Find the equation of a line which is perpendicular to the line joining

1.	Find the equation of a line which is perpendicular to the line joining (4, 2) and (3 5) and cuts off an intercept of length 3 on y – axis.
Answer» Given, A line segment joining (4, 2) and (3, 5) if it cuts off an intercept 3 from y–axis. To Find: The equation of that line. Formula used: The equation of line is y = mx + C Explanation: Here, The required equation of line is y = mx + c Now, c = 3 (Given) Let m be slope of given line = – 1 Slope of line joining (x₁ – x₂) and (y₁ – y₂) ,m = \(\frac{y_2-y_1}{x_1-x_2}\) So, Slope of line joining (4, 2) and (3, 5),m = \(\frac{5-2}{3-4}\) = \(\frac{2}{-1}\) Therefore, m = \(\frac{1}{3}\) Now, The equation of line is y = mx + c y = \(\frac{1}{3}x+3\) x – 3y + 9 = 0 Hence, The equation of line is 2y + 5x + 6 = 0.

Find the equation of a line which is perpendicular to the line joining (4, 2) and (3 5) and cuts off an intercept of length 3 on y – axis.

Answer»

Given, A line segment joining (4, 2) and (3, 5) if it cuts off an intercept 3 from y–axis.

To Find: The equation of that line.

Formula used: The equation of line is y = mx + C

Explanation: Here, The required equation of line is y = mx + c

Now, c = 3 (Given)

Let m be slope of given line = – 1

Slope of line joining (x₁ – x₂) and (y₁ – y₂) ,m = \(\frac{y_2-y_1}{x_1-x_2}\)

So, Slope of line joining (4, 2) and (3, 5),m = \(\frac{5-2}{3-4}\) = \(\frac{2}{-1}\)

Therefore, m = \(\frac{1}{3}\)

Now, The equation of line is y = mx + c

y = \(\frac{1}{3}x+3\)

x – 3y + 9 = 0

Hence, The equation of line is 2y + 5x + 6 = 0.