hence $AB\perp BC$AB BC and the triangle is right-angled at $B$B.

Gradient of $AB$AB: $m_{AB}=\dfrac{5-1}{3-1}=\dfrac{4}{2}=2.$m_AB=5-1/3-1=4/2=2.

Gradient of $BC$BC: $m_{BC}=\dfrac{6-5}{5-3}=\dfrac{1}{2}.$m_BC=6-5/5-3=1/2.

Since $m_{AB}=2$m_AB=2 and $m_{BC}=\dfrac12$m_BC=12 are not equal, the line $AB$AB and the line $BC$BC have different directions even though they share the point $B$B.

Therefore the three points do not lie on one straight line, i.e. $A$A, $B$B, $C$C are not collinear.

Using $m=\dfrac{y_2-y_1}{x_2-x_1}$m=y_2-y_1/x_2-x_1 with $m=3$m=3:
$3=\dfrac{4-(-2)}{k-3}=\dfrac{6}{k-3}.$3=4-(-2)/k-3=6/k-3.

$\Rightarrow 3(k-3)=6$3(k-3)=6

$\Rightarrow k-3=2$k-3=2

$\Rightarrow k=5.$k=5.

Therefore $B=(5,4)$B=(5,4).

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

(ii)
$m_{OA}=\dfrac{0-0}{4-0}=0$m_OA=0-0/4-0=0.
$m_{AB}=\dfrac{3-0}{4-4}=$m_AB=3-0/4-4= undefined.
$m_{BC}=\dfrac{3-3}{0-4}=0$m_BC=3-3/0-4=0.
$m_{CO}=\dfrac{0-3}{0-0}=$m_CO=0-3/0-0= undefined.

(iii) $OA$OA and $BC$BC have gradient $0$0, so they are horizontal (parallel to the $x$x-axis). $AB$AB and $CO$CO have undefined gradient, so they are vertical (parallel to the $y$y-axis). Adjacent sides are perpendicular, confirming a rectangle.

(i)
$m_{AB}=\dfrac{3-1}{2-(-2)}=\dfrac{2}{4}=\dfrac12.$m_AB=3-1/2-(-2)=2/4=12.
$m_{BC}=\dfrac{5-3}{6-2}=\dfrac{2}{4}=\dfrac12.$m_BC=5-3/6-2=2/4=12.

(ii) Since $m_{AB}=m_{BC}=\dfrac12$m_AB=m_BC=12 and $B$B is common to both segments, the three points $A$A, $B$B, $C$C are collinear (they lie on one straight line of gradient $\tfrac12$12).

(iii) For $D(p,7)$D(p,7) to be on the same line, the gradient of $CD$CD must also be $\dfrac12$12:
$\dfrac{7-5}{p-6}=\dfrac12\Rightarrow \dfrac{2}{p-6}=\dfrac12\Rightarrow p-6=4\Rightarrow p=10.$7-5/p-6=12 2/p-6=12 p-6=4 p=10.
So $D=(10,7)$D=(10,7).

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

(ii) $m_{SL}=\dfrac{7-3}{2-2}=\dfrac{4}{0}$m_SL=7-3/2-2=4/0, which is undefined. Road $SL$SL is vertical (parallel to the $y$y-axis).

(iii) $m_{SP}=\dfrac{3-3}{8-2}=\dfrac{0}{6}=0$m_SP=3-3/8-2=0/6=0. Since the gradient is $0$0, road $SP$SP is horizontal (parallel to the $x$x-axis). Thus $SL\perp SP$SL SP, meeting at right angles at the school.