blakkazn924e4y

2022-02-22

Given a linear equation pass through the point(6,4) and two axes, and formed a triangle with area 6. I want to find the equation of that line.
My attempt is letting the equation be $\frac{y-4}{x-6}=m$. Then using the fact that the area of triangle is 6, forming the equation $\frac{1}{2}\text{base}\star \text{height}=6$. I got the base and height of the triangle from the axes, $\text{base}=6-\frac{4}{m}$ and $\text{height}=4-6m$
Then substitute the equation $\frac{1}{2}\text{base}\star \text{height}=6$ and get $36{m}^{2}-36m+16=0$ which leads to no solution. Can anyone let me know what's wrong?

Jowiszowy9zb

You get if calculated correctly $36{m}^{2}-60m+16=0$

shotokan0758s

The equation is given by
$y=mx+n$
so
$y=m\left(x-6\right)+n$
since
$P\left(6;4\right)$
is situated on the line. An both intersection Points are
$A\left(0;4-6m\right)$
and
$P\left(\frac{6m-4}{m},0\right)$
The area our triangle is giveb by
$A=\frac{1}{2}|4-6m||\frac{6m-4}{m}|$

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