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?

My attempt is letting the equation be

Then substitute the equation

Jowiszowy9zb

Beginner2022-02-23Added 5 answers

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

shotokan0758s

Beginner2022-02-24Added 8 answers

The equation is given by

$y=mx+n$

so

$y=m(x-6)+n$

since

$P(6;4)$

is situated on the line. An both intersection Points are

$A(0;4-6m)$

and

$P(\frac{6m-4}{m},0)$

The area our triangle is giveb by

$A=\frac{1}{2}|4-6m|\left|\frac{6m-4}{m}\right|$

so

since

is situated on the line. An both intersection Points are

and

The area our triangle is giveb by

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