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(i) $ x=-5 $

(ii) $ y=0 $

(a) (-5, - 4) and (3, 0)

(b) (-4, -5) and (3, 0)

(c) (-4, 3) and (5, 0)

(d) (-5, 0) and (-4, 0)

Answer
Verified

Here, we have to find the graph of the given line equation $ x-2y=3......................(i) $ . Now, we will replace the value of x and y with ‘zero’, one by one, and find the required two coordinates, i. e., point A and point B (let us say).

Replacing the value of x with ‘zero’ in equation (i), we get,

$ \Rightarrow 0-2y=3 $

$ \Rightarrow -2y=3 $

$ \Rightarrow 2y=-3 $

$ \therefore y=\dfrac{-3}{2} $

Hence when $ x=0 $ in the above equation (i), the value of y is $ -\dfrac{3}{2} $ . So, the point is $ \text{A}\left( 0,\dfrac{-3}{2} \right) $ .

Similarly, now, we place the value of y with ‘zero’ in the above equation (i), we get,

$ \Rightarrow x-2\times 0=3 $

$ \Rightarrow x-0=3 $

$ \therefore x=3 $

Hence when $ y=0 $ in the above equation (i), the value of x is 3. So, the point is $ \text{B}\left( 3,0 \right) $ .

Now, we need to plot these points $ \text{A}\left( 0,\dfrac{-3}{2} \right) $ and $ \text{B}\left( 3,0 \right) $ in the x – y plane and join these points, in order to obtain the required graph for the equation $ x-2y=3 $ .

Now, to find the coordinates of the point under a given condition, we need to simply replace that value in the given equation (i), so

(i) Replace the value of x with -5 in the given equation (i), we get,

$ \Rightarrow -5-2y=3 $

$ \Rightarrow -2y=3+5 $

$ \Rightarrow -2y=8 $

$ \Rightarrow 2y=-8 $

$ \Rightarrow y=\dfrac{-8}{2} $

$ \therefore y=-4 $

Hence when $ x=-5 $ in the above equation (i), the value of y is -4. So, the required coordinates of point are $ \left( -5,-4 \right) $ .

(ii) Replace the value of y with 0 in the given equation (i), we get,

$ \Rightarrow x-2\times 0=3 $

$ \Rightarrow x-0=3 $

$ \therefore x=3 $

Hence when $ y=0 $ in the above equation (i), the value of x is 3. So, the required coordinates of point are $ \left( 3,0 \right) $ .

(i) when $ x=-5 $ , we simply need to solve this equation and equation (i),

$ x-2y=3............(ii) $

$ x=-5........(iii) $

Now, subtracting equation (iii) from (ii), we get,

$ \begin{align}

& x-2y=3 \\

& -x-0y=5 \\

& \underline{\text{ }} \\

& -2y=8 \\

\end{align} $

$ \Rightarrow 2y=-8 $

$ \Rightarrow y=\dfrac{-8}{2} $

$ \Rightarrow y=-4 $

Hence when $ x=-5 $ in the above equation (i), the value of y is -4. So, the required coordinates of point are $ \left( -5,-4 \right) $ .

(i) when $ y=0 $ , we simply need to solve this equation and equation (i),

$ x-2y=3............(ii) $

$ y=0........(iii) $

Now, multiplying equation (iii) by 2, we get,

$ 2y=0..............(iv) $

Adding equation (ii) from (iv), we get,

$ \begin{align}

& x-2y=3 \\

& 0x+2y=0 \\

& \underline{\text{ }} \\

& x=3 \\

\end{align} $

Hence when $ y=0 $ in the above equation (i), the value of x is 3. So, the required coordinates of point are $ \left( 3,0 \right) $ .