\huge \dag \: \sf{ Question} \: \dag Find the number of straight lines passing through (2 , 4) and forming a triangle of area 16 cm² with the two coordinate axis ! You have to d…

Question

`\huge \dag \: \sf{ Question} \: \dag`

Find the number of straight lines passing through (2 , 4) and forming a triangle of area 16 cm² with the two coordinate axis !

You have to draw the figure by yourself ~

Thanks for your help •••​

Answer 1

One Line

Explanation:

Important Information Given:

• Area (△AOB)= 16sq.units

Let coordinate axes at the straight line be (a,0) and (0,b).

Thus, the equation of the straight line is,

`(x)/(a)+(y)/(b)=1.Hence : (2)/(a)+{4}{b}=1`

Equation: 4a + 2b =ab

Area of the triangle formed =
`16 cm ^2`

`(1)/(2)*ab=16`

`ab=32`

`b=(32)/(a)`

Substitute this value of b in the equation:

`4a+(64)/(a)=32`

`4a^2+64=32a`

`a^2-8a+16=0`

`a=4`

`b=8`

Hence, Only one straight line.

[RevyBreeze]

Answer 2

• One line

Explanation:

Since two sides lie on the x- or y-axis, let the side lengths are x and y.

The area is:

• A = 1/2xy

We need to find x and y such that:

• 1/2xy = 16
• xy = 32

The third side passes through the point (2, 4), which adds limitations:

• x > 2 and y > 4

The slope of the line is:

• m = y/x

It is same as slope of the line with the point (2, 4):

• m = (y - 4)/(x - 2)

Solve the equation:

• (y - 4)/(x - 2) = y/x
• xy - 4x = xy - 2y
• 2y = 4x
• y = 2x

Substitute this into area equation:

• xy = 32
• 2x² = 32
• x² = 16
• x = 4

Then:

• y = 8

So the vertices are:

• (0, 0), (0, 4), (8, 0)