the point (- 7, 4) is reflected over the line x = - 3 . Then , the resulting point is reflected over the line y = x . Where is t
1 answer:
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Please, see the offered decision:
1) common equation for lines is y=kx+b. If k₁=k₂ (for line 1 and line 2) ⇒ 'line 1' || 'line 2'.
2) for line 3x+5y=6 k= -3/5. It means (according to item 1) for unknown line k is the same (-3/5).
3) using points (0;3) it is easy to find parameter b (x=0, y=3) via y=kx+b:
3=0*(-3/5)+b ⇔ b=3.
4) finaly (k=-3/5; b=3):
1) common equation for lines is y=kx+b. If k₁=k₂ (for line 1 and line 2) ⇒ 'line 1' || 'line 2'.
2) for line 3x+5y=6 k= -3/5. It means (according to item 1) for unknown line k is the same (-3/5).
3) using points (0;3) it is easy to find parameter b (x=0, y=3) via y=kx+b:
3=0*(-3/5)+b ⇔ b=3.
4) finaly (k=-3/5; b=3):