# GATE2016 EC-3: GA-9

1.6k views

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

1. $14.95$
2. $15.25$
3. $15.70$
4. $20.35$
1
i am getting 183/12=15.25sq units...what is the ans?
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
0
what is the answer by d way @niharika
0
It was actually asked in virtualgate test series and the answer given was 15.25 sq units .That is why I was mentioning that axis should also be mentioned.
0
15.25 sq units
0
solve for x and y, observe that two given lines are perpendicular. so it will form something like rectangle in first quadrant.

now coordinates of rectangle are (0,0) (x,0) (0,y) (x,y)

find area = length * breadth

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

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 $(B).$

edited by
1

@ why u didn't consider triangle DEC, as it is also satisfying question condition? 0

@Shubhm because $DEC$ is not bounded by two lines and $y$-axis.

0

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. ans may be like this

0
Hello can the picture be more clear please?
Bt how it will come 15.25 square units can someone expkain plz

## Related questions

1
1.2k views
A straight line is fit to a data set (ln $x, y$). This line intercepts the abscissa at ln $x = 0.1$ and has a slope of $-0.02$. What is the value of y at $x = 5$ from the fit? $-0.030$ $-0.014$ $0.014$ $0.030$
Two straight lines are drawn perpendicular to each other in $X-Y$ plane. If $\alpha$ and $\beta$ are the acute angles the straight lines make with the $\text{X-}$ axis, then $\alpha + \beta$ is_______. $60^{\circ}$ $90^{\circ}$ $120^{\circ}$ $180^{\circ}$
Two points $(4, p)$ and $(0, q)$ lie on a straight line having a slope of $3/4$. The value of $( p – q)$ is $-3$ $0$ $3$ $4$
If $y=5x^2+3$, then the tangent at $x=0$, $y=3$ passes through $x=0,y=0$ has a slope of $+1$ is parallel to the $x$-axis has a slope of $-1$