Question

Determine the point on the graph of the linear equation $2x+5y=19$, whose ordinate is $1\frac{1}{2}$ times its abscissa.

Answer 1

Let the abscissa be $a$

Then ordinate $=\frac{3}{2a}$

New coordinates of point is $(a,\frac{3}{2a})$

Given that is a point on line $2x+5y=19$

now

$2\left(a\right)+5\left(\frac{3}{2a}\right)=19$

$2a+\frac{15a}{2}=19$

$4a+15a=38$

$19a=38$

$a=2$

$\therefore$ abscissa $=2$

ordinate $=\frac{3\left(2\right)}{2}=3$

$\therefore$ the required point$(2,3)$