Question
Determine the point on the graph of the linear equation $2x + 5y = 19$ whose ordinate is $1\frac{1}{2}$ times its abscissa.
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Answer
Let x be the abscissa of the given line $2x + 5y = 19$, then by given condition, Ordinate $\text{(y)}=1\frac{1}{2}\times\text{Abscissa}$
$\Rightarrow\text{y}=\frac{3}{2}\text{x}\ ....(\text{i})$ On putting $\text{y}=\frac{3}{2}\text{x}$ in given equation, we get $2\text{x}+5\Big(\frac{3}{2}\Big)\text{x}=19$
$\Rightarrow4\text{x}+15\text{x}=19\times2$
$\Rightarrow4\text{x}+15\text{x}=38$
$\Rightarrow 19\text{x}=38$
$\Rightarrow\text{x}=\frac{38}{19}$
$\therefore\text{x}=2$ On substituting the value of x in Eq. $(i)$ we get $\text{y}=\frac{3}{2}\times2=3$
$\Rightarrow\text{y}=3$ Heanse, the required point is $(2, 3).$
$\Rightarrow\text{y}=\frac{3}{2}\text{x}\ ....(\text{i})$ On putting $\text{y}=\frac{3}{2}\text{x}$ in given equation, we get $2\text{x}+5\Big(\frac{3}{2}\Big)\text{x}=19$
$\Rightarrow4\text{x}+15\text{x}=19\times2$
$\Rightarrow4\text{x}+15\text{x}=38$
$\Rightarrow 19\text{x}=38$
$\Rightarrow\text{x}=\frac{38}{19}$
$\therefore\text{x}=2$ On substituting the value of x in Eq. $(i)$ we get $\text{y}=\frac{3}{2}\times2=3$
$\Rightarrow\text{y}=3$ Heanse, the required point is $(2, 3).$
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