[5 marks sum]

Question 15 Marks

Show that $A(3, 2), B (6, −2)$ and $C (2, −5)$ can be the vertices of a square.
(i) Find the co-ordinates of its fourth vertex $D,$ if $ABCD$ is a square
(ii) Without using the co-ordinates of vertex $D,$ find the equation of side $AD$ of the square and
also the equation of diagonal $BD.$

Answer

Using distance formula, we have:
$A B=\sqrt{(6-3)^2}+(-2-2)^2=\sqrt{9+16}=5$
$B C=\sqrt{(2-6)^2+\left(-5+2^2\right)}=\sqrt{16+9}=5$
Thus, $AC = BC$
Also, Slope of AB$=\frac{-2-2}{6-3}=\frac{-4}{3}$
Slope of $B C=\frac{-5+2}{2-6}=\frac{-3}{-4}=\frac{3}{4}$
Slope of $AB \times$ Slope of $BC = - 1$
Thus, $AB \perp BC$
Hence, $A, B, C$ can be the vertices of a square…..
(i) Slope of $A B=\frac{-2-2}{6-3}$ Slope of $C D$
Equation of the line $CD$ is
$y - y_1 = m (x -x_1)$
$\Rightarrow y + 5 = (-4)$
$\Rightarrow 3y + 15 = -4x +$
$\Rightarrow 4x + 3y = -7 .........(1)$
Slope of $B C=\left(\frac{-5+2}{2-6}\right)=\frac{-3}{-4}=\frac{3}{4}$ slope of $A D$
Equation of the line $AD$ is
$y - y_1 = m (x - x_1)$
$\Rightarrow y-2=\frac{3}{4}(x-3)$
$\Rightarrow 4 y-8=3 x-9$
$\Rightarrow 3 x-4 y=1......(2)$
Now, D is the point of intersection of $CD$ and $AD.$
$(1) \Rightarrow 16x + 12y = -28$
$(2) \Rightarrow 9x -12y = 3$
Adding the above two equations we get,
$25x = -25$
$\Rightarrow x =-1$
so, $4y = 3x -1 = -3 -1= -4$
$\Rightarrow y = − 1$
Thus, the co-ordinates of point $D$ are $(−1, −1).$
(ii) The equation of line AD is found in part (i)
It is $3x – 4y =1$ Or $4y = 3x – 1$
Slope of $B D=\frac{-1+2}{-1-6}=\frac{1}{-7}=\frac{-1}{7}$
The equation of diagonal $BD$ is
$y - y_1 = m(x - x_1 )$
$\Rightarrow y+1=\frac{-1}{7}(x+1)$
$\Rightarrow 7y + 7 = -x -1$
$\Rightarrow x + 7y + 8 = 0$

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Question 25 Marks

Given equation of line $L_1$ is $y = 4.$
(i) Write the slope of line $L_2$ if $L_2$ is the bisector of angle $O.$
(ii) write the co-ordinates of point $P.$
(iii) Find the equation of $L_2$

Answer

(i) Equation of line $L_1$ is $y = 4$
$\because L_2$ is the bisector of $\angle O$

$\therefore \angle POX = 45^\circ$
Slope $= tan 45^\circ = 1 ...(i)$
Let coordinates of $P$ be $(x, y)$
$\because P$ lies on $L_1$
(ii) $\therefore$ Slope of $L _2=\frac{ y _2- y _1}{x_2-x_1}$
$1=\frac{4-0}{x-0}$
$\Rightarrow 1=\frac{4}{x} .$
from equation (i) & (ii)
$1=\frac{4}{x}$
$\Rightarrow x = 4$
$\therefore$ Coordinates of $P$ are $(4, 4)$
(iii) Equation of $L_2$ is
$y – y_1 = m (x – x_1)$
$\Rightarrow y – 4 = 1 (x - 4)$
$\Rightarrow y – 4 = x – 4$
$\Rightarrow x = y$

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Question 35 Marks

$ABCD$ is a parallelogram where $A(x, y), B (5, 8), C (4, 7)$ and $D (2, -4).$ Find :
(i) co-ordinates of $A.$
(ii) equation of diagonal $BD$

Answer

In parallelogram $ABCD, A (x, y), B(5, 8), C(4, 7)$ and $D(2, -4).$
The diagonals of the parallelogram bisect each other.
$O$ is the point of intersection of $AC$ and $BD$ Since $O$ is the midpoint of $BD,$ its coordinates will be
$O \left(\frac{2+5}{2}, \frac{-4+8}{2}\right)$
i.e. $O \left(\frac{7}{2}, \frac{4}{2}\right)$ i.e. $O \left(\frac{7}{2}, 2\right)$
Now, O $\left(\frac{x+4}{2}, \frac{ y +7}{2}\right)$
(i) Since O is the midpoint of AC also,
$\therefore \frac{x+4}{2}=\frac{7}{2}$
$\Rightarrow x+4=7$
$\Rightarrow x=7-4$
$\therefore x=3$
$\frac{y+7}{2}=2$
$\Rightarrow y+7=4$
$\Rightarrow y=4-7$
$\therefore y=-3$
Thus, Coordinates of $A$ are $(3, 7)$
$\text { (ii) } \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$
$\Rightarrow y-y_1=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)} \times\left(x-x_1\right.$
$\Rightarrow y+4=\frac{8+4}{5-2} \times(x-2)$
$\Rightarrow y+4=\frac{12}{3} \times(x-2)$
$\Rightarrow y+4=4(x-2)$
$\Rightarrow y+4=4 x-8$
$\Rightarrow 4 x-y=12$

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Mathematics [5 marks sum]

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