If each of the points (x1 , 4), (-2 , y1) lies on the line joining the points (2, -1) and (5, -3), then the point P (x1, y1) lies on the line

Question

If each of the points (x1 , 4), (-2 , y1) lies on the line joining the points (2, -1) and (5, -3), then the point P (x1, y1) lies on the line (A) 6(x + y) - 25 = 0 (B) 2x + 6 y + 1 = 0 (C) 2x + 3y - 6 = 0 (D) 6(x+ y) + 25 = 0

Tardigrade Admin · Accepted Answer

The equation of the line joining the points

(2, - 1)

and

(5, - 3)

is given by

y + 1 = \frac{- 1 + 3}{2 - 5} (x - 2)

or

2 x + 3 y - 1 = 0 ... (i)

Since

(x_{1}, 4)

and

(- 2, y_{1})

lie on

2 x + 3 y - 1 = 0

, we have

2 x_{1} + 12 - 1 = 0

or

x_{1} = - \frac{11}{2}

and

- 4 + 3 y_{1} - 1 = 0

or

y_{1} = \frac{5}{3}

Thus,

(x_{1}, y_{1})

satisfies

2 x + 6 y + 1 = 0

.

Q. If each of the points $(x_{1}, 4), (- 2, y_{1})$ lies on the line joining the points $(2, - 1)$ and $(5, - 3)$ , then the point $P (x_{1}, y_{1})$ lies on the line

$6 (x + y) - 25 = 0$

$2 x + 6 y + 1 = 0$

$2 x + 3 y - 6 = 0$

$6 (x + y) + 25 = 0$

Q. If each of the points (x1​,4),(−2,y1​) lies on the line joining the points (2,−1) and (5,−3), then the point P(x1​,y1​) lies on the line

6(x+y)−25=0

2x+6y+1=0

2x+3y−6=0

6(x+y)+25=0

The equation of the line joining the points (2,−1) and (5,−3) is given by y+1=2−5−1+3​(x−2) or 2x+3y−1=0...(i) Since (x1​,4) and (−2,y1​) lie on 2x+3y−1=0, we have 2x1​+12−1=0 or x1​=−211​ and −4+3y1​−1=0 or y1​=35​ Thus, (x1​,y1​) satisfies 2x+6y+1=0.

Q. If each of the points $(x_{1}, 4), (- 2, y_{1})$ lies on the line joining the points $(2, - 1)$ and $(5, - 3)$ , then the point $P (x_{1}, y_{1})$ lies on the line

$6 (x + y) - 25 = 0$

$2 x + 6 y + 1 = 0$

$2 x + 3 y - 6 = 0$

$6 (x + y) + 25 = 0$