Question

find the equation the stright line passing through the points (3,-5) and parllel to the line X-7y+15=0

Answer 1

Answer:

Step-by-step explanation:

The equation of a line perpendicular to x - 7y + 5 = 0 is

7x+y+λ = 0

Its x-intercept is 3. This means that the line cuts x-axis at a distance of 3 units from the origin.

Consequently, it passes through the

point (3,0) on x-axis. Therefore, 21+0+λ = 0

λ = -21

Thus, required equation of the line is

7x + y 21 = 0.

Answer 2

The equation of any line parallel to the y-axis is of the form x=a

The two given lines are x−7y+5=0 and 3x+y

On solving equations (2) and (3) we obtain,  x=−225andy=2215

Therefore,{−225,2215} is the point of intersection of lines (2) and (3)

since line x=a passes through point {−225,2215}, a=−225

Thus, the required equation of line is x=−225

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