Since the tangent of B is 4/3, then:

On the other hand, since ABC is a right triangle with hypotenuse BC, then:

Substitute AC=3/4 BA and BC=15 to find BA:
![\begin{gathered} (BA)^2+(\frac{4}{3}BA)^2=15^2 \\ \Rightarrow BA^2+\frac{16}{9}BA^2=15^2 \\ \Rightarrow(1+\frac{16}{9})BA^2=15^2 \\ \Rightarrow\frac{25}{9}BA^2=15^2 \\ \Rightarrow BA^2=\frac{9}{25}\cdot15^2 \\ \Rightarrow BA=\sqrt[]{(\frac{9}{25}\cdot15^2)} \\ \Rightarrow BA=\frac{3}{5}\cdot15 \\ \Rightarrow BA=9 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%28BA%29%5E2%2B%28%5Cfrac%7B4%7D%7B3%7DBA%29%5E2%3D15%5E2%20%5C%5C%20%5CRightarrow%20BA%5E2%2B%5Cfrac%7B16%7D%7B9%7DBA%5E2%3D15%5E2%20%5C%5C%20%5CRightarrow%281%2B%5Cfrac%7B16%7D%7B9%7D%29BA%5E2%3D15%5E2%20%5C%5C%20%5CRightarrow%5Cfrac%7B25%7D%7B9%7DBA%5E2%3D15%5E2%20%5C%5C%20%5CRightarrow%20BA%5E2%3D%5Cfrac%7B9%7D%7B25%7D%5Ccdot15%5E2%20%5C%5C%20%5CRightarrow%20BA%3D%5Csqrt%5B%5D%7B%28%5Cfrac%7B9%7D%7B25%7D%5Ccdot15%5E2%29%7D%20%5C%5C%20%5CRightarrow%20BA%3D%5Cfrac%7B3%7D%7B5%7D%5Ccdot15%20%5C%5C%20%5CRightarrow%20BA%3D9%20%5Cend%7Bgathered%7D)
Substitute BA=12 into the expression for AC to find its value:

On the other hand, we know that:

Substitute BA=12 and DA=3 to find BD:

Finally, since BDE and BAC are similar triangles, we know that:

Substitute AC=12, BA=9 and BD=6 to find the length DE:

Therefore, the length of DE is:
1 is really b, I just learned this
2 is really a
This iis easie guise, why don't you do it? so
point slope=y-y1=m(x-x1)
so just subsitute
first find m
m=(y2-y1)/(x2-x1)=(1/3-(-1/2))/(-2/3-3/2)=(1/3+1/2)/(-2/3-3/2)=(5/6)/(-13/6)=-5/13
m=-5/13
subsitute
y-y1=-5/13(x-x1)
subsitute
y-(-1/2)=-5/13(x-3/2)
y+1/2=-5/13x+15/26
subtract 1/2 from both sides
y=-5/13x+15/26-13/26 (1/2=13/26)
y=-5/13x+2/26
y=-5/13x+1/13
Plz help explain i think you have info missing