Question

coordinates of a point which is at 3 units distance from point (1,-3) on the line 2x+3y+7=0

Answer 1

2x+3x+7=0
5x+7=0
5x=0-7
x=-7/5
x=-1.4

Answer 2

Answer:

the coordinates of point are (1 ± $\frac{9}{\sqrt{13} }$ , -3 ± $\frac{6}{\sqrt{13} }$)

Step-by-step explanation:

Let point P(m,n) lie on line 2x+3y+7

So this point will satisfy the line equation:

2m+3n+7=0     (Let this be Eqn 1)

Now, given distance of point = 3

As we know distance d = √ ( x - x₁  )² + ( y - y₁  )²

Putting the points (m,n) and (1,-3) in this equation,

√ ( m- 1 )² + (n + 3 )² = 3

taking square on both the sides,

( m- 1 )² + (n + 3 )² = 9

m² + 1 - 2m + n² + 9 + 6n = 9

m² + n² - 2m + 6n + 1 = 0  (Let this be Eqn 2)

From Eqn 1 ,

m= $\frac{-3n - 7}{2}$ (Let this be Eqn 3)

Taking sqaure on both the sides,

m² =  $\frac{9n^{2} + 49 + 42n }{4}$

Substituting value of m² and m in Eqn 2 we get,

$\frac{9n^{2} + 49 + 42n }{4}$ + n² + 3n + 7 + 6n + 1 = 0

On solving this equation we get,

13n² + 78n + 81 = 0

Applying the qaudratic formlua to find roots,

n = - 78 ± √( 78 )² - 4 (13 ) ( 81 ) ÷ ( 2 * 13 )

n = -78 ± √ 6084 - 4212 ÷ 26

n = -78 ± √1872 ÷ 26

n = -78 ± 12√ 13 ÷ 26

n = -3 ± $\frac{6}{\sqrt{13} }$

Substituting value of n in eqn 3,

m = $\frac{-3n - 7}{2}$

m = - 3 ( -3 ± $\frac{6}{\sqrt{13} }$ )

m = 1 ± $\frac{9}{\sqrt{13} }$

Hence the coordinates of Point P (m,n ) = (1 ± $\frac{9}{\sqrt{13} }$ , -3 ± $\frac{6}{\sqrt{13} }$)