Q. 535.0( 1 Vote )

# State whether the statements are true or false.

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).

Answer :

Given two lines are:

4x + y - 1 = 0 …(i)

and 7x - 3y -35 = 0 …(ii)

Now, point of intersection of these lines can be find out as:

Multiplying eq. (i) by 3, we get

12x + 3y – 3 = 0 …(iii)

On adding eq. (ii) and (iii), we get

7x – 3y – 35 + 12x + 3y – 3 = 0

⇒ 19x – 38 = 0

⇒ 19x = 38

⇒ x = 2

On putting value of x in (i), we get

4(2) + y – 1 = 0

⇒ 8 + y – 1 = 0

⇒ y = -7

So, the point of intersection of given two lines is:

(x, y) = (2, -7)

Now, we have to find the equation of the line joining the point (3, 5) and (2, -7).

Again, equation of line is given by:

⇒ y – 5 = 12(x – 3)

⇒ y – 5 = 12x – 36

⇒ 12x – y – 36 + 5 = 0

⇒ 12x – y – 31 = 0 …(iv)

Now, the distance of eq. (iv) from the point (0, 0) is

Now, the distance of eq. (iv) from the point (8, 34) is

Hence, the equation of line 12x – y – 31 = 0 is equidistant from (0, 0) and (8, 34)

Hence, the given statement is **TRUE**

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