$△ A B C$ has vertices $(0, 0), (11, 60)$ and $(91, 0)$ . If the line $y = k x$ cuts the triangle into two triangles of equal area, then $k$ is equal to

\frac{30}{51}

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\frac{4}{7}

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\frac{7}{4}

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\frac{30}{91}

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Solution

The correct option is A $\frac{30}{51}$
Equation of $B C$ is
$9 x + 8 y = 819$
Solving this with $y = k x$ we get intersection point as $(\frac{819}{8 k + 9}, \frac{k 819}{8 k + 9})$
Equation area of triangle $A B D$ and triangle $A D C$
we get $k = 30 / 51$

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△ABC has vertices (0,0),(11,60) and (91,0). If the line y=kx cuts the triangle into two triangles of equal area, then k is equal to

The correct option is A 3051Equation of BC is 9x+8y=819Solving this with y=kx we get intersection point as (8198k+9,k8198k+9)Equation area of triangle ABD and triangle ADCwe get k=30/51

$△ A B C$ has vertices $(0, 0), (11, 60)$ and $(91, 0)$ . If the line $y = k x$ cuts the triangle into two triangles of equal area, then $k$ is equal to

The correct option is A $\frac{30}{51}$
Equation of $B C$ is
$9 x + 8 y = 819$
Solving this with $y = k x$ we get intersection point as $(\frac{819}{8 k + 9}, \frac{k 819}{8 k + 9})$
Equation area of triangle $A B D$ and triangle $A D C$
we get $k = 30 / 51$