In an orthonormal system the two lines with equations y=2x+3 and 2y+x=1 are ( interesting at the points 1,5)

(perpendicular)

(parallel)

justify your answer​

The parallel lines have the same a slope.

We have:

y=-\dfrac{3}{2}x+4y=−

2

3

x+4

Therefore, the searched line has the equation:

y=-\dfrac{3}{2}x+by=−

2

3

x+b

We know, the line passes through point (4,0). Substitute the coordinates of the point to the equation of the line:

\begin{gathered}0=-\dfrac{3}{2}(4)+b0=-(3)(2)+b0=-6+b\qquad|+66=b\to b=6\end{gathered}

0=−

2

3

(4)+b

0=−(3)(2)+b

0=−6+b∣+6

6=b→b=6

y=-\dfrac{3}{2}x+6y=−

2

3

x+6

2.

Let k:y=m_1x+b_1k:y=m

1

x+b

1

and l:y=m_2x+b_2l:y=m

2

x+b

2

The line l is perpendicular to the line k if and only if m_1m_2=-1m

1

m

2

=−1

We have k:y=-\dfrac{1}{3}x-1k:y=−

3

1

x−1

\begin{gathered}l:y=mx+bl\ \perp\ k\iff-\dfrac{1}{3}m=-1\qquad|\cdot(-3)m=3\end{gathered}

l:y=mx+b

l ⊥ k⟺−

3

1

m=−1∣⋅(−3)

m=3

The searched line has the equation: y=3x+by=3x+b

We know, the line passes through point (1,5). Substitute the coordinates of the point to the equation of the line:

\begin{gathered}5=3(1)+b5=3+b\qquad|-32=b\to b=2\end{gathered}

5=3(1)+b

5=3+b∣−3

2=b→b=2

y=3x+2y=3x+2