Question

Does the point $P(\frac{3}{2},5)$ lies on the line $x+y=\frac{13}{2}$? If so, then draw the graph and check it.

Answer 1

Given,
$x+y=\frac{13}{2}$
$=>y=\frac{13}{2}-x$
Substituting different values of $x$ to get corresponding values of $y$, we get
$(0,\frac{13}{2}),(1,\frac{11}{2}),(-1,\frac{15}{2}),(2,\frac{9}{2}),(-2,\frac{17}{2})$
Plotting these points on the graph, we get the graph of the equation, $x+y=\frac{13}{2}$
From the graph drawn we can see that, $p(\frac{3}{2},5)$ lies on the graph of $x+y=\frac{13}{2}$