Answer:
Neither of the options are correct.
Step-by-step explanation:
Option A :
Line parallel to the equation [tex]y = \frac{7}{5}x - 5[/tex] will have equation
[tex]y = \frac{7}{5}x + c[/tex] ........... (1)
If it passes through the point (5,2) then this will satisfy the equation (1).
Hence, [tex]2 = \frac{7}{5} \times 5 + c[/tex]
⇒ c = - 5
So, the equation (1) becomes [tex]y = \frac{7}{5}x - 5[/tex] (Answer)
Option B :
Line perpendicular to the equation y = - x - 4 will have equation - x + y = c ........... (2)
{Because slope of the original line is - 1 and that of the perpendicular line is 1 and hence the product is (- 1) × (1) = - 1}
Now, if the equation (2) passes through the point (1,-5), then we get,
- 1 + (- 5) = c
⇒ c = - 6
Therefore, the equation (2) becomes, -x + y = - 6. (Answer)
Option C :
Line perpendicular to the equation [tex]y = \frac{9}{2} x - \frac{19}{2}[/tex] will have equation [tex]y = - \frac{2}{9}x + c[/tex] ........... (3)
{Because slope of the original line is [tex]\frac{9}{2}[/tex] and that of the perpendicular line is [tex]- \frac{2}{9}[/tex] and hence the product is ([tex]\frac{9}{2}[/tex]) × ([tex]- \frac{2}{9}[/tex]) = - 1}
Now, if the equation (3) passes through the point (3,4), then we get,
[tex]4 = - \frac{2}{9}\times 3 + c[/tex]
⇒ [tex]c = \frac{14}{3}[/tex]
Therefore, the equation (3) becomes, [tex]y = - \frac{2}{9}x + \frac{14}{3}[/tex]. (Answer)
Option D :
Line parallel to the equation [tex]y = \frac{7}{3}x + 2[/tex] will have equation
[tex]y = \frac{7}{3}x + c[/tex] ........... (4)
If it passes through the point (-3,-5) then this will satisfy the equation (1).
Hence, [tex]- 5 = \frac{7}{3} \times (-3) + c[/tex]
⇒ c = 2
So, the equation (1) becomes [tex]y = \frac{7}{3}x + 2[/tex] (Answer)
Therefore, neither of the options are correct. (Answer)
Answer:
A and B
Step-by-step explanation:
The line y = 7/5 x + 4 passing through the point (5, 2) is parallel to the line y = 7 /5x − 5.
and
The line −x + y = −1 passing through the point (1, −5) is perpendicular to the line y = −x − 4.
