Find the equation of the line perpendicular to the line x – 7y + 5 = 0 and having x-intercept 3

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The equation of the line is given as x – 7y + 5 = 0.

The above equation can be written in the form y = mx+c

Thus, the above equation is written as:

y= (1/7)x + (5/7)

From the above equation, we can say that,

The slope of a line, m = 5/7

The slope of the line perpendicular to the line having a slope of 1/7 is

m = -1/(1/7) = -7

Hence, the equation of a line with slope -7 and intercept 3 is given as:

y = m (x – d)

⇒ y= -7(x-3)

⇒ y=-7x + 21

7x+ y = 21

Hence, the equation of a line which is perpendicular to the line x – 7y + 5 = 0 with x-intercept 3 is 7x+ y = 21.

Thanks