The line through the points (h, 3) and (4, 1) intersects the line 7x − 9y −19 = 0. At right angle. Find the value of h.

Let the slope of the line passing through (h, 3) and (4, 1) be m₁

Then, m₁ = \( \dfrac{1-3}{(4-h)}\) = \( \dfrac{-2}{(4-h)}\)

Let the slope of line 7x – 9y – 19 = 0 be m₂

7x – 9y – 19 = 0

So, y = \( \dfrac{7}{9}x\) – \( \dfrac{19}{9}\)

m₂ = \( \dfrac{7}{9}\)

Since, the given lines are perpendicular

m₁ × m₂ = -1

\( \dfrac{-2}{(4-h)}\) × \( \dfrac{7}{9}\) = -1

\( \dfrac{-14}{(36-9h)}\) = -1

-14 = -1 × (36 – 9h)

36 – 9h = 14

9h = 36 – 14

h =\( \dfrac{22}{9}\)

The value of h is\( \dfrac{22}{9}\)