2x+3y+k=0
if x=2, and y=1 is the solution of the linear equation 2x+3y+k=0 then
2(2)+3(1)+k=0
k=-7

Rectangle

Put x = 4 and y =3,
then 3y-2x=3(3)-2(4)=1
So, (4,3) is the solution of the equation
Again put \(x=2\sqrt{2}\)
and \(y=3\sqrt{2}\) then 
\(3y-2x=3(3\sqrt{2})-2(2\sqrt{2})\)
\(=5\sqrt{2}\ne1\)
So, \((2\sqrt{2},3\sqrt{2})\) is not a solution of the given equation.

Here, point 0(0, 0) is the origin. A (3, 0) lies on positive direction of X-axis, B(3, 4) lies in J quadrant and C(O, 4) lies on positive direction of Y-axis. On joining OA, AB, BC and CO, the figure obtained is a rectangle

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

y=8+5(x-1) ⇒ y=5x+3; Rs.318

y=3x

(0,3), (4, 6), (-4, 0)

(i)

(ii) Rectangle
(iii) 3 : 5

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

We have 3x - 4y + 9 = 0
Putting x = 1, y = 3, we get
L.H.S. = 3(1) - 4(3) + 9 = 3 - 12 + 9
= 12 - 12 = 0 = R.H.S.
Since, L.H.S. = R.H.S.
∴ x = and y = 3 satisfy the given linear equation.

(i) For rectangle ABCD:
AB = 8 - 2 = 6 units ← Length
AD = 12 - 8 = 4 units ← Breadth
∴ Area of rectangle ABCD = AB x AD
= 6 units x 4 units
= 24 sq. units.
For triangle PQR:
QR = 12 - 4 = 8 units ← Base
SP = 6 - 0 = 6 units ← Height
∴ Area of ΔPQR = \(\frac{1}{2}\) base x height
= \(\frac{1}{2}\) x 8 x 6 sq. units
= 24 sq. units.
⇒ \(\left[\begin{array}{l} \text { area of } \\ \text { rect. } \mathrm{ABCD} \end{array}\right]=\left[\begin{array}{l} \text { area of } \\ \Delta \mathrm{PQR} \end{array}\right]\)
(ii) Co-ordinate Geometry.
(iii) Community or social upliftment.

From the figure, we have
(i) The co-ordinates of B are (-5, 2).
(ii) The co-ordinates of C are (5, -5).
(iii) The point E is identified by the co-ordinates (-3, -5)
(iv) The point G is identified by the co-ordinates (2, -4).
(v) The abscissa of the point D is 6.
(vi) The ordinate of the point H is -3.
(vii) The co-ordinates of the point L are (0, 5).
(viii) The co-ordinates of the point M are (-3, 0).
Plotting a point in the plane when coordinates of the point are given
To plot a point in the coordinate plane we draw the coordinate axes and choose our units such that we can mark equal distances on the x-axis or on the y-axis or on both the axes. We take origin as zero for x-axis as well as on y-axis. The distances marked along OX and OY are taken as positive and those along OX and OY are taken as negative.

Given equation is 2x + y = 7 ... (i)
On putting x = 0 in Eq. (i), we get y=7
So, (0, 7) is a solution of given equation. Similarly, putting y = 0 in Eq. (i), we get
2x = 7 \(\Rightarrow\) =\(-\frac{7}{2}\)
So, \((\frac{7}{2},0)\)is a solution of given equation.
On putting x = 1 in Eq. (i), we get
2(1) + y = 7
\(\Rightarrow\) y=5
Therefore, (1, 5) is also a solution of the given equation.
Again, putting x = 2 in Eq. (i), we get
2(2)+y=7
\(\Rightarrow\) y=7-4=3
So, (2,3) is a solution of given equation.
Hence, four out of the infinitely many solutions of the
given equation, are (0, 7), \((\frac{7}{2},0)\) , (1, 5) and (2, 3).