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प्रश्न
The equation of the line joining the point (3, 5) to the point of intersection of the lines 4x + y – 1 = 0 and 7x – 3y – 35 = 0 is equidistant from the points (0, 0) and (8, 34).
विकल्प
True
False
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उत्तर
This statement is True.
Explanation:
Given equations are
4x + y – 1 = 0 ......(i)
And 7x – 3y – 35 = 0 ......(ii)
From equation (i) y = 1 – 4x ......(iii)
Putting the value of y in equation (ii) we get
7x – 3(1 – 4x) – 35 = 0
⇒ 7x – 3 + 12x – 35 = 0
⇒ 19x – 38 = 0
⇒ x = 2
From equation (iii) we get,
y = 1 – 4 × 2
⇒ y = – 7
The point of intersection is (2, – 7).
Equation of line joining the point (3, 5) to the point (2, – 7) is
y – 5 = `(-7 - 5)/(2 - 3) (x - 3)`
⇒ y – 5 = 12(x – 3)
⇒ y – 5 = 12x – 36
⇒ 12x – y – 31 = 0 .......(iv)
Distance of equation (iv) from the point (0, 0)
= `|(-31)/sqrt((12)^2 + (-1)^2)|`
= `31/sqrt(145)`
Distance of equation (iv) from the point (8, 34) is
= `|(12 xx 8 - 34 - 31)/sqrt((12)^2 + (-1)^2)|`
= `|(96 - 65)/sqrt(145)|`
= `31/sqrt(145)`
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