7 votes

Find the area bounded by the lines $3x + 2y=14, 2x - 3y = 5$ in the first quadrant.

- $14.95$
- $15.25$
- $15.70$
- $20.35$

0

Yes it is 15.25 provided we take y axis as well for binding as well which is not mentioned in the question

1

but it z not mentioned in the question if we will take x-axis for binding the area would b different

so if we will take the given two line and 1st quadrant the answer would be 1/2*(14/3)*(7)= 49/3

so if we will take the given two line and 1st quadrant the answer would be 1/2*(14/3)*(7)= 49/3

7 votes

Best answer

- $3x+2y=14\rightarrow(1)$
- $2x−3y=5\rightarrow(2)$

To get the intersection point, we solve the equations and get $x=4,y=1$

We can draw the required area like this

From the above diagram,

Area $=\triangle ABC +\Box OBCE-\triangle CDE$

Area $=\left(\dfrac{1}{2}\times 4\times 6\right)+\left(1\times 4\right)-\left(\dfrac{1}{2}\times\dfrac{3}{2}\times 1\right)$

Area $=12+4-\dfrac{3}{4}=16-0.75=15.25$

So, the correct answer is $(B).$

1

@Lakshman Patel RJIT why u didn't consider triangle DEC, as it is also satisfying question condition?

0

sorry! I made mistake on asking question.

My question is about triangle DCF, say F is point where line AC meets the x-axis. we can find area of that DCF by heron's formula.

My question is about triangle DCF, say F is point where line AC meets the x-axis. we can find area of that DCF by heron's formula.

2

My question is about triangle DCF, say F is point where line AC meets the x-axis. we can find area of that DCF by heron's formula.

We can only apply the Heron's formula when all the sides are given in the triangle.

Here, $\triangle DCF = \dfrac{1}{2}(\text{ Base} \times \text{Height}) = \dfrac{1}{2}\bigg(\dfrac{13}{6} \times 1\bigg) = \dfrac{13}{12} $

Required area $ = \triangle AOF - \triangle DCF = \dfrac{1}{2}\times \dfrac{14}{3}\times 7 - \dfrac{13}{12} = \dfrac{49}{3}-\dfrac{13}{12} $

$= \dfrac{196-13}{12} = \dfrac{183}{12}=15.25\:\text{Unit}^{2}.$

0

To understand how to draw that graph in the answer -

The useful equation of line is

$\frac{x}{a} + \frac{y}{b} = 1$

where, $a = \text{x-intercept}, b = \text{y-intercept}$

So, we now convert given two equations of lines into this above mentioned form.

For first line, i.e. $3x +2y = 14$

we have $\frac{x}{(14/3)} + \frac{y}{(14/2)} = 1$

Hence first line intercepts x-axis at $14/3$ and y-axis at $7$.

For second line, i.e. $2x-3y = 5$

we have $\frac{x}{(5/2)} + \frac{y}{(-5/3)} = 1$

Hence first line intercepts x-axis at $5/2$ and y-axis at $-5/3 $.

The rest of the answer should be self-explanatory.

The useful equation of line is

$\frac{x}{a} + \frac{y}{b} = 1$

where, $a = \text{x-intercept}, b = \text{y-intercept}$

So, we now convert given two equations of lines into this above mentioned form.

For first line, i.e. $3x +2y = 14$

we have $\frac{x}{(14/3)} + \frac{y}{(14/2)} = 1$

Hence first line intercepts x-axis at $14/3$ and y-axis at $7$.

For second line, i.e. $2x-3y = 5$

we have $\frac{x}{(5/2)} + \frac{y}{(-5/3)} = 1$

Hence first line intercepts x-axis at $5/2$ and y-axis at $-5/3 $.

The rest of the answer should be self-explanatory.