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.

Asked by Abhisek | 1 year ago |  121

Solution :-

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$$\dfrac{22}{9}$$

Answered by Pragya Singh | 1 year ago

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