Either (5,0) or (-5, 0).

x=-2

y=10+6(x-1) ⇒ y=4+6x

x+y=7; y=x+1; infinitely many

Yes, no, no

2x+3y-4.37=0;
a=2, b=3 and c=-4.37

\(\left( 0,4\sqrt { 3 } \right) \)

(2,3), (2,0), (0,3)

(a,a) satisfies y=x

3(4)+2(1)=k ⇒ k=14

3(3)-k(2)=5 ⇒ k=2

0x+3y+4=0

Equation of the sides are,
AB: Y=0
BC: X=-1
CD: Y=-4
DA; X=-4
Area=4 X 3
=12 sq. units

(i) Let us draw mutually perpendicular axes XOX' and YOY' and choose a suitable units of distance on the axes.
Let 1 crn = 1 unit. In point P(1, 0), y-coordinate is zero, so it lies on X-axis at a distance of 1 unit from Y-axis.
In point Q (4, O),y-coordinate is zero, so it lies on X-axis at a distance of 4 units from Y-axis. The point S(1, 3) is at a distance of 1 unit from Y-axis and 3 units from X-axis.
On plotting these points, we get the below graph

Now, we need to take a point R on the graph such that PQRS is a square. For this, draw a line passing through Q and parallel to PS and draw a line passing through S and parallel to PQ. Both lines intersect each other at a point, say R. Thus, we get a square PQRS. Clearly, abscissa of R will be equal to abscissa of Q, i.e. 4 and ordinate of R will be equal to ordinate of S, i.e. 3. Hence, the coordinates of point R
are (4, 3).
(ii) We have, S (1, 3) and R (4,3)
Here, we see that y-coordinate of both points S and R are same.
\(\therefore\)  Length of the line segment SR
= Difference of x-coordinate of both points S and R
= 4 - 1 = 3 units

Firstly, draw two mutually perpendicular lines on a graph paper, which intersect at point O. Now, plot all the given points on the graph paper and join them to get a quadrilateral and then find the area of the quadrilateral.
Let us draw the coordinates axes XOX' and YOY', and choose a suitable unit of distance on the axes.
Let 1 cm = 1 unit. Then, the points A (-4,4), B( -6,0), C (-4, - 4) and D (-2, 0) can be plotted as shown below.
Now, join all these points in order to form the quadrilateral.
It is clear from the figure that, AB = BC = CD = DA and sides are not perpendicular to each other.  So, it forms a rhombus.

\(\therefore \) Area of rhombus = \(\frac { 1 }{ 2 } \times d_{ 1 }\times d_{ 2 }\)
\(=\frac { 1 }{ 2 } \times 8\times 4\)
\([\therefore \ d_{ 1 }=AC=4+4=8\ and\ d_{ 2 }=BD=4]\)
= 16 sq units

x+y+1=0, 2x+x=2, 3y+x=9; infinitely many

\(\pi x+y=9\)
\(y=9-\pi x\)
Put x=0, we get y=9-\(\pi\)(0)=9-0=9
Put x=1, we get y=9-\(\pi\)(1)=9-\(\pi\)
Put x=-1, we get y=9-\(\pi\)(-1)=9+\(\pi\)
Put \(x={9\over \pi}\), we get \(y=9-\pi\left(9\over\pi\right)=9-9=0\)
Four solution are (o, 9), (1, 9-\(\pi\)), (-1, 9+\(\pi\)) and \(\left({9\over \pi},0\right)\)

\(x-{y\over 5}-10=0\)
Comparing with ax+by+c=0, we get
a=1, b=\(-{1\over 5 }\), c=-10