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[tex] \huge \dag \: \sf{ Question} \: \dag[/tex]


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 •••​

Respuesta :

mhanifa
  • One line

Step-by-step 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)
RevyBreeze

Answer:

One Line

Step-by-step 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,

[tex]\frac{x}{a}+\frac{y}{b}=1.Hence : \frac{2}{a}+{4}{b}=1[/tex]

Equation: 4a + 2b =ab

Area of the triangle formed =[tex]16 cm ^2[/tex]

[tex]\frac{1}{2}*ab=16[/tex]

[tex]ab=32[/tex]

[tex]b=\frac{32}{a}[/tex]

Substitute this value of b in the equation:

[tex]4a+\frac{64}{a}=32[/tex]

[tex]4a^2+64=32a[/tex]

[tex]a^2-8a+16=0[/tex]

[tex]a=4[/tex]

[tex]b=8[/tex]

Hence, Only one straight line.

[RevyBreeze]

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