Question

A line hhas slope -3/4,positive y intercept and forms a triangle of area 24 sq.units with coordinate axes.Then,the equation of the line is

Answer 1

Answer:

Let the equation of the line be y = -3/4 x +c

At x =0 the value of y =c

At y=0 the value of x= 3c/4

Given the area of triangle = 24

             = base×height÷2

          24    =3c²÷8

          c² = 8×8

c=8

The equation of the line y = -3x÷4 + 8

Answer 2

Answer:

$\frac{x}{8}+\frac{y}{6}=1$

Step-by-step explanation:

The slope of the line is -(3/4) and it has a positive Y-intercept.

So, tanФ=-(3/4)

⇒ Ф= -36.87° with positive X-axis

⇒Ф= 36.87° with negative X-axis.

So, we can draw the line on the coordinate plane (See diagram).

Let us assume that the line intersects the X-axis at A(a.0) point and the Y-axis at B(b,0) point and the origin is O(0,0).

So, ΔAOB is right-angled, where ∠O=90°.

Given that the area of ΔAOB is 24 square units.

Hence, $\frac{1}{2} ab=24$

⇒ ab=48 ..... (1)

Again, we have $\frac{b}{a}$=$\frac{3}{4}$

⇒ $b=\frac{3}{4}a$ ..... (2)

So, putting the value of b in equation (1), we get,

$\frac{3}{4}a^{2}=48$

⇒ $a^{2}=64$

⇒ $a=8$ ( Neglecting negative value, as a can not be negative.)

Hence, from equation (2), b=6.

Therefore the equation of the line will be,

$\frac{x}{8}+\frac{y}{6}=1$ (Answer)