Question

Find the equation of st . Line passing through the point(2,-6)and the point of intersection of the lines 5x-2y+14=0 and 2y=8-7x

Answer 1

$\textbf{- Answer -}$

The two lines are

5x - 2y + 14 = 0 ...(i)

2y = 8 - 7x ...(ii)

Now, putting the value 2y = 8 - 7x in (i), we get

5x - (8 - 7x) + 14 = 0

⇒ 5x - 8 + 7x + 14 = 0

⇒ 12x + 6 = 0

⇒ 12x = - 6

⇒ x = - 6/12

⇒ x = - 1/2

Putting x = - 1/2 in (ii), we get

2y = 8 - 7 (- 1/2)

⇒ 2y = 8 + 7/2

⇒ 2y = (16 + 7)/2

⇒ 2y = 23/2

⇒ y = 23/4

So, the intersection of the lines (i) and (ii) is
(- 1/2, 23/4).

Therefore, the line passing through the points (2, - 6) and (- 1/2, 23/4) be

{y - (- 6)}/(- 6 - 23/4) = (x - 2)/{2 - (- 1/2)}

⇒ (y + 6)/(- 47/4) = (x - 2)/(5/2)

⇒ 5/2 (y + 6) = (- 47/4) (x - 2)

⇒ (5y + 30)/2 = -47 (x - 2)/4

⇒ 4 (5y + 30)/2 = - 47 (x - 2)

⇒ 2 (5y + 30) = - 47x + 94

⇒ 10y + 60 = - 47x + 94

⇒ 47x + 10y = 94 - 60

⇒ 47x + 10y = 34

So, the required line is

47x + 10y = 34.

#$\textbf{MarkAsBrainliest}$

Answer 2

Answer:

The equation of the straight line is $47x+10y-34=0$

Explanation:

Step 1:

We need to find the intersection point of the lines $5x-2y+14=0$ and $2y=8-7x$

From the equation of the second line, we get $y=\frac{8-7x}{2}$

Substituting the value of y obtained in the equation of the first line.

$5x-2y+14=0$

$5x-2(\frac{8-7x}{2} )+14=0$

$5x-(8-7x )+14=0$

$5x-8+7x +14=0$

$12x+6=0$

$x=-\frac{6}{12}$

$x=-\frac{1}{2}$

Step 2:

Substituting the value of x in the equation of the second line.

$2y=8-7x$

$2y=8-7(-\frac{1}{2} )$

$2y=8+\frac{7}{2}$

$2y=\frac{16+7}{2}$

$y=\frac{23}{4}$

Therefore, $(-\frac{1}{2},\frac{23}{4})$ is the point of intersection of the two lines.

Step 3:

We find the equation of the line passing through points $(2,-6)$ and $(-\frac{1}{2},\frac{23}{4})$ using the two-point form formula.

$\frac{y-y_{1} }{y_{1}-y_{2} }=\frac{x-x_{1} }{x_{1}-x_{2} }$

Substituting the two points, we get

$\frac{y-(-6) }{-6-\frac{23}{4} } =\frac{x-2 }{2-(-\frac{1}{2} ) }$

$\frac{y+6 }{-6-\frac{23}{4} } =\frac{x-2 }{2+\frac{1}{2} }$

$\frac{y+6 }{\frac{-24-23}{4} } =\frac{x-2 }{\frac{4+1}{2} }$

$\frac{4(y+6) }{{-47} } =\frac{2(x-2) }{{5} }$

$5(4y+24)=-47(2x-4)$

$20y+120=-94x+188$

$94x+20y+120-188=0$

$94x+20y-68=0$

$47x+10y-34=0$

Therefore, the equation of the line passing through the given points is $47x+10y-34=0$

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