Question

What is the ratio of the area of triangle formed along x-axis and y-axis in the given figure,
​

Answer 1

$i)Triangle \:formed \: along \: X -axis \:is \:\triangle CED ,$

$Here , C(-1,0) = (x_{1}, y_{1} ), \\E(2,3) = (x_{2}, y_{2} )\: and \: D(4,0) = (x_{3}, y_{3} )$

$Area \: of \: \triangle CED\\= \frac{1}{2}|x_{1}(y_{2}-y_{1})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|$

$=\frac{1}{2}| (-1)(3-0)+2(0-0)+4(0-3)| \\= \frac{1}{2}|-3+0-12|= \frac{1}{2} \times 15\\= \frac{7}{2} \: square \:units \: --(1)$

$ii)Triangle \:formed \: along \: Y -axis \:is \:\triangle ABE ,$

$Here , A(0,6) = (x_{1}, y_{1} ), \\B(0,1) = (x_{2}, y_{2} )\: and \: E(2,3) = (x_{3}, y_{3} )$

$Area \: of \: \triangle ABE \\= \frac{1}{2}|x_{1}(y_{2}-y_{1})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|$

$=\frac{1}{2}| 0(1-3)+0(3-6)+2(6-1)| \\= \frac{1}{2}| 2\times 5|= 5 \: square \:units \: --(2)$

$\therefore \red{Required \:ratio} \\= \frac{Area \: \triangle CED}{Area \: \triangle ABE}\\= \frac{\frac{7}{2}}{5} \\= \frac{7}{2} \times \frac{1}{5} \\=\frac{7}{10} \\= 7 : 10$

Therefore.,

$\red { Required \:ratio } \green { = 7 : 10}$

•••♪