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

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

Let the slope of line 7x – 9y – 19 = 0 be m_{2}

7x – 9y – 19 = 0

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

m_{2} = \( \dfrac{7}{9}\)

Since, the given lines are perpendicular

m_{1} × m_{2} = -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\( \dfrac{22}{9}\)

Answered by Pragya Singh | 1 year agoFind the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.

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